DIFFERENTIAL EQUATIONS AND ITS APPLICATIONS€¦ · Examples 1 to 5 are ordinary differential...
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ALAGAPPA UNIVERSITY Accredited with ’A+’ Grade by NAAC (CGPA: 3.64) in the Third Cycle and Graded as Category-I University by MHRD-UGC KARAIKUDI - 630 003 DIRECTORATE OF DISTANCE EDUCATION B.SC 11333 DIFFERENTIAL EQUATIONS AND ITS APPLICATIONS COPY RIGHT RESERVED FOR PRIVATE USE ONLY
Transcript of DIFFERENTIAL EQUATIONS AND ITS APPLICATIONS€¦ · Examples 1 to 5 are ordinary differential...
ALAGAPPA UNIVERSITY
Accredited with ’A+’ Grade by NAAC (CGPA: 3.64) in the Third Cycle and Graded as
Category-I University by MHRD-UGC
KARAIKUDI - 630 003
DIRECTORATE OF DISTANCE EDUCATION
B.SC
11333
DIFFERENTIAL EQUATIONS AND ITS
APPLICATIONS
COPY RIGHT RESERVED FOR PRIVATE USE ONLY
Contents
List of Figures i
1 EXACT, HOMOGENEOUS AND LINEAR DIFFERENTIAL EQUATIONS 1
1.1 INTRODUCTION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
1.2 EXACT DIFFERENTIAL EQUATIONS . . . . . . . . . . . . . . . . . . . 2
1.3 CONDITIONS FOR EQUATION TO BE EXACT . . . . . . . . . . . . . . 3
1.4 WORKING RULE FOR SOLVING IT . . . . . . . . . . . . . . . . . . . . 4
1.5 PROBLEMS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
1.6 EQUATIONS OF THE FIRST ORDER, BUT OF HIGHER DEGREE . . . 6
1.7 Clairaut’s form. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8
1.8 Equations that do not contain (i) x explicitly (ii) y explicitly . . . . . . . . 9
1.9 EQUATIONS HOMOGENEOUS IN x and y ................................................... 10
1.10 LINEAR EQUATION WITH CONSTANT COEFFICIENTS ................... 12
1.11 PROBLEMS ............................................................................................. 13
1.12 LINEAR EQUATIONS WITH VARIABLE COEFFICIENTS ................... 13
1.13 EQUATIONS REDUCIBLE TO THE LINEAR EQUATIONS.................... 15
2 SIMULTANEOUS AND TOTAL DIFFERENTIAL EQUATIONS 16
2.1 INTRODUCTION ............................................................................................ 16
2.2 Simultaneous equations of the first order and first degree ................................... 17
2.2.1 Methods for solving dx = dy = dz ......................................................... 17
P Q R
2.3 Simultaneous linear differential equations ......................................................... 20
2.4 Complete solution given a known integral ......................................................... 22
2.5 Reduction to normal form.................................................................................. 25
2.6 Change of the independent variable ................................................................... 28
2.7 Method of variation of parameters ..................................................................... 30
2.8 Total differential equations ................................................................................ 34
2.8.1 Methods of solving the differential equation Pdx + Qdy + Rdz = 0 34
2.8.2 Solved problems ................................................................................... 35
3 PARTIAL DIFFERENTIAL EQUATIONS 36
3.1 Necessary and Sufficient condition of integrability of Pdx+Qdy+Rdz=0 ............ 36
3.2 Rules for integrating Pdx+Qdy+Rdz=0 .............................................................. 39
3.3 Examples .......................................................................................................... 39
3.4 Partial Differential Equations of the First Order ................................................. 41
3.5 Classification of integral .................................................................................... 41
3.5.1 Singular integral ................................................................................... 42
3.5.2 General integral .................................................................................... 42
3.6 Derivation of partial differential equations: ........................................................ 44
3.6.1 By elimination of constants ................................................................... 44
3.6.2 By elimination of an arbitrary function .................................................. 45
3.7 Special Methods; Standard forms ...................................................................... 46
3.7.1 Standard 1 46
3.7.2 Standard II ............................................................................................ 48
3.7.3 Standard III ........................................................................................... 50
3.7.4 Standard IV; Clairant’s form. ................................................................ 51
3.8 Questions .......................................................................................................... 52
3.8.1 2 Marks 52
3.8.2 5 Marks 52
3.8.3 10 Marks 52
4 STANDARD FORMS OF PARTIAL DIFFERENTIAL EQUATIONS AND
TRAJECTORIES 53
4.1 Standard Form of Partial Differential Equations(PDE) ....................................... 53
4.1.1 Example 54
4.1.2 Example 55
4.1.3 Example 56
4.1.4 Example 57
4.1.5 Excercises 57
4.2 CHARPIT’S METHOD .................................................................................... 57
4.2.1 Solution Procedures for Charpit’s Method ............................................. 59
4.2.2 Examples 59
4.2.3 Excercises 60
4.3 The Brachistochrone Problem ........................................................................... 60
4.4 FALLING BODIES........................................................................................... 62
4.4.1 Retarded Fall ........................................................................................ 63
4.5 ORTHOGONAL TRAJECTORIES ................................................................... 64
4.6 Procedure for finding an Orthogonal Trajectories .............................................. 65
4.6.1 If the curves are in Cartesian Coordinates .............................................. 65
4.6.2 If the curves are in Polar Coordinates .................................................... 65
4.7 Examples .......................................................................................................... 66
4.7.1 Problem 66
4.7.2 Problem 66
4.7.3 Problem 67
4.8 Excercises ......................................................................................................... 67
4.9 TAUTOCHRONOUS PROPERTY IN CYCLOID ............................................. 68
i
List of Figures
4.1 Diagram representation of Brachistochrone problem ......................................... 61
4.2 Geometrical view of Tautochronous .................................................................. 68
1
√
∂x ∂y
Chapter 1 EXACT, HOMOGENEOUS AND LINEAR
DIFFERENTIAL EQUATIONS
Unit- I
1.1 INTRODUCTION
A differential equation is an equation which involves derivatives. The following are
some examples of differential equations.
1. yJ = sin x.
2. yJJ
+ 3yJ + 2y = e
x.
3. (yJJ )2 + (y
J )3 + 3y = x
2.
4. yJJJ
+ 3(yJJ )2 + y
J = e
x.
5. y = xyJJ
+ r [1 + (yJ )2].
6. ∂z − x ∂z = z.
7. ∂2z +
∂2z = x2 + y.
∂x2 ∂y2
2
∂y ∂x
In a differential equation if there is a single independent variable and the derivatives are
ordinary derivatives then it is called an ordinary differential equation.
Examples 1 to 5 are ordinary differential equations.
If there are two or more independent variables and the derivatives are partial derivatives
then it is called a partial differential equation.
Examples 6 and 7 are partial differential equations.
The order of a differential equation is the order of the highest derivative appearing in it.
Examples 1 and 6 are of first order; and 2, 3, 5, 7 are of order two and 4 is of order 3.
The degree of the differential equation is the degree of the highest ordered derivative
occurring in it when the differential coefficients are free from radicals and fractions.
In the above examples all except 3 and 5 are of degree one; examples 3 and 5 are of
degree two.
1.2 EXACT DIFFERENTIAL EQUATIONS
Definition.
Consider the differential equation
M (x, y)dx + N (x, y)dy = 0. (1.1)
Suppose there exists a function f (x, y) such that
∂f = M and
∂f = N.
Then the differential equation takes the form
(∂f )dx + (
∂f )dy = 0 (i.e.,) df = 0.
∂x ∂y
On integration the general solution is given by f (x, y) = c. In this case the expression
Mdx + Ndy is said to be an exact differential and 1.1 is called an exact differential
equation.
3
∂y = .
∂x ∂y
= ∂x = .
∂y = .
∂y
∂x ∂y ∫
∂x
∂y
∂y
∂y
Σ∫ Σ
∂y
.
N − ∂y
(f Mdx) = ∂x
− ∂x∂y
(
∂x −
∂y∂x(
∂x
∂x
∂x
∫ Σ
1.3 CONDITIONS FOR EQUATION TO BE EXACT
Theorem 1
The differential equation Mdx + Ndy = 0 is exact if and only if ∂M ∂N
Proof: Suppose the equation is exact.
Then there exists a function f (x, y) such that ∂f
= M and ∂f
= N .
∂M ∂2f
∂y ∂y∂x and ∂N
∂2f
∂x∂y
Hence ∂M ∂N
Conversly:
Suppose ∂M
∂N ....................................... (1)
We have to construct a function f (x, y) such that ∂f
= M and ∂f
= N .
Let f = Mdx + g(y) ..................... (2)
where g is an arbitrary function of y.
By definition of f , we have ∂f
= M .
Hence the problem is to determine g(y) in such a way that ∂f
= N .
(ie.,) ∂ Mdx + g(y) = N .
(ie.,) ∂ .∫
MdxΣ
+ gJ(y) = N .
∴ gJ
(y) = N − ∂ .∫
MdxΣ ................
(3). If we prove that the right hand side of (3) is a function of y only then we can integrate (3) to
obtain g(y).
Now,
∂ Σ
∂
Σ ∂N ∂
2 ∫
∂N ∂2
∫
∂N ∂
= ∂x
− ∂y
∂ ( Mdx)
∂x ∂N ∂M
= ∂x
− ∂y
= 0. by (1).
∂x
∴
=
Mdx)
= Mdx)
4
∂y
∂y = .
∂y ∂x
∂x
This shows that the right hand side of (3) is a function of y only.
Hence (3) ⇒ g(y) = ∫ Σ
N − ∂ ∫
MdxΣ
dy.
Substituting this value of g(y) in (2) we get the required function f .
Hence the equation is exact and the general solution is given by f (x, y) = c.
1.4 WORKING RULE FOR SOLVING IT
1. Verify whether the given equation Mdx + Ndy = 0 is exact.
(i.e., ) verify ∂M ∂N
2. If exact, integrate M w.r.t x keeping y as constant.
3. Find out those terms in N which are free from x and integrate those terms w.r.t. y.
4. The sum of these two expressions equated to an arbitrary constant is the required
general solution of the given exact equation.
Note:(Grouping Method)
In certain cases the function f (x, y) can be guessed by grouping the terms of the equation
properly.
For example, the equation (y + sin y)dx + (x + x cos y)dy = 0 can be grouped as (ydx +
xdx) + (sin ydx + x cos ydy) = 0.
(ie.,) d(xy) + d(x sin y) = 0
Hence the solution is xy + x sin y = c.
1.5 PROBLEMS
.
Problem 1. Verify whether eydx + (xe
y + 2y)dy = 0 is exact, if so solve.
Solution:
Here M = ey and N = xe
y + 2y and ∂M = e
y =
∂N .
5
∂y ∂x
∫Now, Mdx
= x − 2x y − 2y x. 3
x2+y2
∫ −Now,
Mdx = x + tan (x/y). 2
Hence the equation is exact.
The given equation can be grouped as
(eydx + xe
ydy) + 2ydy = 0
d(xey) + d(y
2) = 0.
xey + y
2 = c.
Problem 2. Solve (x2 − 4xy − 2y
2)dx + (y
2 − 4xy − 2x2)dy = 0.
Solution:
Here M = x2 − 4xy − 2y
2 and N = y2 − 4xy − 2x
2 and ∂M = −4x − 4y = ∂N
.
Hence the equation is exact.
1 3 2 2 3
In N , the term free from x is y2 whose integral w.r.t y is 1 y
3.
Hence the complete solution is 1 x3 − 2x
2y − 2y
2x +
1 y3 = cJ.
3 3
(ie.,) x3 + y
3 − 6xy(x + y) = 3cJ = c.
Problem 3. Solve xdx + ydy − .
xdy−ydx Σ
= 0.
Solution:
The equation can be written as .
x + y
Σ
dx +
.
y − x
dy
Σ
= 0.
x2 + y2
Here M = x +
y and N = y − x and ∂M
x2 + y2
= x2−y2
= ∂N .
x2+y2
Hence the equation is exact.
x2+y2 ∂y (x2+y2)2 ∂x
1 2 1 2
In N the term free from x is y whose integral w.r.t y is 1 y2.
Hence the complete solution is 1 x2 +
1 y2 + 2 tan−1
(x/y) = cJ
2 2
(i.e.,) x2 + y
2 + 2 tan−1
(x/y) = c.
Exercises.
1. Verify whether the following differential equations are exact.
(i) ydx − xdy = 0.
(ii) (x2 − y)dx + (y2 − x)dy = 0.
6
dx
dx
x x
2. Show that the following equations are exact differential equations and solve them.
(i) (x2 − 2xy
2 + y4)dx − (2x
2y − 4xy
3 + sin y)dy = 0.
(ii) (x2 − ay)dx + (y
2 − ax)dy = 0.
Unit- II
1.6 EQUATIONS OF THE FIRST ORDER, BUT OF
HIGHER DEGREE
.
Equations solvable for dy
We shall denote dy
hereafter by p.
Let the equation of the first order and of the nth degree be
p
n + P1p
n−1 + P2pn−2 + ... + Pn = 0, .........(1)
where P1, P2, .. , Pn denote functions of x and y.
Suppose the first member of (1) can be resolved into factors of the first degree of the form
(p − R1)(p − R2)(p − R3) ... (p − Rn).
Any relation between x and y which makes any of the these factors vanish is a solution of
(1). Let the primitive of p − R1 = 0, p − R2 = 0, etc., be φ1(x, y, c1) = 0, φ2(x, y, c2) = 0,
..., φn(x, y, cn) = 0 respectively, where c1, c2, .. , cn are arbirary constants. Without any loss
of generality, we can replace c1, c2, .. cn by c, where c is an arbitrary constant.
Hence the solution of (1) is φ1(x, y, c).φ2(x, y, c) ... φn(x, y, c) = 0.
Problems:
Problem 1. Solve x2p
2 + 3xyp + 2y2 = 0.
Solution:
Solving for p, p = − y or p = − 2y
.
7
x
x
dx
dx x
dx x
x x
y x dx y
dy = − y
gives xy = c.
dy = − 2y
gives yx2 = c.
The solution is (xy − c)(yx2 − c) = 0.
Problem 2. Solve p2 +
.x + y − 2y
Σ p + xy +
y2
− y − y2
= 0.
Solution:
x x2 x
Solving for p, p = y − y or p =
y − x.
dy =
. 1 − 1
Σ dx or
dy − x = −x.
The first equation gives
log y = −x + log c
therefore, y = cxe−x
.
The second equation is linear in y. Hence the solution is y = −x + c i.e., y + x
2 − cx = 0.
The general solution is (y − cxe−x
)(y + x2 − cx) = 0.
Equations solvable for y
f (x, y, z) = 0 can be put in the form y = F (x, y) − − − − − − − (1)
Differentiating with respect to x, p = φ(x, p, dp
).
This being an equation in the two variables p and x, can be integrated by any of the foregoing
methods. Hence we obtain,
ψ(x, p, c) = 0 − − − − − − − − − − − − − (2)
Eliminating p between (1) and (2), the solution is got.
Equations solvable for x
Let f (x, y, z) = 0 be in this case put in the form
x = F (y, p) − − − − − − − (1)
Differentiating with respect to y, 1 = φ(y, p, dp
). p dy
Integrating leads to ψ(y, p, c) = 0 − − − − − − − − − − − −(2)
Eliminating p between (1) and (2), the solutions of (1) is got.
Examples:
Example 1. Solve xp2 − 2yp + x = 0.
8
2p
dy
e dy
dx
dx
Solution:
Given the equation is xp2 − 2yp + x = 0.
Solving for y, y = x(p2+1)
.
Differentiating with respect to x, p2−1
= dp
. p2−1
x.
∴ dx = dp . p dx p2
x p
Integrating p = cx.
Eliminating p between this and the given equation, the solution is 2cy = c2x
2 + 1.
Example 2. Solve x = y2 + log p
Solution:
Given the equation is x = y2 + log p − − − − − − − − − − − −(1)
Differentiating with respect to y, 1 = 2y + 1 dp
. p p dy
dp + 2py = 1. This is linear in p and hence
pey2
=
∫ y2
+ c − − − − − − − − − −(2)
The eliminate of p between (1) and (2) gives the solution.
1.7 Clairaut’s form.
The equation known as Clairaut’s is of the form
y = px = f (p) − − − − − − − − − −(1)
Differentiating with respect to x,
p = p + {x + fJ(p)} dp
. i.e., {x + fJ(p)} dp
dx dx
Either dp
= 0 or x + fJ(p) = 0.
dp = 0 gives p = c, a constant.
The solution of (1) is y = cx + f (c) − − − − − − − − − −(2)
We have to replace p in Clairaut’s equation by c. The other factor x + fJ(p) = 0 taken
along with (1) give, on elimination of p, a solution of (1). But this solution is not included in
the general solution of (2). Such a solution as this is called a singular solution.
9
dX
Examples:
Example 1. Solve y = (x − a)p − p2.
Solution:
The given equation is y = (x − a)p − p2.
This is Clairaut’s equation; hence the solution is
y = (x − a)c − c2.
Example 2. Solve y = 2px + y2p
3
The given equation is y = 2px + y2p
3
Putting X = 2x and Y = y2, the equation transforms into
Y = XP + P 3, where P =
dY = py.
This is Clairaut’s equation; hence Y = cX + c3.
The solution is y2 = 2xc + c
3.
1.8 Equations that do not contain (i) x explicitly (ii) y
explicitly
(i) Equations that do not contain x explicitly
Let the equation is of the form
f (y, p) = 0 − − − − − − − − − −(1)
If this is solvable for p, then p = φ(y) and hence is immediately integrable.
If (1) is solvable for y, so that y = φ(p), then the equation solvable for y method is applied.
(ii) Equations that do not contain x explicitly
Let the equation is of the form
f (x, p) = 0 − − − − − − − − − −(1)
If this is solvable for p, then p = φ(x), it is directly integrable.
If (1) is solvable for x, so that y = φ(p), then the equation solvable for x method is applied.
10
√
dx f2(x,y)
√
2
Examples:
Example 1: Solve x2 = (1 + p
2).
Solution:
Given the equation is x2 = (1 + p
2).
x = ± 1 + p2. Here y is explicitly absent.
Differentiate with respect to y 1 p dp
p =
(1 + p2) du
p2
Hence
∴ dy = √(1 + p2)
dp. − − − − − − − − − − − (1)
∫ p2
y + c =
√(1 + p2)
dp
1 √ y + c = (p 1 + p2 − sinh
−1 p) − − − − − − − − − (2)
Eliminating p between (1) and (2), the solution is got.
Example 2: Solve xyp2 + p(3x
2 − 2y2) − 6xy = 0.
Solution:
Given the equation is xyp2 + p(3x
2 − 2y2) − 6xy = 0.
This is homogeneous in x and y and solvable for p.
p = 2y
or - 3x . x y
dy =
2dx or ydy + 3xdx = 0
y x
∴ y = cx2 or y
23x
2 = c.
The solution is (y − cx2)(y
2 + 3x2 − c) = 0.
Unit- III
1.9 EQUATIONS HOMOGENEOUS IN x and y
. Consider dy
= f1(x,y)
, where f1 and f2 are homogeneous functions of the same degree
in x and y.
11
dx = .
∫
x
√
√
√
f1(x, y) can be written as xnφ(
y ) and f2(x, y) as x
nψ(
y ).
x x
If we put y = vx, dy
= v + xdv
. dx dx
The given equation becomes v + xdν
φ(ν)
ψ(ν)
The variables can be separated; the equation is
dx ψ(ν)dv +
x νψ(ν) − φ(ν)
= 0.
Integrating, log x + ψ(ν)dv = c.
νψ(ν)−φ(ν)
The solution is got by substituting y for ν after the integration is over.
Examples:
Example 1. Solve y2 + x
2 dy = xy
dy .
Solution:
dx dx
Given the equation is y2 + x
2 dy = xy
dy .
dx dx
Put y = vx; dy
= v + xdv
. dx dx
The equation reduces to dv(1−v) +
dx = 0.
v x
Integrating, log v − v + log x = log c.
The solution is y = cey/x
.
Example 2. Solve xdy − ydx = x2 + y2dx.
Solution:
Given the equation is xdy − ydx = x2 + y2dx.
Put y = vx; dy = vdx + xdv.
The equation reduces to √ dv = dx
. 1+v2 x
Integrating, log(v + √
1 + v2) = log x + log c.
The solution is y + x2 + y2 = cx2.
Exercises:
Solve the following equations:-
1. dy = x−y . dx x+y
2. (y2 − 2xy)dx = (x
2 − 2xy)dy.
3. (x + y)2dx = 2x
2dy.
12
2 n−2dx
1.10 LINEAR EQUATION WITH CONSTANT
COEFFICIENTS
Introduction:
A linear equation of nth
order with constant coefficients is of the form
dny
dxn
dn−1y
1 dxn−1 dn−2y
+ ... + any = X − − − − − − − − − − − − − (1)
where a1, a2, a3, ....an are constants and X is a function of x. This equation can also be
written in the form
(Dn + a1D
n−1 + a2Dn−2 + ..... + any = X
where D = d ; D
2 = d2
........ Dn =
dn
.
dx
Result 1:
dx2 dxn
Consider (Dn + a1d
n−1 + a2Dn−2 + ...... + an)y = 0 − − − − − − − − − −(2).
If y = y1(x), y = y2(x).......y = yn(x) are solutions of (2) then c1y1 + c2y2 + ....... + cnyn is
also a solution of (2) where c1, c2, ....., cn are arbitrary constants.
Result 2:
If c1y1 + c2y2 + .... + cnyn is the general solution of (2) and if y = u is a particular
solution of (1) then c1y+c2y2 + ...... + cnyn + u be the general solution of (1).
Definition:
The general solution of (1) is of the form y = Y + u where Y = c1y1 + c2y2 + ...... + cnyn
is the general solution of (2). Y is called the complementary function (C.F) and u is called a
particular integral (P.I).
Methods of finding complementary functions:
Consider the differential equation yJJ + ay
J + by = 0 − − − − − −(1) where a, b are
constants.
case i: Roots of the auxiliary equation are real and distinct say m1 and m2. Then the general
solution of (1) is y = c1em1x + c2e
m2x.
case ii: Roots of the auxiliary equation are imaginary. Then the solution of (1) is y =
eαx
(c1 cos βx + c2 sin βx).
+ a + a
13
2
case iii: Roots of the auxiliary equation are real and equal say m1 = m2.Then the general
solution is y = em1x(c1x + c2).
1.11 PROBLEMS
Problem 1. Solve (D2 − 5D + 6)y = 0
Solution:
The auxiliary equation is m2 − 5m + 6 = 0.
(ie.,) (m − 3)(m − 2) = 0.
Hence m = 2, 3
The complementary function is C.F= c1e3x
+ c2e2x
.
The general solution is given by y = c1e3x
+ c2e2x
.
Problem 2. Solve (D2 + D + 1)
2y = 0.
Solution:
The auxiliary equation is (m2 + m + 1)
2 = 0.
(ie.,) (m2 + m + 1)(m
2 + m + 1) = 0.
Hence m = −1±i
√3 (twice)
The complementary function is C.F =
e−x/2
[(c1 + c2x) cos(√
3/2)x + (c3 + c4x) sin(√
3/2)x].
The general solution is given by e−x/2
[(c1 + c2x) cos(√
3/2)x + (c3 + c4x) sin(√
3/2)x].
Unit- IV
1.12 LINEAR EQUATIONS WITH VARIABLE
COEFFICIENTS
A homogeneous linear equation is of the form
n dny
x dxn
+ p1x dn−1
dxn−1 + ... + pny = X, n−1
14
2
2
2
D3+1
dz
x
where p1, p2, ..., pn are constants and X a function of x.
Examples
Example 1.
Solve x3 d3y
+ 3x2 d2y
+ xdy
+ y = x + log x.
dx3
Solution:
dx2 dx
Given the equation is x3 d3y
+ 3x2 d2y
+ xdy
+ y = x + log x. Putting z = log x and D = d
the equation becomes
(D3 + 1)y = e
z + z.
dx3 dx2 dx dz
The auxiliary equation is m3 + 1 = 0.
∴ m = −1 or 1±√
3i .
C.F = C1e−z
+ ez/2
(C2 cos √
3 z + C3 sin √
3 z). 2 2
C.F = C1x−1 +
√x[C2 cos(
√3 log x) + C3 sin(
√3 log x)].
2 2
P.I. = 1 (ez
+ z)
= 1 ez + (1 − D3
)z
= x + log x.
y=C.F+P.I.
Example 2.
Solve x2 d2y
+ 3xdy
+ y = 1 .
dx2 dx
Solution:
(1−x)2
Given the equation is x2 d2y
+ 3xdy
+ y = 1 .
dx2 dx (1−x)2
Putting z = log x and D = d the equation becomes
(D2 + 2D + 1)y =
1 . (1−x)2
The auxiliary equation is (m + 1)2 = 0
∴ m = −1 (twice)
C.F. = e−
z(A + Bz) = 1 (A + B log x).
P.I. = 1 1 changing D to the operator θ = x
d
(θ+1)2 (1−x)2 dx
= 1 1 1 θ+1 x 1−x
= 1 log x
x 1−x
y=C.F+P.I.
15
du
−
4 8
1.13 EQUATIONS REDUCIBLE TO THE LINEAR
EQUATIONS
Consider an equation of the form
n dny
n−1
dn−1
(a + bx) + (a + bx) dxn p1
dxn−1 + ... + pny = X,
where p1, p2, ... , pn are constants and X is any function of x.
Putting z = a + bx, the equation transforms into
n dny
p1zn−1 dn−1y
pn 1
z − a
z dzn
+ b dzn−1
+ .... + bn
y = bn
X( b
.
This is linear homogeneous equation.
Example: Solve (5 + 2x)2 d2y − 6(5 + 2x)
dy + 8y = 6x.
Solution:
dx2 dx
Given the equation is (5 + 2x)2 d2y − 6(5 + 2x)
dy + 8y = 6x.
dx2 dx
Putting z = 5 + 2x, the equation becomes
2 d2y dy
4z dz2
− 12zdz
+ 8y = 3(z − 5)
.
Putting u = log z and D = d , the equation is now transformed into
(4D2 − 16D + 8)y = 3(e
u − 5).
The auxiliary equation is m2 − 4m + 2 = 0.
∴ m = 2 ± √
2. C.F= (5 + 2x)
2[A(5 + 2x)
√2 + B(5 + 2x
√2 .
) ]
P.I= 3 (eu 5)
4(D2−4D+2)
P.I= −3 z − 15
P.I= −3 x − 45
2 8
y=C.F+P.I.
16
Chapter 2
SIMULTANEOUS AND TOTAL
DIFFERENTIAL EQUATIONS
UNIT-V
2.1 INTRODUCTION
In this chapter, we shall discuss differential equations is which there is one independent
variable and two or more than two dependent variables. To solve such equations completely,
there must be as many equations as there are dependent variables. Such equations are called
its ordinary simultaneous differential equations
17
P = = Q R
2.2 Simultaneous equations of the first order and first
degree
Taking z as the independent variable and x and y as the pair of dependent variables, a
pair of simultaneous differential equations of the first order and first degreemay be written as
dx dy P1
dz + Q1
dz + R1 = 0
dx dy P2
dz + Q2
dz + R2 = 0 (2.1)
where P1, Q1, R1, P2, Q2 and R2 are the functions of x, y and z. These equations can be
written as
P1dx + Q1dy + R1dz = 0
P2dx + Q2dy + R2dz = 0 (2.2)
The ratios of the differentials dx, dy, dz can be obtained as
dx
Q1R2 − Q2R1
dy =
R1P2 − R1P1
dz =
P1Q2 − P2Q1
or dx dy dz
= = (2.3) P Q R
where P , Q, R are the functions of x, y and z.
We shall take (2.3) as the standard form for a pair of ordinary simultaneous equations of
the first order and first degree.
Two independent relations between the variables x, y and z each involving an arbitrary
constant constitute the general solution of equations (2.2) and so of equations (2.3) if and
only if dx and dy
deducible from them satisfy equations (2.2) and so (2.3). dz dz
2.2.1 Methods for solving dx dy dz
Case i. Whentwo of the ratios in equation (2.3) involve only the two corresponding out
of the three variables x, y and z.
18
P =
P =
Q
Q
Suppose that the equation dx dy does not involve z or that it can be reduced to a
differential equation in x and y only by the cancellation of a common factor in P and Q.
In that case the equation dx dy is of the first order and first degree and the equation
may be solved. Let its solution be u = c1. It will be a part of the general solution of equations
(2.3) since
∂u ∂u P + Q
∂x ∂y
∂u = 0, = 0
∂z
and therefore P ∂u
+ Q∂u
+ R∂u
= 0 ∂x ∂y ∂z
We might, either by making use of the relation u = c1 or otherwise independently, be
similarly able too obtain another relation v = c2 which is also a part of the general solution
of equations (2.3). These two relations, if they are independent of each other, will constitute
the general solution of equations (2.3).
Case ii. A part or the whole of the general solution of equations (2.3) can also be found
by the following method:-
Each of the ratios in equations (2.3) is equal to
ldx + mdy + ndz
lP + mQ + nR
where l, m and n are any multipliers; and hence, if the multipliers l, m and n are such that
lP + mQ + nR = 0, we must then also have ldx + mdy + ndz = 0.
If the left-hand side of this equation can be expressed as the differential du of some
function u of x, y and z then u = c1 is a solution of the equation ldx + mdy + ndz = 0 if l,
m and n are proportional to
∂u ∂u ∂u , , .
∂x ∂y ∂z
Since lP + mQ + nR = 0, P ∂u
+ Q∂u
+ R∂u
= 0. ∂x ∂y ∂z
Hence u = c1 is a part of the general solution of equation (2.3).
A very simple case in which u can be obtained immediately from equation ldx + mdy +
ndz = 0 that where l is a function of x alone, m of y alone and n of z alone. In that case
u =
∫
(ldx + mdy + ndz)
19
z
= .
If two essentially different sets of such multipliers l, m and n can be obtained, then the
general solution of equation (2.3) can be deduced by this method.
Examples:
1. Solve the equations dx = dy = dz
.
Solution:
yz xz xy
Taking the first two equations, we get dx = dy
, i.e., dx = dy
yz xz y x
i.e., xdx − ydy = 0 i.e., x2 − y2 = c1
Taking the first and the last equations, we get
dx = dz , i.e., dx = dz yz xy z x
i.e., xdx − zdz = 0 i.e., x2 − z2 = c2
∴ The general solution is φ(x2 − y2
, x2 − z2
) = 0.
2. Solve the equations dx = dy
= dz .
Solution:
−y2−z2 xy xz
The equation dy
= dz
gives dy
= dz
xy xz y z
i.e., log y = log z + log c1, i.e., y = c1z
Considering each of the ratios in the given equation
dx = dy = dz , we get xdx+ydy+zdz = xdx+ydy+zdz −y2−z2 xy xz x(−y2−z2)+xy2+xz2 0
∴ xdx + ydy + zdz = 0, ∴ x2 + y2 + z
2 = c2.
Hence the general solution is φ .
y , x
2 + y2 + z
2Σ
= 0
Exercises
Solve the following equations:
1. dx = dy = dz x2+y2 2xy (x+y)z
2. dx = dy = dz . y2−z2+x2 2xy 2xz
dx y+z
dy
x+z dz x+y
4. dx = dy = dz . x2−yz y2−zx z2−xy
5. dx = dy = dz . y+zx −x−yz x2−y2
3. =
20
dt
2.3 Simultaneous linear differential equations
We now treat systems of equations. There is only one independent variable usually
denoted by the symbol t and a number of dependent variables equal to the number of
equations. We shall further suppose that the equations are linear.
Let D stands for d . Taking the simplest case of two dependent variables x and y, the
equations can be written in the form
f1(D)x + φ1(D)y = T1 (2.4)
f2(D)x + φ2(D)y = T2 (2.5)
where we shall suppose that f1, f2, φ1 and φ2 are rational integral functions of D with
constant coefficients and T1, T2 explicit functions of t. To eliminate y as in solving
simultaneous algebraic equations we operate on (2.4) by φ2(D) and (2.5) by φ1(D) and
subtract. Then we have
{f1(D)φ2(D) − f2(D)φ1(D)}x = φ2(D)T1 − φ1(D)T2
Substituting for x in (2.4) or (2.5), y can be found.
It must be noted that the number of arbitrary constants in the complete solution is the
exponent of the highest index in the operator D of f1(D)φ2(D) − f2(D)φ1(D).
The above method can be extended to equations of more than two dependent variables.
The following examples illustrate the process clearly.
Examples:
1. Solve the system
and
dx
dt + 2x − 3y = t (2.6)
Solution
The equations can be written as
dy 2t
dt − 3x + 2y = e
(2.7)
(D + 2)x − 3y = t and (D + 2)y − 3x = e2t
21
. Σ
1 2 25 5 7
1 2 25 5 7
To eliminate y, operate on (2.6) by D + 2 and (2.7) by 3 and add. We get
((D + 2)2 − 9)x = (D + 2)t + 3e
2t
i.e.,(D2 + 4D − 5)x = 1 + 2t + 3e
2t
The auxillary equation is m2 + 4m − 5 = 0.
∴ m = 1 or −5.
C.F. is C1et + C2e
−5t.
1 P.I. =
D2 + 4D − 5 (1 + 2t + 3e
2t)
1 4D = 1 +
5 5 (1 + 2t) +
3e2t
7
13 =
25 −
2t 3e2t
+ .
5 7
∴ x = C et + C e
−5t +
13 − 2t + 3e2t
Thus y = C et − C e
−5t − 12 − 3t +
4e2t
.
2. Solve the system 4
dx + 9
dy + 2x + 31y = e
t (2.8)
and
dt dt
Solution
dx dy 3 + 7
dt dt + x + 24y = 3 (2.9)
The equations can be written as
2(2D + 1)x + (9D + 31)y = e
t and (3D + 1)x + (7D + 24)y = 3.
Eliminating y, we get (D2 + 8D + 17)x = 31(e
t − 3).
The auxillary equation is m2 + 8m + 17 = 0.
∴ m = −4 ± i.
∴ C.F. = e−4t
(C1 sin t + C2 cos t).
22
D2+8D+17 26 17
3 4 17 13
2 1 2 1 13 17
dt2 dt dt2 dt
P.I. = 1 31(e
t − 3) =
31 et − 93 .
∴ x = e−4t
(C1 sin t + C2 cos t) + 31 et
− 93 . 26 17
Similarly, eliminating x, we have
(D2 + 8D + 17)x = 5 − 4e
t.
t
∴ y = e−4t
(C cos t + C sin t) + 6 − 2e .
The relations between the constants are got by substituting these values in the two
equations. Substituting, we get
sin t(−14C1 − 4C2 − 5C4 + 9C3) + cos t(−14C3 + 4C1 + 9C4 − 5C3) = 0,
whence C3 = −(C1 + C2) and C4 = C2 − C1. t
∴ y = [(C − C ) sin t − (C + C ) cos t] e−4t
− 2e + 6 .
Exercises
Solve the following equations:
1. dx = ax + by + c; dy = a
Jx + b
Jy + c
J.
dt dt
2. d2x + 2 dx − x + sin t = 0;
d2y − dy − y + cos t = 0.
3. 3(1 − D)x + 4y = 3t + 1; 3(D + 1)y + 2x = et
4. (D + 5)x + y = et; (D + 3)y − x = e
2t.
5. (D + 8)x + (2D + 1)y = 0; (6D − 2)x − (3D − 11)y = 0.
UNIT-VI
2.4 Complete solution given a known integral
If an integral included in the complementary function of the given equation be known,
the complete solution can be found in terms of this known integral.
Let
d2y
dx2
+ P dy
+ Qy = R (2.10) dx
23
dx
1
∫ ,
1
y1 dx2
+ dx 2
dx + Py1
dx2 dx
be the given equation, where P , Q and R are the functions of x.
Let y = y1 be a known integral in the C.F. of (2.10).
i.e., d2y dy
dx2 + P
dx + Qy = 0
∴ d
2y1 dy1
dx2 + P
dx + Qy1 = 0 (2.11)
Putting y = y1v in (2.10), where v is a function of x, we get
d2v
dv .
dy1 Σ
in virtue of (2.11). This is linear in dv ; Hence
dv c1
− ,
P dx e− ,
P dx ∫
, P dx
Integrating,
dx =
y2 e + 2 Ry1e dx 1
v = c2 +
c1
e− P dx
y2
dx +
∫ .e−
, P dx
∫
Ry1e
, P dx dx
Σ
dx.
The solution of (2.10) is y = vy1, where v has the above value.
It must be noted that this solution includes the given solution and that there are two
arbitrary constants.
Examples:
1. Solve x d2y − (2x − 1) dy
+ (x − 1)y = ex.
dx2 dx
Solution
As the sum of the coefficients of the first member is zero, ex is a solution of
x d2y − (2x − 1)
dy + (x − 1)y = e
x.
Putting y = vex, the given equation reduces to x
d2v + dv
= 1.
Solving, dv x = x + c1. i.e., dv = 1 + c1 .
dx2 dx
dx dx x
Integrating v = x + c1 log x + c2.
Hence y = ex (x + c1 log x + c2).
y2
y 1
= R
24
dx2 dx
dx2 dx
dx2 dx
2. Solve x2 d
2y − (x2 + 2x) dy + (x + 2)y = x
3e
x.
dx2 dx
Solution
y = x is a solution of this equation without the second member,
Putting y = vx, the given equation reduces to v2 − v1 = ex.
Hence, v1ex = x + c1 or v1 = (x + c1)e
x.
Integrating v = (c1 − 1)ex + xe
x + c2.
∴ y = c2x + (c1 − 1)xex + x
2e
x.
Exercises
Solve the following equations:
1. x d2y
+ 2(4x − 1) dy
− (9x − 2)y = x3e
x.
2. xy2 + (1 − x)y1 − y = ex.
3. x2 d
2y − 2x(1 + x) dy
+ 2(1 + x)y = x3.
4. (x sin x + cos x) d2y − x cos x
dy + y cos x = 0
dx2 dx
5. d2y − 2x
dy + x
2y = 0.
6. x d2y
+ 2 dy + xy = 0.
dx2 dx
25
,d v P dx+ Iv = Re
(2.13)
2 dx
P 2
x2
2
2 dx − dx
2
1
4
UNIT VII
2.5 Reduction to normal form
Consider
Putting y = y1v, this becomes
d2y
dx2
+ P
dy + Qy = R (2.12)
dx
d2v
dv .
dy1 Σ .
d2y1
dy1 Σ
y1. dx2
+ dx
If y1 be chosen to satisfy
2 dx
+ py1 + v dx2
+ P dx
+ Qy1 = R
2 dy1
+ py = 0
− 1 , Pdx
dx 1
2 1 2
Where I = Q − 1 dP
− 4 .
dx2
Equation 2.13 is immediately integrable if I be either a constant or a , where a is a constant.
This method is either called reducible equation 2.12 to the normal form or removing the first
derivative.
Examples:
Example 1: 2
Solve y2 − 4xy1 + (4x2 − 3)y = e
x .
Here P = −4x, Q = 4x2 − 3 and R = e
x2
y = e− 1 ,
Pdx = ex2
putting y = vy1 I = Q − 1 dP
P 2
= −1 and the equation reduces to d2v − v = 1.
Hence v = Aex + Be
−x − 1
Therefore y = ex2
(Aex + Be
−x − 1).
Example 2:
i.e., y1 = e , then the above equation becomes,
26
2
. Σ
2
2 dx −
dx2
1 1 2
. Σ . Σ
dx2 dx
dx2 dx
2
4
Solve 4x2 d2y
+ 4x5 dy
+ (x8 + 6x
4 + 4)y = 0.
dx2 dx
Here P = x3; Q =
1 (x6 + 6x2 + 4 ).
Hence
4 x3
y =e− 1 ,
Pdx = e− x
1 2
8
1 dP P 2 1
I =Q − 2 dx
− 4
= x2
.
Hence the equation in v, where y = vy1, is
d2v 1
dx2 +
x2 v = 0
This is a homogeneous linear equation whose solution is
√xAcos
.√3
logx + B
Σ
Example 3:
Solve d2y − ( dy
)2tanx + 5y = 0
Here P = −2tanx; Q = 5 and R = 0.
To remove the first derivative we choose u = e
Put y = vy1 = vsecx.
− 1 , Pdx = e−
, (−2tanxdx)dx
= elogsecx
= secx.
I = Q − 1 dP P = 5 + sec2x − tan
2x = 6
∴ The given equation reduces to d2v + 6v = 0 which is a linear equation with constant
coefficients.
The auxillary equation is m2 + 6 = 0. Hence m = ±i
√6.
∴ y1 = c1cos√
6x + c2sin√
6x.
∴ The general solution of the given equation is
y = vy = secx(c cos√
6x + c sin√
6x)
Exercise
Solve the following by removing the first derivative:
(i) d2y + 4x
dy + 4x
2y = 0
4
27
dx2 x dx x2
dx2 dx
dx2 dx
(ii)
. d2y
Σ
− 2
. dy
Σ
+
.
1 + 2 y
Σ
= xex
(iii)
. d2y
Σ
+ 4x
. dy
Σ
+ 4(x2 − 1)y = −3e
x2 sin2x
(iv)
. d2y
Σ
+ (2/x)
. dy
Σ
+ n2y = 0
28
dz
. Σ
. Σ
+
Q = µ dz dx
, where µ is constant. Then equation 2.15 reduces to
dx
2.6 Change of the independent variable
Let z be the new independent variable.
Consider
d2y
dx2
+ P dy
+ Qy = R (2.14) dx
dy dy dz
= . ; dx d
y d
2y dz
2 2
= +
dy d2z
2
dz dx dx
The equation 2.14 transforms into
dz dx dz dx
d2y .
dz Σ2
dy .
d2z dz
Σ
+ P + Qy = R (2.15)
dz2 dx dz dx2 dx
We may choose z such that the coefficient of dy
,
Therefore z = ∫
e−
, Pdxdx
i.e., d
2z
dx2
+ P dz
vanishes (2.16) dx
Using this value of z, equation 2.15, may become integrable. One particular case where
equation 2.15 becomes immediately integrable is when . Σ2
dy
dz2
+ µy = R
dz 2 dx
The second case, when equation 2.15 becomes integrable, is when
. dz
Σ2
Then equation 2.15 becomes a homogeneous linear equation.
Example 2:
Solve d2y
+ dy
tanx + ycos2x = 0.
dx2 dx
Here P = tanx and Q = cos2x.
Choosing z = ∫
e,
−tanxdxdx = sinx
2
Qz2 = µ
29
. Σ . Σ
dz2
. Σ . Σ
The equation transforms to d2y
+ y = 0 as here Q = cos2x =
. dz
Σ2.
dz2 + P1
dz + Q1y = R1whereP1 =
dx2 + P
dx
dx2 dx
dx2 dx
dz2 dx
Thereforey = Acosz + Bsinz.
= Acos(sinx) + Bsin(sinx).
Example 2:
Solve d2y
+ dy
cotx + 4ycosec2x = 0 by changing the independent variable x to z.
dx2 dx
Here P = cotx; Q = 4cosec2x; R = 0.
To change the independent variable x to z choose z such that
z =
∫
e−
, Pdxdx
∴ z =e−,
cotxdxdx
=
∫
elogcosecxdx
=
∫
cosecxdx
=logtan(x/2).
With this choice of z the given equation reduces to
d2y dy
. dz
Σ−2. d
2z
dz Σ
Q1 = Q
dz −2
dx
= 4; R1 = R
dz −2
dx
= 0.
Hence the equation is transformed to d2y
+ 4y = 0 which is a linear second order differential
equation with constant coefficients, whose solution is t = c1cos2z + c2sin2z.
∴ The general solution is
y = c1cos(2logtan(x/2)) + c2sin(2logtan(x/2)).
Exercise
(i) d2y
+ (2/x) dy
+ (a2/x
4)y = 0
(ii)
. d2y
Σ
− cotx
. dy
Σ
+ ysin2x = 0
dx = 0
30
dx2 dx
dx2 dx
(iii)
. d2y
Σ
− (1/x)
. dy
Σ
+ 4x2y = x
4
(iv)
. d2y
Σ
+
. dy
Σ
tanx − 2ycos2x = cos
4x.
UNIT VIII
2.7 Method of variation of parameters
The method of variation of parameters is another method for solving a linear differential
equation, either of the first order or of the second order. If the given equation is of the form
f (D)y = x, this method can be applied to get the general solution, provided the
corresponding homogeneous equation, viz., f (D)y = 0 can be solved by earlier methods.
The procedures to solve linear equations of the first and second orders are the following.
Solution of the equation
d2y + P
dy + Qy = R (2.17)
dx2 dx
Where P , Q and R are functions of x or constants.
The homogeneous equation corresponding to equation 2.17 is
d2y
dx2
+ P dy
+ Qy = 0 (2.18) dx
Let the general solution of equation 2.18 be
y = Au + bv (2.19)
Where A and B are arbitrary constants (parameters) and y = u(x) and y = v(x) are
independent particular solutions of equation 2.18.
Now, we treat A and B as functions of x and assume equation 2.19 to be the general
solution of equation 2.17. Differentiating equation 2.19 with respect to x, we have
dy = (Au
J + BvJ) + (A
Ju + B
Jv) (2.20)
dx
31
dx x2+1
We choose A and B such that
Then equation 2.20 becomes
A
Ju + B
Jv = 0 (2.21)
dy = Au
J + BvJ (2.22)
dx
Differentiating equation 2.22 with respect to x, we have
d2y
dx2
= AuJJ
+ BvJJ
+ AJu
J
+ BJv
J
(2.23)
Since equation 2.19 is a solution of equation 2.17, 2.19, 2.22 and 2.7 satisfy equation 2.17.
(AuJJ + Bv
JJ + AJu
J + BJv
J) + P (Au
J + BvJ) + Q(Au + Bv) = R
A(uJJ + Pu
J + Qu) + B(vJJ + P v
J + Qv) + AJu
J + BJv
J = R (2.24)
since y = u is a solution of equation 2.18
uJJ + Pu
J + Qu = 0
similarlyvJJ + P v
J + Qv = 0
Inserting these values in equation 2.24, it reduces to
AJu
J + BJv
J = R (2.25)
Solving equation 2.21 and 2.25, we get the values of AJ and B
J integrating which, we get
the values of A and B as functions of x. Using these values in equation 2.19, we get the
required general solution of equation 2.17. Examples
Example 1:
Solve the equation (x2 + 1)
dy + 4xy =
1 , by using the method of variation of
parameters.
The homogeneous equation corresponding to the given equation is
(x2 + 1)
dy + 4xy =0 (2.26)
dx dy 4x
y +
x2 + 1 dx =0
32
dx
−
Integrating, we get logy + 2log(x2 + 1) = logc
c y =
(x2 + 1)2 is the solution of equation 2.26 (2.27)
Treating c as a function of x and differentiating equation 2.27 with respect to x, we have
dy (x2 + 1)
2c
J c.2(x2 + 1)2x
dx =
(x2 + 1)4
(2.28)
Using equation 2.27 and 2.28 in the given equation, we have
(x2 + 1)
2c
J − 4cx(x2 + 1) 4cx 1
(x2 + 1)2 +
(x2 + 1)2 =
x2 + 1
i.e.(x2 + 1)
2c
J − 4cx(x2 + 1) + 4cx(x
2 + 1) =(x2 + 1)
2
i.e.cJ =1
∴ c =x + k (2.29)
using equation 2.29 in 2.27, the required solution of the given equation is (x
2 +1)2y = x+k,
where k is an arbitrary constant.
Example 2:
Solve the equations dy
+ xsin2y = x3cos
2y, by the method of variation of parameters.
The method of variation of parameters can be applied to solve only a linear differential
equation. The given equation is not linear. We shall convert the given equation into a linear
equation and then apply the method of variation of parameters. Dividing the given equation
by cos2y, we get,
sec2y
dy + 2xtany = x
3 (2.30) dx
Putting tany = z , equation 2.30 becomes
dz + 2xz = x
3, which is linear (2.31)
dx
The homogeneous equation corresponding to equation 2.31 is
dz + 2xz = 0or
dz
dx z + 2xdx = 0 (2.32)
33
∫
t
−
dx
dx
dx2
dx2
−
Integrating, we get logz = logc − x2
i.e. the solution of equation 2.32 is
z = ce−x2
(2.33)
Treating c as a function of x and differentiating equation 2.33 with respect to x,
= 2cxe−x2
+ cJe−x2
(2.34) dx
Using equation 2.34 and 2.33 in 2.31,
−2cxe−x2
+ cJe−x2
+ 2cxe−x2
=x2
cJ =x
3e
x2
c =
∫
x3e
x2
dx + k
= 1
tetdt + k, on puttingx
2 = t 2 1
= (tet
2 − e ) + k
1 = (x
2 2
− 1)e + k (2.35)
Using equation 2.35 in 2.33, the required general solution of equation 2.31 is
1 z = (x
2 2
− 1) + ke −x2
Therefore the general solution of the given equation is
tany = 1
(x2 1) + ke
−x2
2
where k is an arbitrary constant.
Exercise
(i) dy − ytanx = exsecx
(ii) xdy + (1 + x)y = e
−x
(iii) d2y
+ y = xcosx
(iv) d2y
+ y = xsinx
dz
x2
34
. Σ . Σ . Σ
∂z ∂y ∂x ∂z ∂y ∂x
∂Q ∂R ∂R ∂P ∂P ∂Q ∂z
− ∂y ∂x
− ∂z ∂y
− ∂x
2.8 Total differential equations
The equation
Pdx + Qdy + Rdz = 0 (2.36)
where P , Q and R are functions of x, y, z is called a total differential equation. If there
exists a function u such that du = Pdx + Qdy + Rdz then u = c is a solution of equation
2.36 and in this case we say that equation 2.36 is integrable.
2.8.1 Methods of solving the differential equation Pdx+Qdy +Rdz = 0
Method 1 (By taking one variable constant)
Suppose the condition of integrability is satisfied. Consider any one of the variables z as
constant. Hence dz = 0. Now integrate the resulting equation Pdx + Qdy = 0 where the
arbitrary constant of integration is taken as an arbitrary function of z. On differentiation
of this integral with respect to x, y and z and comparing it with the given equation we can
determine the arbitrary function of z.
Method 2 (P ,Q, R are homogeneous functions)
If P , Q, R are homogeneous in x, y, z then the given equation can be solved by effecting the
substitution x = zu and y = zv.
Method 3(By forming auxiliary equations)
Consider Pdx + Qdy + Rdz = 0 which satisfies the condition of integrability
P
.∂Q
− ∂R
Σ
+ Q
.∂R
− ∂P
Σ
+ R
.∂P −
∂Q Σ
= 0
Comparing these two equations we have
dx dy dz = =
These equations are called auxiliary equations and can be solved like simultaneous equation.
If u = a and v = b are two solutions of the auxiliary equations then by comparing adu +
bdv = 0 with the given equation we get the values of a and b and get the complete solution.
35
2.8.2 Solved problems
1. Solve z2dx + (z
2 − 2yz)dy + (2y2 − yz − xz)dz = 0
We can easily verify that the conditions of integrability is satisfied since the given equation
is homogeneous we put x = uz and y = vz such that dx = zdu + udz and dy = zdv + vdz.
Hence the given equation becomes
z2(zdu + udz) + z
2(1 − 2v)(zdv + vdz) + (2v
2 − v − u)z2dz =0
∴ z3du + z
3(1 − 2v)dv + [u + v(1 − 2v) + 2v
3 − v − u]z2dz =0
∴ z3du + z
3(1 − 2v)dv =0
∴ du + (1 − 2v)dv =0
∴ u + v − v2 =c
∴ x
+ y
− (y
)2 =c
z z z
∴ (x + y)z − y2 =cz2is the solution
2. Solve yz2(x
2 − yz)dx + zx2(y
2 − xz)dy + xy2(z
2 − xy)dz = 0
Dividing by x2y
2z2 and regrouping we get
ydx − xdy +
ydz − zdy +
zdx − xdz =0
y2 z2 x2
∴ d
.x
Σ
+ d
.y Σ
+ d
. z
Σ
=0
y z x
Hence the required solution is x + y +
z = 0.
Exercise
Solve:
y z x
(i) 3x2dx + 3y
2dy − (x3 + y
3 + e2z
)dz = 0
(ii) (y + z)dx + (z + x)dy + (x + y)dz = 0
(iii) (x − y)dx − xdy + zdz = 0
(iv) (yz + z2)dx − xzdy + xydz = 0
36
Chapter 3
PARTIAL DIFFERENTIAL
EQUATIONS
Unit- IX
3.1 Necessary and Sufficient condition of integrability of
Pdx+Qdy+Rdz=0
In a total differential equation, we have the differential coefficients of several dependent
Pdx + Qdy + Rdz = 0 (3.1)
Criterion of integrability
Let (3.1) have an integral
u(x, y, z) = a (3.2)
where a is an arbitrary constant.
Then differentiating (3.2) totally,
δu
δx
dx +
δu dy +
δy
δu dz = 0.
δz
37
δx δz δy δx
δz δy δx δz δy δx
This is identical with (3.1). Hence,
δu = µP,
δx
δu = µQ and
δu
δy δz
= µR.
δ (µP ) =
δy
δy
δz
δ2u
δyδx
δx
δy
δ2u
= δxδy
δx
δ = (µQ).
δx
(3.3)
(3.4)
µ
.δR
− δP
Σ
=P δµ
− Rδµ
(3.5)
Multiplying (3.3), (3.4), (3.5) by R, P and Q respectively and adding, we obtain
P
.δQ
− δR
Σ
+ Q
.δR
− δP
Σ
+ R
.δP
− δQ
Σ
= 0. (3.6)
The above condition is apoken of as the condition of integrability of equation (3.1).
We have just shown that the above condition (3.6) is necessary for the existence of the
integral of equation (3.1).
We shall show that the condition is sufficient by proving that an integral of (3.1) can be
found when (3.6) holds good.
1.1 If, relation (3.6) exists among P, Q, R a similar relation holds good between the
coefficients of
µPdx + µQdy + µRdz = 0 (3.7)
where µ is any function of x, y, z.
If Pdx + Qdy (where z is a parameter) is not an exact differential, we can always find
Similarly,
µ
.δP
− δQ
Σ
=Qδµ
−
δµ P
δy
µ
.δQ
−
Σ δR
δµ δµ =R
δy − Q
δx
and
38
.
.
.
δz δz
δz
δx δy δz δz
=
an integration factor µ such that µPdx + µQdy is exact. Hence, without loss of generality,
Pdx+Qdy can be regarded as an exact differential and (3.7) as the equation to be considered.
δP δQ =
δy δx
i.e.,
Let V =
∫
(Pdx + Qdy).
Hence
δV dx +
δx
δV
δV dy = Pdx + Qdy
δy
δV
(3.6) now reduces to
= P and δx δx
δP δ2V
and δQ
δz δzδx δz
= Q
δ2V
= δzδy
δV .
δ2V δR
Σ δV
.δR δ
2V
Σ
δx δzδy −
δy
δV
+ δy
δ .
δV
δx −
δzδx = 0
− RΣ
.
i.e., δx
δx δz . Σ.
δV δ
δy δy
δz − R .
i.e., δ(V, δV
− R) = 0.
δ(x, y) ∴ A relation independent of x and y exists between V and δV
− R; hence δV − R can be
expressed as a function of z and V .
Let δV − R = φ(z, V ).
Since Pdx + Qdy + Rdz
= δV
dx + δV
dy + δV
dz +
.
R − δV
Σ
dz
= dV − φ(z, V )dz.
This, being an equation in two variables z and V , leads on integration to F (V, z) = 0.
Hence (3.6) is the necessary and sufficient criterion for integrability of (3.6).
δz
= 0 δV
∴
39
3.2 Rules for integrating Pdx+Qdy+Rdz=0
• When the condition of integrability is satisfied, consider one of the variables, say z, as
constant, i.e., dz = 0 and integrate the equation Pdx + Qdy = 0.
• Put the arbitrary constant of integration that occurs in this integral as an arbitrary
function of z.
• This is justified as the arbitrary constant in this integral is a constant only with respect
to x and y.
• Differentiating this integral with respect to x, y and z and comparing the result with
(3.1), we can determine the arbitrary function of z.
The following examples will illustrate the process.
3.3 Examples
Ex.1. Solve (y
2 + yz)dx + (xz + z2)dy + (y
2 − xy)dz = 0 (3.8)
Here the condition of integrability is satisfied. Neglecting(y2 − xy)dz, we get
(y2 + yz)dx + (xz + z
2)dy = 0
dx
x + z
zdy +
y(y + z)
= 0.
Integrating, on the assumption that z is a constant,
log(x + z) + log y
y + z
= logc
i.e., (x + z)y
y + z = c = f (z), (3.9)
40
where the arbitrary constant c of integration is put as f (z), an arbitrary function of z.
Differentiating (3.9) totally with respect to x, y, z,
(y + z)[(dx + dz)y + (x + z)dy] − y(x + z)(dy + dz) = f J(x)dz
(y + z)2
i.e., (y2 + yz)dx + (xz + z
2)dy + dz[y
2 − xy − f J(z)(y + z)
2] = 0.
comparing this with (3.8), we have fJ(z)(y + z)
2dz = 0.
As (y + z)2dz ƒ= 0, f
J(z) = 0...f (z) = constant c.
Hence the integral of (3.8) is y(x + z) = c(y + z).
Exercises
Verify the condition of integrability in the following equations and solve them: 1.
(y + z)dx + (z + x)dy + (x + y)dz = 0.
2. (y + z)dx + dy + dz = 0.
41
Unit-X
3.4 Partial Differential Equations of the First Order
We now consider equations in which the number of independent variables is two or
more and only one dependent variable. We usally denote this by z and the independent
variables by x and y, if there be two; if there be n independent variables, we shall call them
x1, x2, ...xn. The partial derivatives δz , δz
are denoted by p and q while, in the latter case, δx δy
δz , δz ... δz are represented by p1, p2, ... pn respectively. δx1 δx2 δxn
Partial differential equations are those which involve oe or more partial derivatives. The
order of a partial differential equation is determined by the highest order of the partial
derivative occuring in it. We consider only partial differential equations of the first order.
3.5 Classification of integral
Let the partial differential equation be
F (x, y, z, p, q) = 0 (3.10)
Let the solution of this be
φ(x, y, z, a, b) = 0 (3.11)
where a and b are arbitrary constants.
The solution (3.11) which contains as many arbitrary constants as there are independent
variables is called the complete integral of (3.10).
A particular integral of (3.10) is that got by giving particular values to a and b in (3.11).
42
3.5.1 Singular integral
The elimation of a and b between
φ(x, y, z, a, b) = 0;
δφ δφ = 0 and = 0
δa δb
when it exists, is called the singular integral.
Geometrically, this includes the envelope of the two parameter surfaces represented by
the complete integral (3.11) of (3.10). The two parameters occuring in (3.11) are a and b.
The locus of all points whose coordinates with the corresponding values of p and q satisfy
(3.10) is the doubly infinite system of surfaces represented by (3.11). As the envelope of
these surfaces touches at each point one member of the system (3.11) the coordinates of any
point of the envelope and the associated p and q satisfy (3.10) and is thus a solution of (3.10).
This is a singular solution as we cannot deduce thus from (3.11) by giving any values of a
and b.
In any actual example, one should test whether what is apparently a singular integral
really satisfies the differential equation.
3.5.2 General integral
In (3.11), we shall assume an arbitrary relation between a and b of the form b = f (a).
Then (3.11) becomes
φ[x, y, z, a, f (a)] = 0
Differentiating this partially with respect to a,
δφ +
δφf
J(a) = 0
δa δb
The eliminant of a between these two equations is called the general integral of (1).
The above two equations represent a curve, viz., the curves of intersection of two
consecutive surfaces of the system φ[x, y, z, a, f (a)] = 0 for a particularvalues of a. The
43
envelope of the family of the surfaces touches them along this curve, which is called the
characteristic of the envelope. Thus the gerenal integral represents the envelope of a family
of surfaces, considered as composed of its characteristics.
44
Unit-XI
3.6 Derivation of partial differential equations:
Partial differential equations can be derived either by the elimination of (i) arbitrary
constants from a relation between x,y,z or (ii) of arbitrary function of these variables.
3.6.1 By elimination of constants
Let,
φ(x, y, z, a, b) = 0 (3.12)
be a relation between x, y, z involving two arbitrary constans a and b.
Differentiating (3.12) with respect to x and y partially, we get,
∂φ ∂φ +
∂x ∂z
∂φ ∂φ +
∂y ∂z
p = 0 (3.13)
q = 0 (3.14)
Eliminating a and b, we have a partial differential equation of the first order of the form,
F (x, y, z, p, q) = 0
Here the number of constants to be eliminated is equal to the number of independent
variables and an equation of the first order results. If the number of constants to be
eliminated is greater than the number of independent variables, equations of the second and
higher derivatives are deducted.
Example: Eliminate a and b from
z = (x + a)(y + b)
Differentiating with respect to x and y partially,
p = y + b
45
∂u ∂x ∂z ∂v ∂x ∂z
∂u ∂y ∂z ∂v ∂y ∂z
and
Eliminating a and b , z = pq.
q = x + a
Excercise: Eliminate a and b from z = ax + by + a
3.6.2 By elimination of an arbitrary function
Let u and v be any two functions of x,y,z an be connected by an arbitrary relation
φ(u, v) = 0 · · · (1a)
By eliminating φ, we shall form a partial differential equation and show that this is linear,
i.e., of the first degree in p and q.
Differentiating (1a) partially with respect to x and y,
∂φ .
∂u +
∂up
Σ
+ ∂φ .
∂v +
∂vp
Σ
= 0 (3.15)
∂φ .
∂u +
∂uq
Σ
+ ∂φ .
∂v +
∂vq
Σ
= 0 (3.16)
Eliminating ∂φ
and ∂φ
, we have ∂u ∂v
(ux + uzp)(vy + vzq) = (uy + uzq)(vx + vzp)
where ux =
∂u, uy =
∂u, etc. This equation can be put in the form Pp + Qq = R where
∂x ∂y
P = uyvz − uzvy, Q = uzvx − uxvz, R = uxvy − uyvx
Example: Eliminate the arbitrary function from
z = f (x
2 + y2)
Differentiating partially with respect to x and y,
p = f
J
(x2 + y
2)2x and q = f
J
(x2 + y
2)2y.
46
Eliminating fJ (x
2 + y2) between latter two equations
py = qx
Exercise: Eliminate the arbitrary function z = e
yf (x + y) Consider the equations u = a
and v = b, where a and b area arbitrary constants. By eliminating a and b , we form the
differential equations corresponding to them
3.7 Special Methods; Standard forms
3.7.1 Standard 1
Equations, in which the variables do not occur explicitly, can be written in the form
F (p, q) = 0.
A solution of this is z = ax + by + c, where a and b are connected by F (a, b) = 0.
Solving this for b, b = f (a).
Hence the complete integral is,
z = ax + yf (a) + c
The singular integral is obtained by eliminating a and c between
z = ax + yf (a) + c
0 = x + yfJ
(a)
0 = 1
The last equation is absurd and shows that there is not singular integral in the case.
To obtain the general integral, we assume an arbitrary relation c = φ(a). Then
z = ax + yf (a) + φ(a)
Differentiating partially with respect to a,
0 = x + yfJ
(a) + φJ
(a)
47
2
−
−
The elimination of a between these equations is the general integral.
Note:
The singular and general integrals must be indicated in every equation besides the
complete integral. Then only the equation besides is said to be the completely solved.
Example: p
2 + q2 = npq
Let the solution be z = ax + by + c, where
Solving,
a2 + b
2 = nab.
√
The complete integral is
b = a(n ± n2 − 4)
2
z = ax + ya
(n ± √
n2 − 4) + c
Differentiating partially with respect to c, we find that there is no singular integral as we
get an absurd result.
To find the general integral, put c = f (a).
z = ax + ay
(n + √
n2 4) + f (a). 2
Differentiating partially with respect to a,
0 = x + y (n +
√n2 4) + f
J
(a) 2
The elimination of a between the equation is the general integral.
Exercise:
Solve the following: (i) z = 3p2 − 2q
2 = 4pq (ii) p2 + q
2 = 4
48
3.7.2 Standard II
Only one of the variables x,y, z occurs explicitly. Such equations can be written in
one of the forms,
F (x, p, q) = 0; F (y, p, q) = 0; F (z, p, q) = 0
(i) Let us consider the form F (x, p, q) = 0.
dz =
∂z ∂z dx + dy
∂x ∂y
Let us assume that q = a.
= pdx + qdy
The equation becomes F (x, p, a) = 0.
Solving this for p, we get p = φ(x, a).
∴ dz = φ(x, a)dx + ady.
∴ z =
∫
φ(x, a)dx + ay + b.
This contains two arbitrary constants a and b and hence it is a complete integral.
(ii) Let us consider the form F (y, p, q) = 0.
Let us assume that p = a.
∴ F (y, a, q) = 0.
∴ q = φ(y, a).
Hence dz = adx + φ(y, a)dy.
∴ z = ax +
∫
φ(y, a)dy + b
which is a complete integral.
(iii) Let us consider the equation F (z, p, q) = 0.
Let us assume that q = ap.
49
. √
√
y
y
4
Then this equation becomes F (z, p, ap) = 0
i.e., p = φ(z, a)
Hence dz = φ(z, a)dx + aφ(z, a)dy
i.e., dz
φ(z, a)
= dx + ady
Examples:
i.e., dz
φ(z, a)
= x + ay + b which is a complete integral.
Solve: (i) q = xp + p2. (ii) p = y
2q
2 (iii) p(1 + q2) = q(z − 1)
(i) q = xp + p2 Let q = a. Then a = xp + p
2 i.e., p2 + xp − a = 0
∴ p = −x ± √
x2 + 4a
2
Hence
dz = −x ± √
x2 + 4a x
dx + ady
∴ z = −x ± √
x2 + 4a 2
dx + ay + b
x2
= − ± x
4a + x2 + asinh −1
. x
ΣΣ
+ ay + b.
(ii) p = y2q
2. Let p = a
2.Then q = ±
.a Σ
Hence dz = a2dx ±
.a
Σ
dy.
z2 = a
2x ± alogy + b
(iii) p(1 + q2) = q(z − 1).
Letq = ap. Thenp(1 + a2p
2) = ap(z − 1)
.
∴ 1 + a2p
2 = a(z − 1).
a 2
∫
4
50
a
a
a
a a
(i.e.,) p = ±
√(az − a − 1)
.
Hence dz = ±
√(az − a − 1)
dx ±
√(az − a − 1)
dy.
Exercise:
(ie.,) ±
√(az − a − 1)
= dx + ady.
(ie.,) ±
∫ √(az − a − 1)
= x + ay + b.
(ie.,) ± 2√
(az − a − 1) = x + 2y + b.
Solve: (i) z(p2 + q
2 + 1) = a2 (ii) p = 2qx (iii) p
2 + q2 = z
3.7.3 Standard III
Equations of the form f1(x, p) = f2(y, q). In this form the equation is of the first
order and the variables are separable. In this equation z does not appear. We shall assume a
tentative solution that each of these quantities is equal to a.
Solvingf1(x, p) = a, p = φ1(a, x).
Solvingf2(y, q) = a, q = φ2(a, y).
Hence dz = φ1(a, x)dx + φ2(a, y)dy.
∴ z =
∫
φ1(a, x)dx +
∫
φ2(a, y)dy + b
which is the complete integral.
Example: Solve the equation p + q = x + y.
We can write the equation in the form p − x = y − q.
Let p − x = a. Then y − q = a.
Hence p = x + a, q = y − a.
∴ dz = (x + a)dx + (y − a)dy.
51
−
∴ z =
(x + a)2
x
(y a)2
+ 2
+ b.
There is no singular integral and the general integral is found as-usual.
Exercise: (i) p2 + q
2 = x + y (ii) p2 + q
2 = x2 + y
2
3.7.4 Standard IV; Clairant’s form.
This is of the form z = px + qy + f (p, q).
The solution of the equation is z = ax + by + f (a, b) for p = a and q = b can easily be
verified to satisfy the given equation.
Example: Solve z = px + qy + √
1 + p2 + q2.
The complete integral is obviously
z = ax + by + √
1 + a2 + b2.
To find the singular integral, differentiating partially with respect to a and b.
a
x + √1 + a2 + b2
= 0.
b
and y + √1 + a2 + b2
= 0.
Eliminating a and b the singular integral is
x
2 + y2 + z
2 = 1.
To find the general integral, assume b = f (a) where f is arbitrary.
Differentiating partially with respect to a and eliminate the a between two equations.
Exercise: (i) z = px + qy + pq (ii) z = px + qy + 2√
pq.
52
Chapter 4
STANDARD FORMS OF PARTIAL
DIFFERENTIAL EQUATIONS AND
TRAJECTORIES
UNIT-XII
4.1 Standard Form of Partial Differential Equations(PDE)
In generally, a PDE having three kinds of standard forms.
Type 1 :
If the equation of the form f(p, q) = 0 (4.1)
Clearly z = ax + by + c where a,b are such that f (a, b) = 0 satisfies (4.1).
Now, solving for b from f (p, q) = 0 , we get b = ga.
The complete integral is
z = ax + yg(a) + c (4.2)
53
a+1
. Σ
−
a + 1
. Σ Jx −
y + ϕ (a)
To find the general solution, put c = ϕ (a) in (4.2).
We get z = ax + yg(a) + ϕ (a) (4.3)
Differentiating (4.3) w.r.to a we get x + ygJ(a) + ϕ J
(a) = 0 (4.4)
Eliminating a from (4.3) and (4.4) we get the general solution.
To find the singular integral we have to eliminate a, c from the following three equations.
z = ax + g(a)y + c; x + ygJ(a) = 0; 0 = 1.
The equation, 0 = 1 assures that there is no singular integral in this case.
4.1.1 Example
Solve : pq + p + q = 0
Solution : z = ax + by + c where, ab + a + b = 0 is a complete integral.
ab + a + b = 0 ⇒ b = − a
∴The complete integral is
z = ax a
y + c (4.5) a + 1
There is no singular integral. To find the general integral put c = ϕ(a) in 4.5
∴ We get, z = ax −
. a
Σ
y + ϕ (a) (4.6)
Differentiating w.r.to a we get
1 (a+1)2
Eliminating a from (4.5) & (4.6) we get the general solution.
Type 2 :
If the equation of the form f(p, q) + qy + px = z (4.7)
The complete solution is given by z = ax + by + f (a, b).
The general and singular integrals are obtained by the usual method.
54
p
a
a2
y y2
a2 a
du du
4.1.2 Example
Solve : px + qy + ( q − p) = 0
Solution :
b
z = ax + by + (a
− a) (4.8)
To find a singular integral we differentiate (4.8) w.r.to a, b we get,
x −
. b
Σ
− 1 = 0 (4.9)
y +
. 1
Σ
= 0 (4.10)
(4.9) ⇒ a = −
. 1
Σ
and (4.10) ⇒ b= (x−1)
Substituting the values of a, b in (4.8) we get, yz = 1 − x. is the singular solution.
To find general integral put b = ϕ(a) in (4.8) and we get
ϕ (a) z = ax + ϕ (a)y + − a (4.11)
a
Differentiating w.r.to a we get,
x + ϕ J(a)y −
ϕ (a) +
ϕ J(a) − 1 = 0 (4.12)
Eliminating a from (4.11) and (4.12) we get the general solution.
Type 3 :
If the equation of the form f(z, p, q) = 0 (4.13)
Put z = F (x + ay) = F (u) where a is a constant.
dz ∂u p =
du ∂x
; q = dz ∂u dz
= a du ∂y du
Substituting the values of p, q in (4.13) we get
f
.
z, dz
, adz
Σ
= 0 (4.14)
55
du ϕ (z,a)
. Σ
√.
√
Which is an ordinary differential equation of order.
A solution of (4.14) gives an expression of the form dz
= ϕ (z, a) ∴ dz = du.
Hence, it’s solution is given by g(z, a) = u + b = x + ay + b is the complete integral of
(4.13).
The general and singular integrals are found by the usual methods.
4.1.3 Example
Solve : 4(1 + z3) = 9z
4pq
Solution :
Put z = F (x + ay) = F (u) where a is a constant.
dz ∂u p =
du ∂x
; q = dz ∂u
du ∂y = a
dz ; 4(1 + z
3) = 9z
4 du
dz 2 .
du
dz 2 .√
(1 + z3)Σ
3z2√
a
∴ du
= √a
3z2 Hence, dz = 2du.
(1 + z3)
Integrating, √
a (1 + z2) = u + b.
Hence, a(1 + z3) = (u + b)
2. ∴ The complete integral is a(1 + z
3) = (x + ay + b)
2.
Differentiating the complete integral w.r.to a, b.
we get 1 + z3 = 2(x + ay + b)y and 2(x + ay + b) = 0.
Hence, 1 + z3 = 0 which is the singular solution.
Type 4 :
If the equation of the form f1(x, p) = f2(y, q) (4.15)
Put f1(x, p) = f2(y, q) = a where, a is a constant. Solving this equation (4.15) w.r.to p, q
we get, p = g1(x, a) and q = g2(y, a).
We know that dz = pdx + qdy. ⇒ 4 dz = g1(x, a)dx + g2(y, a)dy.
Hence,
z =
∫
g1(x, a)dx +
∫
g2(y, a)dy + b
56
is the complete integral.
The general and singular solutions are found as usual.
4.1.4 Example
Solve : qex = pe
y.
Solution :
The given equation can be written as pe−x
= qe−y
.
pe−x
= qe−y
= a ⇒ p = aex., q = ae
y.
∴ z = a(ex + e
y) + b is the complete integral.
We can’t find easily the singular integral but as usual we found the general solution.
4.1.5 Excercises
Find the complete integral value for the following by using four types of standard forms.
1. p + q = pq
2. z = px + qy + pq
3. p3 + q
3 = 27z
4. p2 + q
2 = x + y
4.2 CHARPIT’S METHOD
We now give a general method of solving a partial differential equation(PDE) of first
order due to Charpit.
Consider a partial differential equation f(x, y, z, p, q) = 0 (4.16)
We know that dz = pdx + qdy (4.17)
57
If we find the relation between x, y, z, p, q then
F (x, y, z, p, q) = 0 (4.18)
Find p, q from (4.16), (4.18) and substitute the values of p, q in (4.17).
We get the ordinary differential equation(ODE) and the solution of this ODE satisfies (4.16).
Next to find the relation of (4.18), differentiating (4.16) and (4.18) w.r.to x, y we get,
fx + fzp + fppx + fqqx = 0 (4.19)
Fx + Fzp + Fppx + Fqqx = 0 (4.20)
fy + fzq + fppy + fqqy = 0 (4.21)
Fy + Fzq + Fppy + Fqqy = 0 (4.22)
Eliminating px from (4.19), (4.20) we get,
(fx + fzp + fqqx)Fp − (Fx + Fzp + Fqqx)fp = 0
(ie., ) (fxFp − Fxfp) + (fzFp − Fzfp)p + (fqFp − Fqfq)qx = 0 (4.23)
Similar way to eliminate (4.21), (4.22)
(ie., ) (fyFq − Fyfq) + (fzFq − Fzfq)q + (fpFq − Fpfq)py = 0 (4.24)
Since, qx ∂2z
∂x∂y = py adding (4.23), (4.24) and rearranging we get,
(fx + pfz)Fp + (fy + fzq)Fq + (−pfp − qfq)Fz + (−fp)Fx + (−fq)Fy = 0 (4.25)
This (4.25 is a linear PDE of first order and its integral is obtained by solving the auxillary
equations
dp
fx + pfz
dq =
fy + qfz
dz =
−pfp − qfq
dx =
−fp
dy =
−fq (4.26)
We can use (4.26) to obtain the function F (x, y, z, p, q) = 0
=
58
0 0
4.2.1 Solution Procedures for Charpit’s Method
• Express the given equation as f (x, y, z, p, q) = 0 and take the partial derivatives of
fp, fq, fx, fy, fz
• Write the auxillary equation and obtain a simple relation involving atleast one of p or
q.
• Use the given equation to get p, q.
• Follow the result dz = pdx + qdy and integrate to found the complete integral, general
integral solutions.
4.2.2 Examples
Problem 1. Find the complete integral for z = px + qy + p2 + q
2
Solution:
Let f (x, y, z, p, q) = z − px − qy − p2 − q2 (4.27)
∴
fp = −x − 2p; fq = −y − 2q; fx = −p; fy = −q; fz = 1
The charpit’s auxillary equations are,
dp
fx + pfz
dq =
fy + qfz
dz =
−pfp − qfq
dx =
−fp
dy =
−fq
∴ dp
= dq
= dz dx dy
= + 0 0 p(x + 2p) x + 2p y + 2q
Taking dp
= dq
we get, dp = 0 ⇒ p = a.
Similarly, dq = 0 ⇒ q = b. Substitute these values in (4.27).
We get, z = ax + by + a2 + b
2 is the complete integral.
59
Problem 2. Find the complete integral for z = px + qy + f (p, y)
Solution :
The charpit’s auxillary equations are given by
dp dq = =· · · ·
0 0
∴ p = a and q = b.
Hence, the complete integral is z = ax + by + f (a, b).
4.2.3 Excercises
Find the complete integral value for the following by using Charpit’s Method
1. q = (z + px)2
2. p2 = q + xp
3. p = (qy + z)
2
UNIT-XIII
4.3 The Brachistochrone Problem
If A, B are two point in a space but not in the same vertical lines. Let A, B be connected
by a plane curve C and suppose A is higher level than B. Consider the smooth particle which
falls from A to B under gravity along the curve C starting from rest. The problem of finding
the shape of C which ensures that the time of fall from A to B is minimum is called the
Brachistochrone problem. We now give the Bernouli’s solution to this problem. In Figure
1 we construct a horizontal and downward vertical lines through A as the x and y axes. Let
P (x, y) be the position of the particle at time t. Let s be the arc length AP . Let α be the
60
−
−
. Σ √
. Σ
. Σ
⇒
.. Σ
Figure 4.1: Diagram representation of Brachistochrone problem
angle made by the tangent at P with y axis. Let v be the velocity of the particle at P .
Taking the x − axis as the standard level initially potential energy V = 0 and kinetic energy
T = 0.
Hence, by the principle of conservation of energy is
1 mv
2 mgy = 0. 2
Further,
∴ v = √
2gy (4.28)
by Snell’s Law, sinα
= c(a constant) (4.29) v
yJ = tan(90 α) = cotα =
cosα.
sinα
∴ sinα = cosα
= 1
. yJ yJsecα
1 1 1
using (4.29)
sinα = yJ
√1 + tan2α
= yJ
.
1 +
2 =
1 + (y J)2
1 yJ
cv2 = sin
2α =
1
1 + (y J)2
Using (4.28) we get c(2gy) = 1
1 + (y J)2
y 1 + (yJ)2 =
1 = k(a constant)
2gc
(ie., ) y
.
1 + .
dy Σ2Σ
= k.
Which is the differential equation of the Brachistochrone Problem.
Seperating the variables we get, dx = y
dy (4.30)
k − y
dx
61
−
⇒ −
.. Σ
2
−
2
Let y
= tanϕ (4.31)
k − y
∴
y = tan
2ϕ y = (k y)tan2ϕ .
k − y
∴ ysec2ϕ = k tan
2ϕ ⇒ y = ksin2ϕ .
Hence, dy = 2k sinϕ cosϕ dϕ .
Now, (4.30) ⇒ dx = tanϕ dy = 2k sin2ϕ dϕ .
∴ dx = k(1 − cos2ϕ ) dϕ .
Integrating this we get,x = k(ϕ 1
sin2ϕ ) + c 2
k x =
2 (2ϕ − sin2ϕ ) + c (4.32)
The curve passing through the origin ∴ (4.31) ⇒ ϕ = 0 when x = 0 and hence (4.32)
⇒ c = 0.
Thus we get, x = k .
2ϕ − sin2ϕ
Σ
(4.33)
Also, y = k sin2ϕ =
k (1 cos2ϕ ) (4.34)
2
If we put a = k and θ = 2ϕ in (4.33) and (4.34)
Then we have x = a(θ − sinθ) and y = a(1 − cosθ).
We know that these are the parametric equations of the Cycloid.
∴ The required curve is Cycloid.
4.4 FALLING BODIES
This lesson includes the dynamical problem of a particle falling vertically either freely
under the influence of gravity alone or with air resiance.
Newton’s second law states that the motion of the acceleration and the mass m and the total
62
dt2
2
dt
force F acting on a particle are related by the relation F = ma.
Free Fall is a particle of mass m. Which falls freely under the influence of gravity alone. In
this case the only force acting on the particle is mg. Where g is the acceleration due to
gravity. If s is the distance travelled at time t then the acceleration is d2s .
d
2s
mdt2
= mg ⇒ d
2s
= g Integrating w.r.to t we get velocity v = dt2
ds
dt = gt + c1.
(4.35)
Clearly, the constant c1 is the value of v when t = 0.
∴ v = gt + v0 is the initial velocity.
v = gt + v Integrating (4.36) w.r.to t we get velocity s =
1 gt
2 + v t + c .
0 2
0 2
(4.36)
Clearly, the constant c2 is the value of s when t = 0. say it s0
s = 1
gt2 + v t + s
(4.37)
2 0 0
(4.37) is the general solution of (4.35).
If the body falls from rest starting at s = 0 we have v0 = s0 = 0.
∴ by (4.36) and (4.37) reduce to v = gt and s = 1 gt
2.
Eliminating t between these two equations we get v = √
2gs. Which gives the velocity in
terms of the distance travelled.
4.4.1 Retarded Fall
Assume the that air exerts a resisting force proportional to the velocity of the falling body.
The total force acting on the body is given by mg − kds.
∴
63
c c
c 1
−
The differential equation of the motion is given by
d2s ds
mdt2
= mg − k dt
d2s ds k
Now, put
⇒ dt2
= (g − c) dt
, where c = m
.
ds dv d2
dt = v. ⇒
dt =
dt2 = g − cv.
∴ dv
g − cv = dt.
Hence, −
.1 Σ
log(g − cv) = t + logc .
(ie., ) log(g − cv) = −ct − clogc1.
(ie., ) log(g − cv)c1c = −ct. ⇒ (g − cv) = c2e
−ct.
Assume the initial condition v = 0 when t = 0. We get c2 = g.
g − cv = ge−ct
⇒∴ v = (1 e−ct
). c
If t → ∞, v → g then the limiting value of v is
g is called the
Terminal velocity.
UNIT-XIV
4.5 ORTHOGONAL TRAJECTORIES
Consider a one parameter family of curves given by the equation f (x, y, c) = 0 where c
is a parameter. A family of curves that intersects each curve of the given family
orthogonally is called the family of orthogonal trajectories for the given family.
Now, we described the method of finding orthogonal trajectories.
g
64
. Σ
∴ The orthogonal trajectory is − = F
(x, y).
dr
. Σ
dx dy
dr r dθ
dx
Let f (x, y, c) = 0 be a one parameter family of curves. It’s differential equation(DE) is
dy = F (x, y). These curves can be characterized by the fact that at any point (x, y) on any
one of the curves the slope is given by F (x, y).
dx dy
The solution of this differential equation gives the required orthogonal trajectories. If the
curve in Polar Co-ordinates then the angle Ψ which makes the tangent with the initial line
is given by the formula tanΨ = r dθ
.
To find the orthogonal trajectory we replace r
.dθ
Σ
by − 1
. dr
Σ
to obtain the differential
equation of the orthogonal trajectories.
dr r dθ
4.6 Procedure for finding an Orthogonal Trajectories
4.6.1 If the curves are in Cartesian Coordinates
1. Obtain the differential equation of the given family of curves. Which can be obtained
by differentiating the given equation and eliminating the parameter.
2. Replace dy
by − dx in the differential equation obtained in (1) also obtained the
differential equation of the orthogonal trajectories.
3. Obtain the general solution of the differential equation obtained in (2) which is the
required family of orthogonal trajectories.
4.6.2 If the curves are in Polar Coordinates
1. Obtain the differential equation of the given family of curves. Which can be obtained
by differentiating the given equation and eliminating the parameter.
2. Replace rdθ
by − 1
. dr
Σ
in the differential equation obtained in (1) also obtained the differential equation of the orthogonal trajectories.
65
dy −
y ⇒
y
3. Obtain the general solution of the differential equation obtained in (2) which is the
required family of orthogonal trajectories.
4.7 Examples
4.7.1 Problem
Find the orthogonal trajectories of the family of circles x2 + y
2 = a2
Solution : Differentiating w.r.to x by x2 + y
2 = a2 we get 2x + 2yy
J = 0.
(ie.,) dy
= x which is the differential equation of the given family of circles.
dx y
Hence, the differential equation of the orthogonal trajectories is given by
−
.dx
Σ
= x
dx dy = .
where m is a constant.
∴ y = mx.
∴ logx + logm = logy
Hence the orthogonal trajectories of the given family of circles is the family of straight lines
passing through the origin.
4.7.2 Problem
Show that the family of parabolas y2 = 4c(x + c) is self orthogonal in the sense that
when a curve in the family intersects another curve of the family then it is orthogonal to it.
Solution :
y2 = 4c(x + c) (4.38)
yy
J = 2c (4.39)
Eliminating c from (4.38) (4.39) we get
y = yJ(2x + yy
J) (4.40)
x
66
dr
yJ yJ
(4.40) is the differential equation of the given family of parabolas.
∴ The differential equation of the orthogonal trajectories is given by
y = − 1
Σ
2x + y
.
− 1
ΣΣ
(ie., ) y(yJ)2 = −2xy
J + y.
⇒ y = y
J(2x + yy
J) (4.41)
From (4.40) (4.41) we see that the differential equation of the given family and the
differential equation of the orthogonal trajectories are identical.
Hence, the given family of curves is itself self orthogonal.
4.7.3 Problem
Find the orthogonal trajectories of the family of curves given by r = asinθ
Solution :
dr r = asinθ ⇒
dθ = acosθ (4.42)
Eliminating a from (4.42) we get, rdθ
= tanθ.
The differential equation of the orthogonal trajectories is
− 1 .
dr Σ
= tanθ
r dθ dr
⇒ − r
= tanθdθ
⇒ −logr = log secθ + logc1
⇒ logr = log cosθ + logc
⇒ r = c cosθ is the equation of the orthogonal trajectories.
4.8 Excercises
1. xy = c
67
g
.
√
dt
2. y = cxn
3. y = ce
x
4. r = e
cθ
5. r = c cosθ
4.9 TAUTOCHRONOUS PROPERTY IN CYCLOID
If a bead is realized at the origin and slides down the wire which is in the form of the
cycloid then show that
(i) The time taken to reach the point (aπ, 2a) at the vertex is given by π a.
(ii) If at whatever point of the curve the particle starts, the time taken o reach the bottom
(vertex) remains the same. Proof : We know that the equation
x = a(θ − sinθ) & y = a(1 − cosθ)
Since, the bead is allowed to slide down the wire which in the form of the cycloid in the
Figure 4.2: Geometrical view of Tautochronous
Figure 2. We can assume this cycloid is inverted about the x − axis.
Now, the velocity of the bead at any time is given by
dx v = = dt
dx2 + dy2
dt
⇒ v =
.Σ
a2(1 − cosθ)2 + a2sinθ2
Σ
dθ
dt
v = √
2a√
(1 − cosθ)
.dθ
Σ
(4.43)
68
2
2
2
dt
g
g
dθ 0 g
Now, at θ = 0; v = 0; y = 0
∴ kinetic energy(K.E) = T = 1 mv
2 = 0
∴ potential energy(P.E) = V = −mgy = 0
At a time t, t = 1 mv
2 and V = −mgy.
Hence by the principle of conservation energy 1 mv2 − mgy = 0
∴ v2 = 2gy. ⇒ v =
√2gy =
√2ga(1 − cosθ)
∴ v = √
2ga√
1 − cosθ (4.44)
Comparing (4.43) and (4.44) we get,
√2ga
√1 − cosθ
.dθ
Σ
= √
2ga√
1 − cosθ
∴ dt =
.a
dθ (4.45)
The time taken to reach the vertex (aπ, 2a) is given by
∫ π .a
∴ t = π
.a
Now, suppose the particle starts sliding from y0 corresponding to θ = α.
By principle of conservation of energy v = √
2g(y − y0)
∴ v =
.
2gΣa(1 − cosθ) − a(1 − cosα)
Σ
(4.46)
Comparing (4.43 ) and (4.47) we get,
v = √
2ga(cosα − cosθ) (4.47)
dt =
.a .
1 − cosθ Σ
dθ
g cosα − cosθ
t =
69
. ∫ . 2sin ( )
a
g 2
g
cos2(2 ) − cos2(2 )
cos ( 2 ) − cos ( 2 )sin ϕ .
a ∫ π
.a
. ∫
∴ The time taken to reach the vertex at θ = π from θ = α is given by
∫ π .
a .
1 − cosθ Σ
α g cosα − cosθ π 2 θ
t = 2 dθ
g α (2cos2(α) − 1) − (2cos2(
θ ) − 1)
2 a π
t = .
2 sin(
θ )
dθ θ θ
θ Put cos
2
α = cos
2
θ sinϕ so thatsin
2
α dθ = 2cos
2
cosϕ dϕ
.a ∫ 0
cos(α)dϕ
t = −2 √ 2 α 2 α 2
t = 2 g
dϕ ⇒ 2 g
( 2
)
∴ t = π
.a
(4.48)
From (4.45) and (4.48) we notice that the time taken for the bead to slide from the origin
to the bottom of the cycloid(vertex) is same as the time taken to the slide from any position
of the cycloid to the vertex. This phenomenon is known as Tautochronous property of the
Cycloid.
π 2
π 2
g
t =
2
α
dθ
0
70
2 1
MODEL QUESTION PAPER
B.Sc., DEGREE EXAMINATION, NOVEMBER 2019
Mathematics
DIFFERENTIAL EQUATIONS AND ITS APPLICATIONS
(2018-2019 onwards)
Time: 3 hours Maximum: 75 Marks
PART A (10×2=20)
Answer all questions.
1. Define exact differential equation.
2. Solve y = (x − a)p − p2 by using clairaut’s form.
3. Write the system of simultaneous linear differential equations.
4. Solve y − 4xy + (4x2 − 3)y = e
x2 by using the removal of first derivative.
5. Write the rules for integrating Pdx + Qdy + Rdz = 0?
6. Write the classification of integral ?
7. Eliminate a and b from z = (x + a)(y + b).
8. Define the Brachistochrone problem.
9. Write the Auxillary Equation of Charpit’s method.
10. Define Self Orthogonal.
PART B (5×5=25)
Answer all questions choosing either (a) or (b).
11. (a) Solve (x
2 − 4xy − 2y2)dx + (y
2 − 4xy − 2x2)dy = 0..
71
dx2 dx
x x2 x
dt2 dt dt2 dt
(Or)
(b) Solve (D2 + D + 1)
2y = 0.
12. (a) Solve x
d2y − (2x − 1) dy + (x − 1)y = e
x.
dx2 dx
(Or)
(b) Solve z2dx + (z
2 − 2yz)dy + (2y2 − yz − xz)dz = 0.
13. (a) Solve (y2 + yz)dx + (xz + z
2)dy + (y
2 − xy)dz = 0.
(Or)
(b) Solve z = px + qy + √
1 + p2 + q2.
14. (a) Solve: (i) q = xp + p2. (ii) p = y
2q
2
(Or)
(b) Solve (5 + 2x)2 d2y − 6(5 + 2x)
dy + 8y = 6x.
15. (a) Discussed the impact of Retarded Fall in a Falling Bodies.
(Or)
(b) Solve : q = (z + px)2
PART C (3×10=30)
Answer any three questions.
16. Solve p2 +
.x + y − 2y
Σ p + xy +
y2
− y − y2
= 0.
17. Solve d2x + 2 dx − x + sin t = 0;
d2y − dy − y + cos t = 0.
18. Describe the necessary and sufficient condition of integrability of Pdx+Qdy +Rdz =
0.
72
dx
19. Analyze the concepts of Tautochronuos property in the Cycloid.
20. Solve the equations
dy + xsin2y = x
3cos
2y, by the method of variation of parameters.
*******
73
REFERENCE BOOKS:
1. Differential Equations and its Applications by S. Narayanan & T. K.
Manickavachagom Pillay, S. Viswanathan (Printers & Publishers) Pvt.Ltd., 2015.
2. Differential Equations and its Applications by Dr. S. Arumugam and Mr. A.
Thangapandi Issac, New Gamma Publishing House, Palayamkottai, Edition, 2014.
