Differential Equations Complete Manual

BASIC CONCEPTS OF DIFFERENTIAL EQUATIONS

Dependent Variable and the Independent Variable

Definition:

By a Differential Equation (DE), we shall mean any equation that involves the derivatives

or differential of a function or functions.

The following are examples of differential equations:

Common symbols/notations used to denote the derivative of a function are:

The numerator y of dy

dx indicates the dependent variable and the denominator x is the

independent variable.

Example:

Identify the dependent variable (DV) and the independent variable/s (IV) of the following

equations.

DV IV

22

2

2 2

2 22

2 2

32

2

dycos x

dx

d yk y 0

dx

(x y )dx 2xydy 0

u u uh

t x y

d y dy7 8y 0

dx dx

' "

x x

df dyD f ,D y, , , y , y

dx dx

2

2

2

'

da1) 4ab b a b

db

dx d x2) 5x x y x y

dy dy

dy3) y cos x 4 y x

dx

y y4) a b ab y a,b

a b

v v v5) 0 v x, y, z

x y z

Classification of Differential Equations

Ordinary Differential Equations (ODE) are equations on which the dependent variable

depends on only one independent variable.

The notations usually used to express ODE are:

Examples of ODE are the following:

Partial Differential Equations (PDE) are equations on which the dependent variable depends

on two or more independent variables.

The notation used to denote PDE are:

Examples of PDE are the following:

2' "

2

dy d y df, y , , y , ,etc.

dx dx dw

" 2 '

22

2

dx1) 2y x 0

dy

2) xy 1 x 2x 1 y 0

da3) 5b 4

db

d y dy4) 5x 3x ycos x 0

dx dx

5) ydx 2x xy 3 dy 0

2

2

f y y, .

x x x

2 2

2 2

2 2

z z1) z

x y

u u u2)

a b c

w3) 10xy yz

x

y y4) 4

t x

f f5) x y 3f

x y

Order and Degree of a Differential Equation

The ORDER of a differential equation is the order of the highest ordered derivative

involved in the equation.

Example: Give the order and degree of the following equations.

The DEGREE of a differential equation refers to the exponent of the highest ordered

derivative involved in the equation. If the exponent of the highest ordered derivative is one (1),

the degree of the equation is first degree. If the exponent is two (2), the degree is second degree.

If the exponent is three (3), the degree is third degree and so on.

If the differential equation is written as polynomial, then the highest power/exponent to

which the highest ordered derivative appears in the equation is called the degree of the equation.

Example: State the degree of the following equations.

'

" ' 2

3

3

2

2

3 4'" ' x

1) y 5xy 1 First order

2) y 2y 8y x cos x Second order

d y dy3) 3 2y 0 Third order

dx dx

v v4) 2xy 0 Second order

y x

5) y 5x y e 1 Third order

3 5'" " '

3

2 6

64 22

2

2

2

222

2

2

2

dy1) xy cos x First degree

dx

2) y xy 2y y 0 First degree

dy3) 3t sin t y 0 Third degree

dt

dy d y4) 5x cos x 0 Sixth degree

dx dx

d y dy5) 1

dx dx

d y dy1

dx dx

d y dy dy1 2

dx dx dx

2

First degree

Linearity of Differential Equations

Linear Differential Equations are equations in which the dependent variable and its

derivative appear to the first degree only and the coefficients are either constant or function only

of the independent variable.

Example: Identify whether the equation is linear or non-linear.

PRIMITIVES OR SOLUTIONS

Concepts of Primitives

Definition:

Primitive or solution is any non-derivative relation between the variables of a

differential equation that satisfies the equation.

If a solution of an equation of order n involves n arbitrary constants, it is called the

general solution. Any solution that is obtained from the general solution by assigning values to

the arbitrary constants is called the particular solution.

Example:

2 '" 2 " ' x 2

2 ' 2

" ' x

4" ' x

1) x y x 2 y sin x y e y x 1 Linear

dx x2) 3 2y Linear

dy y

3) x y 2xy x 1 Non linear

4) y 5xy e y Linear

5) y 5x y e y Non linear

"1) Show that y A cos 2x Bsin 2x is the general solution of y 4y 0, where A and B are

arbitrary constants. Also find the particular solution of it.

Solution :

Since y A cos 2x Bsin 2x contains two arbitrary constants, it is the general soluti

'

on

of the second order differential equation, if it is a solution. We can see that it is a solution by

differentiating twice the given non derivative equation.

Differentiating the equation

y A cos 2x Bsin 2x

first derivative

y 2Asin 2x

2Bcos 2x

"

"

"

"

secondderivative

y 4Acos2x 4Bsin2x

or

y 4 Acos2x Bsin2x

y 4y

y 4y 0

Hence, y Acos2x Bsin2xisa solution.

Particular solutioncan beobtainedbyassigning values to thearbitraryconstants A

andB.For instance, lettingA 2

andB 1, then

y 2cos2x sin2x

isa particular solutionof thegivendifferential equation.

2x 3x " '

1 2 1 2

2x 3x

1 2

'

1

2)Showthat y Ce C e is thegeneral solutionof y y 6y 0whereC andC are

arbitraryconstants.

Solution:

Since twoconstantsare tobeeliminated, obtain the twoderivativesof

y Ce C e (1)

first derivative

y 2Ce

2x 3x2

" 2x 3x

1 2

1 2

2x 3x

' 2x 3x

" 2x 3x

2x 3x 2x

3x

3C e (2)

secondderivative

y 4Ce 9C e (3)

eliminatingC andC usingdeterminants

y e e

y 2e 3e 0 (4)

y 4e 9e

sincee ande can not bezero, equation (4) maybe written, with thefactorse

ande r

'

"

' '

" "

emoved, as

y 1 1

y 2 3 0

y 4 9

re arranging

y 1 1 y 1

y 2 3 y 2 0

y 4 9 y 4

Condition:

If a relation between two variables involves "n" arbitrary constants, those constants are

essential if they cannot be replaced by a smaller number of constants.

For the following primitives, identify the number of essential constants, where x and y

are the variables. A, B, and C are the arbitrary constants.

In (1), there is only one essential arbitrary constant since A + B is no more than a single

arbitrary constant and can be replaced by a smaller number of constant, say L, without affecting

the given primitive, and it can be written as

y = L + x2

In (2), again only one arbitrary constant is essential since y = Ae(x+B)

can be written as

y = Aexe

B and Ae

B is no more than a single arbitrary constant, and may be replaced by a smaller

number of constant, say G, then the given primitive can be reduced to

y = Gex

In (3), no constants can be combined and replaced by a single constant, then all of them

are essential.

Obtaining Differential Equation from the General Solution

Rule:

To find the differential equation, differentiate the given relation; differentiate the first

derived equation; differentiate the second derived equation; until the number of derived

equations is equal to the number of essential constants in the given relation.

Eliminate the constants using the given relation and the derived equations.

Reminder:

Before differentiating the given relation, count first the essential constants involved in

order to know the required number of derivatives.

number of constants = number of derivatives

" ' " '" '

" '

fromwhich thedifferential equation

18y 3y 4y 2y 12y 9y 0

30y 5y 5y 0

y y 6y 0

2

x B

2

1) y A B x

2) y Ae

3) y Ax Bx C

Example:

'

1) Solve the differential equation of

y A cos 2x

Solution :

y A cos 2x is a relation with one arbitrary constant, so only the first derivative is

necessary.

y A cos 2x (1)

first derivative

y 2Asin 2x (2)

from (1)

yA

cos 2x

substitute the val

'

'

'

ue of A to (2), then simplify

yy 2 sin 2x

cos 2x

sin 2xy 2y 2y tan 2x

cos 2x

y 2y tan 2x 0 differential equation

4 2

4 2

' 3

" 2

'"


y x Ax Bx C

Solution :

y x Ax Bx C 3 constants, 3 derivatives

first derivative

y 4x 2Ax B

second derivative

y 12x 2A

third derivative

y 24x

The last equation does not contain any con

'"

stant, therefore the differential equation of

the given relation is

y 24x differential equation

x y

x + C

2x 2x

1 2

2x 2x

1 2

' 2x 2x 2x 2x

1 1 2 2

' 2x 2x

1 2


y C e cos3x C e sin 3x

Solution :

y C e cos3x C e sin 3x 2 constants, 2 derivatives

first derivative

y 3C e sin 3x 2C e cos3x 3C e cos3x 2C e sin 3x

y 2 C e cos3x C e sin 3

2x 2x

1 2

' 2x 2x

1 2

' 2x 2x

1 2

" ' 2x 2x 2x 2x

1 1 2 2

" ' 2x 2x 2x

1 2 1

x 3C e sin 3x 3C e cos3x

y 2y 3C e sin 3x 3C e cos3x

y 2y 3C e sin 3x 3C e cos3x

second derivative

y 2y 9C e cos3x 6C e sin 3x 9C e sin 3x 6C e cos3x

y 2y 9 C e cos3x C e sin 3x 2 3C e sin 3x

2x

2

" ' '

" ' '

" '

3C e cos3x

y 2y 9y 2 y 2y

y 2y 9y 2y 4y

y 4y 13y 0 differential equation

'

2 2


y x sin x C

Solution :

y x sin x C 1constant,1derivative

y x sin x C (1)

first derivative

y x cos x C sin x C (2)

from (1)

y oppositesin x C

x hypotenuse

by Pythagorean theorem

x y adjacos x C

x

cent

hypotenuse

Families of Curves

An equation involving a parameter, as well as one or both of the coordinates of a point in

a plane, may represent a family of curves, one curve corresponding to each value of the

parameter.

For instance, the equation

x2 + (y-k)

2 = r

2

may be interpreted as the equation of a family of circles having its center anywhere on the y-axis

and its radius of any magnitude.

Figure below shows several members of this family of circles.

2 2'

' 2 2

' 2 2

' 2 2

22' 2 2

2' ' 2 2 2 2

22 ' ' 2 4 2 2

then

x y yy x

x x

yy x y

x

xy x x y y

xy y x x y

squaring both sides of the equation

xy y x x y

xy 2xyy y x x y

x y 2xyy y x x y differential equation

If k and r in equation x2 + (y-k)

2 = r

2 are to be treated as arbitrary constants and

eliminated, the result will be a differential equation of the family of curves represented by that

equation. We shall eliminate both k and r and obtain a second order differential equation for the

family of circles.

Example:

2) Find the differential equation of the family of circles with centers on the line y = x.

Solution:

22 2

'

'

" ' '

2" '

'

1) Find the differential equation of the family of circles having an equation of

x y k r

Solution :

first derivative

2x 2 y k y 0

x y k y 0 (1)

second derivative

1 y k y y y 0

1 y k y y 0 (2)

from (1)

xk y (3)

y

substitute (3)

2" '

'

3' " '

'

3' " '

int o (2)

x1 y y y y 0

y

y xy y0

y

therefore the differential equation that will satisfy the equation for the family of circles is

y xy y 0 differential equation

y = x

y

x

3) Find the differential equation of the family of central conics with center at the origin and

vertices on the coordinate axes.

2 2 2

'

'

" ' '

The equation of the family of circles is

x h y k r

but h k sin ce y x; h and r being an arbitrary constants. We are dealing with two parameter family.

first derivative

2 x h 2 y h y 0

x h y h y 0 (1)

second derivative

1 0 y h y y y 0

2" '

' '

'

'

'2

" '

'

' '2

" '

'

2' " ' " ' ' '

'

1 y h y y 0 (2)

from (1)

x h yy hy 0

x yyh (3)

1 y

substitute (3) int o (2)

x yy1 y y y 0

1 y

y 1 y x yy1 y y 0

1 y

1 y yy 1 y y x yy y 1 y0

1 y

therefore the desired differential equat

2' " ' " ' ' '

2' ' "

ion is

1 y yy 1 y y x yy y 1 y 0

simplifying the result

1 y 1 y y x y 0 differential equation

2 2

2 2

2 2 2 2 2 2

2 ' 2

2 ' 2

22 " 2 ' 2

Solution :

The equation of the family is

x y1

a b

where a and b being an arbitrary constants. Re arranging the equation

x b y a a b

first derivative

2xb 2yy a 0

xb yy a 0 (1)

second derivative

b yy a y a 0 (2)

from (1

' 22

' 22

" 2 ' 2

2' 2 " 2 ' 2

22 ' " '

2" ' '

)

yy ab (3)

x


yy ayy a y a 0

x

yy a xyy a x y a 0

a yy xyy x y 0

xyy x y yy 0 differential equation

SOLUTION OF FIRST ORDER, FIRST DEGREE

ORDINARY DIFFERENTIAL EQUATION

SEPARATION OF VARIABLES

A first order, first degree differential equation is separable if it can be expressed in the

form

f1(x)dx + f2(y)dy = 0

where f1(x) is a function only of x and f2(y) is a function only of y.

The variables x and y can be replaced by any two other variables without affecting

separability.

32

32

2

32

2

32

2 2

Example 1. Separate the var iables of

yxy dx dy 0

cos x

Solution :

yM xy and N

cos x

cos xTo separate the var iables, multiply the whole equatiion by .

y

y cos xxy dx dy 0

cos x y

cos x y cos xxy 0

cos xy y

x

1 2

cos xdx ydy 0,

var iables are separated where f (x) x cos x and f (y) ydy


5 t dx x 3 dt 0

Solution :

M 5 t and N x 3

To separate the var iables, divide the equation by 5 t x 3

5 t x 3dx dt

5 t x 3 5 t x 3

1 2

0

dx dt0

x 3 5 t

1 1Variables are separated with f (x) and f (y)

x 3 5 t

2

2

2

2

2


4dy ydx x dy

Solution : Before separating the var iables, collect and combine first the coefficients of dx and dy.

4dy ydx x dy

4dy ydx x dy 0

4 x dy ydx 0 resulting equation,

M y and N 4 x

To se

2

2

2 2

2

1 22

3 2

parate the var iables, divide the resulting equation by 4 x y

4 x ydy dx 0

4 x y 4 x y

dy dx0

y 4 x

1 1Variables are separated with f (x) and f (y)

y4 x


x dy xydx x dy 2ydx

Solution : Collecti

3 2

3 2

3 2

3 2

3 2 3 2

3 2

ng the coeefficients of dy and dx, and combining it will give :

x x dy xy 2y dx 0

x x dy y x 2 dx 0 resulting equation

To separate the var iables, divide the resulting equation by y x x .

x x y x 2dy dx 0

y x x y x x

x 2dyd

y x x

1 23 2

x 0

x 2 1Variables are separated with f (x) and f (y)

yx x

The purpose of separating the variables is to make the equation integrable, since the

process of integration is to be used to obtain the solution of the equation. A solution containing

arbitrary constant(s) is called the general solution and solution containing no arbitrary

constant(s) is called the particular solution.

32

Example 1. Obta in the general solution of

yxy dx dy 0

cos x

Solution :

Separating the var iable gives,

x cos xdx ydy 0

Integrating

x cos xdx ydy 0

x cos xdx int egration by parts

u x dv cos xdx

du dx v sin x

udv uv vdu

x cos xdx x sin

2

x sin xdx

x sin x cos x

x cos xdx ydy 0

yx sin x cos x C general solution

2

2 2

2 2

Example 2.Solve thegeneral solution and the particular solution if x 0, y 1.

dy x

dx y 2

Solution:

y 2 dy xdx

y 2 dy xdx 0

y 2 dy xdx 0

y x2y C

2 2

y 4y x C general solution

2 2

2 2

for x 0, y 1

1 4 1 0 C

C 5

then y 4y x 5 particular solution

2

Example 3. Solve the general solution of

dyxy 2x

dx

Solution :

Separating the var iables and factoring the right side of equation gives,

dyx y 2

dx

dy x y 2 dx

dy x y 2 dx 0

dyxdx 0

y 2

int egrating

dy. xdx 0

y 2

xln y 2 C general s

2

olution


5 t dx x 3 dt 0

Solution :

Separating the var iables

dx dt0

x 3 5 t

dx dt0

x 3 5 t

ln x 3 ln t 5 C

applying the properties of natural log arithms

ln x 3 5 t ln C

x 3 5 t C general soluti

on

HOMOGENEOUS EQUATIONS

A differential equation of the first order and first degree,

Mdx + Ndy = 0

is said to be homogeneous if M and N are homogeneous of the same degree in x and y.

We say that f(x,y), defines a homogeneous function of degree n in x and y if and only if

f(x,y) = kn f(x,y)

for all k>0.

A homogeneous equation Mdx + Ndy = 0 can be transformed to separable equation by

changing the variable.

Suggested substitution equations are:

y = vx or x = vy

2

2 2 2 2

2 2 2

Example 1. Deter min e whether f (x, y) xy y is hom ogeneous and if so, find itsdeg ree.

Solution :

f (kx,ky) (kx)(ky) (ky) k (xy) k y

k (xy y ) k f (x, y)

The function is hom ogeneous of deg ree 2.

Example 2. Deter min e whether f (x, y)

2 2

2 2 2 2 2 2 2 2 2

2 2

x

3 2 y

x y is hom ogeneous and if so, find its deg ree.

Solution :

f (kx,ky) (kx) (ky) k x k y k (x y )

k x y kf (x, y)


Example 3. Deter min e whether f (x, y) x xy e is hom ogeneous and if so

kx x

3 2 3 3 2 2ky y

x x

3 3 3 2 3 3 2 3y y

, find its deg ree.

Solution :

f (kx,ky) (kx) (kx)(ky) e k x (kx)(k y )e

k x k xy e k (x xy e ) k f (x, y)


Example 4. Solve the general solution of a given hom ogeneous equation

xdy y x dx

Solution :

Usin g the substitution equation

y vx and dy vdx xdv

then

xdy y x dx

x vdx xdv vx x dx

x vdx xdv x v 1 dx

vdx xdv v 1 dx 0

combining the coeffici

ents of dx

v v 1 dx xdv 0

dx xdv 0 ; var iables are separable

int egrating term by term

dxdv 0

x

ln x v C

ybut v

x

yln x C

x

x ln x y Cx general solution

2 2

2 2

2 2 2

2 2 2

2

2 2


xydy x y dy 0

Solution :


y vx and dy xdv vdx

then

xydy x y dy 0

x vx xdv vdx x x v dx 0

x v xdv vdx x 1 v dx 0

v xdv vdx 1 v dx 0

vxdv v dx 1 v dx 0

2 2

2

2

2

2

2

2 2 2

combining the coefficients of dx

v 1 v dx vxdv 0

dx vxdx 0

separating the var iables

dxvdv 0

x

dxvdv 0

x

vln x C

2

2ln x v 2C C

ybut v

x

y2ln x C

x

y2ln x C

x

y 2x ln x Cx general solution

2


x xy dy ydx 0

Solution :


x vy and dx vdy ydv

then

x xy dy ydx 0

vy vy y dy y vdy ydv 0

vy vy dy y vdy ydv 0

vy y v dy y vdy ydv 0

y v v dy y vdy ydv 0

v v dy vdy y

dv 0

combining the coefficients of dy

v v v dy ydv 0

vdy ydv 0

1/ 2

1/ 2

1/ 2

1/ 2

1/ 2


dy dv dy dv0 or 0

y y vv

dyv dv 0

y

int egrating

dyv dv 0

y

ln y 2v C

xbut v

y

xln y 2 C

y

xor ln y 2 C general solution

y

2 2

2 2 2

2 2

2

2


xdy y x y dx 0

Solution :


y vx and dy vdx xdv

then

x vdx xdv vx x x v dx 0

x vdx xdv vx x 1 v dx 0

x vdx xdv vx x 1 v dx 0

x vdx xdv x v 1 v dx 0

vdx xdv

2

2

2

v 1 v dx 0

combining the coefficients of dx

v v 1 v dx xdv 0

1 v dx xdv 0

2

2

1

1


dx dv0

x 1 v

int egrating

dx dv0

x 1 v

ln x sin v 0

ybut v

x

yln x sin C general solution

x

x x

y y

vy vy

y y

v v

v


x1 2e dx 2e 1 dy 0

y

Solution :


x vy and dx vdy ydv

then

vy1 2e vdy ydv 2e 1 dy 0

y

1 2e vdy ydv 2e 1 v dy 0

1 2e vd

v v

v v v v

v v

v

v

y 1 2e ydv 2e 1 v dy 0


v 2ve 2e 2ve dy 1 2e ydv 0

v 2e dy 1 2e ydv 0


1 2edydv 0

y v 2e

EQUATIONS REDUCIBLE TO HOMOGENEOUS EQUATIONS

Consider the differential equation having the form

(ax + by + c) dx + (x + y + ) dy = 0 (1)

Figure 1 shows two lines

ax + by + c = 0

x + y + = 0 (2) meeting at point (h,k); hence

ah + bk + c = 0

h + k + = 0 (3)

y y ah + bk + c

h + k +

(h,k)

x '

x

v

v

v

v

v

x

y

x

y

x

y

int egrating

1 2edydv 0

y v 2e

ln y ln v 2e ln C

ln y v 2e ln C

y v 2e C

xbut v

y

xy 2e C

y

x 2yey C

y

x 2ye C general solution

If we refer these two lines to parallel axes with origin (h,k), by the translation

x = x + h y = y + k

the constant term must vanish. In fact applying the translation (4) to (1) we get

Equation (6) is homogeneous and can be solved by the method of frame 2. Then we must

use (4) in the result to replace x by x h, and y by y k, where h and k are found by solving (3) for h and k.

Suggested substitution

y ' = vx and x = vy

ax ' by ' ah bk c dx ' x ' y ' h k 0 (5)

and because of (3), (5) reduces to

ax ' by ' dx ' x ' y ' dy ' 0 (6)

Example 1. Reduce the equation to hom ogeneous equation and solve the general solution of it.

2x 3y 4 dx 3x 2y 1 dy 0

Solution :

In this case

2h 3k 4 0, 3h 2k 1 0

from that

3 2h 3k 4 0

2 3h 2k 1 0

9k 4k 12 2 0

k 2

if k 2

2h 3(2) 4 0

h

1

The corresponding substitutions are

x x ' h x ' 1; dx dx '

y y ' k y ' 2; dy dy '

making the substitution

2 x ' 1 3 y ' 2 4 dx ' 3 x ' 1 2 y ' 2 1 dy ' 0

2x ' 3y ' dx ' 3x ' 2y ' dy ' 0

the resulting equation is hom ogeneous of deg ree 1.

2

2

Usin g the substitution equations

y ' vx ' and dy ' vdx ' x 'dv

then

2x ' 3vx ' dx ' 3x ' 2vx ' vdx ' x 'dv 0

x ' 2 3v dx ' x ' 3 2v vdx ' x 'dv 0

2 3v dx ' 3 2v vdx ' 3 2v x 'dv 0

combining the coefficients of dx '

2 3v 3v 2v dx ' 3 2v x 'dv 0

2 2v d

2

2

2

2

2

x ' 3 2v x 'dv 0

2 1 v dx ' 3 2v x 'dv 0


3 2v dv2dx '0

x ' 1 v

3 2v dv2dx '0

x ' 1 v

3 2v dv 3 2v A B

1 v 1 v 1 v 1 v1 v

3 2v A 1 v B 1 v

3 A B; 2 A B

1 5A ; B

2 2

3 2v dv dv 5dv

2 1 v 2 1 v1 v

1 5ln 1 v

2

2

ln 1 v2

then

3 2v dv2dx '0

x ' 1 v

1 52ln x ' ln 1 v ln 1 v ln C

2 2

4ln x ' ln 1 v 5ln 1 v ln C

4 5

4 5

54

5

5

5

u sin g the properties of natural log arithms

x ' 1 vln ln C

1 v

x ' 1 vC

1 v

y 'but v

x '

y 'x ' 1

x 'C

y '1

x '

simplifying

x ' y ' C x ' y '

but x ' x 1 and y ' y 2

x 1 y 2 C x 1 y 2

x y 3 C x y 1 general solution


2x 3y 1 dx 4 x 1 dy 0

Solution :

In this equation

2h 3k 1 0; h 1 0

from that equations

h 1 and k 1

the corresponding substitutions are

x x ' h x ' 1; dx dx '

y y ' k y ' 1; dy dy '

making the substi

2

2

tution

2x ' 3y ' dx ' 4x 'dy ' 0

u sin g the substitution equation

y ' vx ' and dy ' vdx ' x 'dv

2x ' 3vx ' dx ' 4x ' vdx ' x 'dv 0

combining coefficients of dx '

2x ' 3vx ' 4vx ' dx ' 4 x ' dv 0

2x ' vx ' dx ' 4 x ' dv 0

2 v dx ' 4x 'dv 0

SIMPLE SUBSTITUTION

To solve a differential equation by simple substitution:

1. Identify the substitution equation/s. 2. Differentiate the substitution equations. 3. Eliminate all but two of the unknowns from the given differential equation and the results

of (1) and (2).

4. Solve the result from (3). 5. Return to the original variables.

4

4

4

4 3

4 3

4 3


dx ' 4dv0

x ' 2 v

dx ' 4dv0

x ' 2 v

ln x ' 4 ln 2 v ln C

ln x ' 2 v ln C

x ' 2 v C

y 'but v

x '

y 'x ' 2 C

x '

2x ' y ' C x '

but x ' x 1 and y ' y 1

2 x 1 y 1 C x 1

2x y 3 C x 1 general solution


x 2y 1 dx 3 x 2y dy 0 (1)

Solution :

To solve the general solution of

x 2y 1 dx 3 x 2y dy 0

let

a x 2y (2)

da dx 2dy (3)

from (3)

dx da 2dy (4)

substitute (2) and (4) int o (1)

a 1 da 2dy 3ady 0

a 1 da 2 a 1 dy 3ady 0


a 1 da 3a 2a 2 dy 0

a 1 da a 2 dy 0


a 1da dy 0

a 2


a 1da dy 0

a 2

a 2 3da

a 2

dy 0

a 2 3da da dy 0

a 2 a 2

a 3ln a 2 y C

but a x 2y

x 2y 3ln x 2y 2 y C

x 3y 3ln x 2y 2 C general solution

Example 2. Solve the general solution and the particular solution when x 1 and y 0

2 x y dx dy 0

Solution :

let

a x y (1)

da dx dy

dx da dy (2)

substitute (1) and (2) int o the given differential equation

2a da dy dy 0

2ada 2ady dy 0

comb

ining the coefficients of dy

2ada 2a 1 dy 0


2adady 0

2a 1


2adady 0

2a 1

2a 1 1da dy 0

2a 1

2a 1 dada dy 0

2a 1 2a 1

1a ln 2a 1 y C

2

2a ln 2a 1 2y C

but a x y

2 x y ln 2 x y 1 2y C

2x 2y ln 2x 2y 1 2y C

2x ln 2x 2y 1 C generalsolution

3

3

3

3

2 4

2 4

2 4

2 4

Example3.Solve thegeneralsolution of

xy xdy ydx 6y dy

Solution :

xy xdy ydx 6y dy (1)

let a xy (2)

da xdy ydx (3)

substitute (3)and (2) int o (1)

ada 6y dy

int egrating

ada 6y dy

a 6yC

2 4

a 3yC

2 2

a 3y C

but a xy

xy 3y C

generalsolution

2

2

2

2

2 2

2 2

Example 4.Solve thegeneral solution of

ydx dy x y 1 xdy ydx

x

Solution :

ydx dy x y 1 xdy ydx (1)

x

ylet a ; b x y (2)

x

xdy ydxda ; db dx dy (3)

x

x da xdy ydx


db b 1 a x da

db1 a x

b

22

2

22

2

2

3

3

3

2

2

da (4)

from (3)and (2)

ya ; y ax

x

b x y x ax x 1 a

bx

1 a

bx (5)

1 a


db b1 a da

b 1 a

dbb da

b

dbda ; b db da 0

b


b db da 0

b 1a C ; 2a C

2 b

returning to theorigi

2

nal var iables

1 y2 C generalsolution

xx y

2

2

2

2

2 2

Example5.Solve the general solution of

dy x y 1 dx 2dx

Solution :

dy x y 1 dx 2dx (1)

let a x y 1 (2)

da dx dy ; dy da dx (3)


da dx a 2 dx

da a 2 dx dx

da a 2 1 dx a 1 dx


d

2

2

2

adx

a 1

int egrating

dadx

a 1

da 1 a 1ln

2 a 1a 1

1 a 1ln x C

2 a 1

returning to theoriginal var iables

x y 1 11ln x C

2 x y 1 1

x y1ln x C generalsolution

2 x y 2

EXACT DIFFERENTIAL EQUATIONS

A differential equation M(x,y)dx + N(x,y)dy = 0 is exact if there exists a function g(x,y),

such that dg(x,y) = M(x,y)dx + N(x,y)dy.

If M(x,y) and N(x,y) are continuous functions and have continuous first derivative

(partial) on some rectangle of (x,y) plane, then the differential equation M(x,y)dx + N(x,y)dy = 0

is exact if and only if

M N

y x

In solving M

y

set x variable as constant, and in solving

N

x

set y variable as constant.

2 2 2 2

2 2 2 2

Example1. Deter min e whether the differential equation

6x 4xy y dx 2x 2xy 3y dy 0 is exact or not.

Solution : For this equation

M 6x 4xy y and N 2x 2xy 3y

M0 4x 2y 4x 2y

y

N4x 2y 0 4x 2y

x

M Nsin ce 4x 2y ; th

y x

2

2

2 2

2

en the equation is exact.

Example 2. Deter min e whether the differential equation

dx xdy 0 is exact or not.

y y

1 xSolution : M and N

y y

M 1 N 1and

y y x y

M N 1sin ce ; then the equation is exact.

y x y

Example 3. Deter min e whe

ther the equation

sin x cos ydx sin y cos xdy 0 is exact or not.

Solution : M sin x cos y and N sin y cos x

Msin x sin y cos y 0 sin x sin y

y

Nsin y sin x cos x 0 sin x sin y

x

M Nsin ce ; then the equation is not exact.

y x

Example 4. De

3 3

3 3

3 3

x 2 2 x

x 2 2 x

x 2 2 x

ter min e whether the equation

e 3x y x dx e dy 0 is exact or not.

Solution : M e 3x y x and N e

Me 3x 0 3x e

y

3 3

3

3 3

3

x 2 2 x

x

x 2 2 x

x

2 2

Ne 3x 3x e

x

M Nsin ce ; then the equation is exact.

y y

Example 5. Deter min e whether the equation in Example 4 is exact or not after dividing it by e .

Solution :

from Ex. 4

e 3x y x dx e dy 0

dividing it by e gives

3x y x d

3

2 2

2 2

x

x dy 0

now M 3x y x and N 1

M3x 0 3x

y

N0

x

M Nsin ce ; then the equation is not exact if it is divided by e .

y x

NOTE :

To test for the exactness of a diferential equation, it is advisable the

equation.

not to manipulate

To Solve t

'

Method 1:

1) Let F x M

2) Integrate 1 with respect to x y cons tan t

F x M

F M f y

3) Take the partial derivative of 2 with respect to y x cons tan t

F Mf y

y y

he Solution of an Exact Differential Equation

'

'

'

'

4) Equate N to 3 and solve for f y

FN f y

y

Ff y N

y

5) Integrate f y to get f y

6) Substitute 5 int o 2 , the result is the general solution.

Method 2.

1) Let Fy N

2) Integrate (1) with respect to y x cons tan t

Fy N

Fy N f (x)

3) Take partial

'

'

'

'

derivative of (2) with respect to x y cons tan t

F Nf (x)

x x

4) Equate M to (3) and solve for f (x)

NM f (x)

x

Nf (x) M

x

5) Integrate (4) to get f (x)

6) Substitute (5) int o (2), the result is the general solution.

Method 3.

By Formul

x y

a b

a

F M t, y dt N a, t dt C

M Nwhere a and b are the smallest number that will give a definite value for .

y x

NOTE :

Use the above methods (methods1, 2, and 3) only if the differential equation is exact.

Adopt the method that is convinient

to you.

2 2 2 2

2 2

2 2

3 2 2

2 '

Example 6. Solve thegeneral solution of

6x 4xy y dx 2x 2xy 3y dy 0

Solution :

M NThe equation is exact sin ce .

y x

Method 1.

1) Fx M 6x 4xy y

2) Fx 6x dx 4y xdx y dx

F 2x 2x y xy f (y)

F3) 0 2x 2xy f (y)

y

4) N

2 2 2 '

' 2

2

3

3 2 2 3

2 2

2 2

2 2 3

2 '

F; 2x 2xy 3y 2x 2xy f (y)

y

f (y) 3y

5) f '(y) 3y dy

f (y) y

6) F 2x 2x y xy y C general solution

Method 2.

1) Fy N 2x 2xy 3y

2) Fy 2x dy 2x ydy 3y dy

F 2x y xy y f (x)

F3) 4xy y f (x)

x

F4) M ; 6x

x

2 2 2 '

' 2

' 2

3

2 2 3 3

x y

a b

4xy y 4xy y f (x)

f (x) 6x

5) f (x) 6x dx

f (x) 2x

6) F 2x y xy y 2x C general solution

Method 3.

Usin g the formula


x y

2 2 2 2

0 o

x y3 2 2 3

0 0

3 2 2 3

2

M N4x 2y

y x

if a 0; b 0

M N4(0) 2(0) 0

y x

therefore :

F 6t 4ty y dt 2a 2at 3t dt C

F 2t 2t y ty 0 0 t C

F 2x 2x y xy y C general solution

Example 7.Solve the general solution of

y 1dx dy 0

x x

S

2

2 2

2

2

2

'

'

'

'

y 1olution : M and N

x x

M 1 N 1and

y x x x

the equation is exact.

Method 1.

y1) Fx M

x

12) Fx y dx y x dx

x

yF f y

x

F 13) f y

y x

F 1 14) N ; f y

y x x

f (y) 0

5) f y 0 sin ce f y 0

y y6) F C; C

x x

yF C general solution

x

' '

2 2

'

2 2

'

'

x y

a b

2

Method 2.

11) Fy N

x

12) Fy dy

x

yF f y

x

F 1 y3) y f x f x

x x x

F y y4) M ; f x

x x x

f x 0

5) f x 0 sin ce f x 0

y6) F C

x


x

Method 3.

Usin g the formula


M N 1

y x x

x y

21 1

x y2

1 1

xy

11

if x 0 and y 0

M N 1; undefined

y x 0

if x 1 and y 1

M N 11

y x 1

then a 1 and b 1

therefore :

y 1F dt dt C

t a

F yt dt dt C

yF t C

t

y y

F y 1 Cx 1

yF y y 1 C

x

yF C 1

x


x

Example 8.So l ve the general solution of

cos y 1dx sin y ln 5x 15 dy 0

x 3 y

cos y 1Solution : M and N sin y ln 5x 15

x 3 y

M sin y N 5and sin y

y x 3 x 5x

'

'

'

'

sin y0

5 x 3

M N sin ysin ce , the equation is exact.

y x x 3

Method 1.

cos y1) Fx M

x 3

dx2) Fx cos y

x 3

F cos y ln x 3 f y

F3) sin y ln x 3 f y

y

F 14) N ; sin y ln 5x 15 sin y ln x 3 f y

y y

1f y sin y ln 5

y

dy5) f y l

y

n 5 sin ydy

f y ln y ln 5cos y

6) F cos y ln x 3 ln y ln 5cos y C

F cos y ln x 3 ln 5 ln y C

F cos y ln 5 x 3 ln y C

F cos y ln 5x 15 ln y C general solution

' '

'

'

'

Method 2.

11) Fy N sin y ln 5x 15

y

dy2) Fy ln 5x 15 sin ydy

y

F cos y ln 5x 15 ln y f x

F 5 cos y3) cos y f x f x

x 5x 15 x 3

F cos y cos y4) M ; f x

x x 3 x 3

f x 0

5) f x 0 sin ce f x 0

6) F cos y ln 5x 15 ln y C general solution

Metho

x y

a b

x y

0 0

x y

0 0

x

0

d 3.

Usin g the formula


M N sin y

y x x 3

if x 0 and y 0

M N sin 00

y x 0 3

then a 0 and b 0

cos y 1F dt sin t ln 5a 15 dt C

t 3 t

cos y 1F dt sin t ln15 dt C

t 3 t

F cos y ln t 3

y

0ln15cos t ln t C

F cos y ln x 3 ln 0 3 ln15cos y ln y ln15cos 0 ln 0 C

F cos y ln x 3 ln 3 ln15cos y ln y ln15 C

F cos y ln x 3 ln 3 cos y ln 3 ln15 ln y C ln15

F cos y ln x 3 ln 3cos y ln 3cos y ln 5cos y ln y C

F cos y ln x 3 ln 5cos y ln y

C

F cos y ln 5x 15 ln y C general solution

2

2 2

2 2

2 2

2

x 2

x 2 x

x 2 x

x x

x

Example 9.Solve the general solution of

e dy 2xydx 3x dx

Solution :

Combining the coefficients of dx

2xye 3x dx e dy 0

M 2xye 3x and N e

M N2xe and 2xe

y x

M N; the equation is exact.

y x

Method 2.

1) Fy N e

2)

2

2

2

2 2

2

x

x

x '

x 2 x '

' 2

' 2

3

x 3

2

2

2

2

Fy e dy

F e y f x

F3) 2xye f x

x

F4) M ; 2xye 3x 2xye f x

x

f x 3x

5) f x 3 x dx

f x x

6) F ye x C general solution

Example10. Solve the general solution of

y ydy x dx 0

x 2x

ySolution : M x a

2x

2 2

2

ynd N

x

M 2y y

y 2x x

N y

x x

x y

a b

2x y

21 1

2 2x y

1 1

x y2 2 2

1 1

Method 3.

Usin g the formula


if x 0 and y 0

M N 0; undefined

y x 0

if x 1and y 1

M N 11

y x 1

then a 1and b 1

y tF t dt dt C

2t a

y tF t dt tdt C

2

y t tF

t 2 2

2 2 2 2

2 2

2 3

C

y x y 1 y 1F C

2x 2 2 2 2 2

y xF C

2x 2

F y x Cx general solution

If the differential equation

M x, y dx N x, y dy 0

is not exact, it can always be transformed to exact equation by multiplying it by an exp resion

INTEGRATING FACTOR

I x, y . The exp ression I x, y that makes the equation exact is called "int egrating factor".

The equation

I x, y M x, y dx I x, y N x, y dy 0

is exact.

An int egrating factor of a non exact differential equation is an exp ression such that the

equatio

n becomes exact if it is multiplied by that factor.

Example1.Show that the equation is not exact and that the given I x, y is an int egrating factor.

2ydx xdy 0 I x, y x

Solution : M 2y and N x

M N2 and 1

y x

M Nsin ce , then the equation is not exact. Multiplying the given equation by I

y x

2

2

x, y

2xydx x dy 0

this time M 2xy and N x

M N2x and 2x

y x

M N2x because of x, therefore I x, y x is an int egrating factor of the given

y x

non exact equation.

Example 2.Show that the equation is not exact and that the given I x, y is an

int egrating factor.

1ydx x ln xdy 0 I x, y

x

Solution : M y and N x ln x

M N1and 1 ln x

y x

M Nsin ce ; then the equation is not exact. Usin g the given I x, y , then

y x

y x ln xdx dy 0

x x

y x ln xthis time M and N ln x

x x

M 1 N 1and

y x x x

M

N 1 1, therefore I x, y is the int egrating factor of the given non exact

y x x x

equation.

1Example 3. Deter min e whether is an int egrating factor for

xy

ydx xdy 0

1Solution : Multiplying the given differential equation by yields

xy

y xdx dy 0

xy xy

dx dyor 0

x y

1 1M and N

x y

M N0 and 0

y x

M N 1sin ce 0, then the given exp ression I x, y is an int egrating factor for

y x xy

the given non exact differential equation.

11) If

N

Determination of Integrating Factor

f x dx

g y dy

2

M Nf x , a function of x alone, then the int egrating factor is :

y x

I e

1 M N2) If g y , a function of y alone, then the int egrating factor is :

M y x

I e

Example1.Solve the int egrating factor of

y y dx xdy 0

S

2

2

olution : M y y and N x

M N2y 1and 1

y x

M N2y 1 1 2y 2 2 y 1

y x

2 y 11 M Nnot a function of x alone

N y x x

2 y 1 2 y 11 M Nfunction of y alone

M y x y y y y 1

2

2dyg y dy 2ln y ln yy

2

2

2

2 2

therefore, the int egrating factor is :

I e e e e

1I y

y

Example 2.Solve the int egrating factor of

4xy 3y x dx x x 2y dy 0

Solution : M 4xy 3y x and N x x 2y 2x 2xy

M N4x 6y and 2x 2y

y x

M N

y x

2

dxf x dx 2ln xx

2

2

4x 6y 2x 2y 2x 4y 2 x 2y

2 x 2y1 M N 2function of x alone

N y x x x 2y x

therefore, the int egrating factor

I e e e

I x

Example 3.Solve the int egrating factor of

ydx xdy xy dx 0

Solution : Combining the co

2

2

2

g y dy

efficients of dx gives

y xy dx xdy 0

M y xy and N x

M N1 2xy and 1

y x

M N1 2xy 1 2 2xy 2 1 xy

y x

2 1 2xy 2 1 2xy1 M N 2function of y alone

N y x y xy y 1 2xy y

therefore, the int egrating factor is

I e e

2dy

2ln yy

2

2

e

1I y

y

The following examples will illustrate how to solve the general solution of a non exact

differential equation u sin g int egrating factor.

Example1.Show that the equation is not exact; find an int egrating factor and then the general

solu

f x dx dx x

tion of x y dx dy 0

Solution : M x y and N 1

M N1and 0

y x

M Nthe equation is not exact sin ce .

y x

For the int egrating factor

M N1 0 1

y x

1 M N 11

N y x 1

then,

I e e e int egrating factor

multiply the equa

x

x x

x x

x x

x

x

x

x

x '

x x ' x x x '

' x

tion by I e

e x y dx e dy 0

this time, M e x y and N e

M Ne and e

y x

M Ne , the equation is now exact.

y x

Usin g Method 2.

1) Fy N e

2) Fy e dy

F e y f x

F3) ye f x

x

F4) M ; e x y ye f x ; xe ye ye f x ;

x

f x xe

' x x

x x x

5) f x dx xe dx; f x xe x (int egration by parts)

6) F ye xe e C general solution

2

2

2

Example 2. Show that the equation is not exact; find an integrating factor and then the general

solution of y + xy dx xdy 0.

Solution : M y xy and N x

M N1 2xy and 1

y x

M NThe equation is not exact sin ce

y

2

dyg y dy 2ln y 2y

2

.x

Solving for the int egrating factor.

M N1 2xy 1 2 2xy 2 1 xy

y x

2 1 xy1 M N 2; function of y alone

M y x y 1 xy y

the int egrating factor is

1I e e e y

y

multiplying the given non exact equation

2

2

2 2

2

2 2

2 2

2

'

2

1by I gives

y

1 xy xy dx dy 0

y y

1 xthis time M y xy and N

y y

M 1 N 1and

y y x y

M Nthe equation is now exact sin ce .

y x

Usin g Method 1.

11) Fx M x

y

12) Fx dx xdx

y

x xF f y

y 2

F x3) f y

y y

4)

'

2 2

'

F x xN ; f y

y y y

f y 0

'

2

2 2

5) f y 0 simce f y 0

x x6) F C general solution

y 2

Example 3.Show that the equation is not exact; find an int egrating factor and then the general

solution of y x y 1 dx x x 3y 2 dy 0.

Solution : M xy y y and N x 3xy 2x

Mx 2y 1and

y

1

dyg y dy y ln y

N2x 3y 2

x

M Nthe equation is not exact sin ce .

y x

Solving for the int egrating factor,

M Nx 2y 1 2x 3y 2 x y 1

y x

x y 11 M N 1function of y alone

M y x y x y 1 y

then,

I e e e y

mult

2

2 3 2 2 2

2 2

2 3 2

2 3

iplying the given non exact equation by I y gives,

y x y 1 dx xy x 3y 2 dy 0

this time M xy y y and N x y 3xy 2xy

M N2xy 3y 2y and 2xy 3y 2y

y x

the equation is now exact.

Usin g Method 1.

1) Fx M xy y y

2) Fx y xdx y d

2

2 23 2

2 2 '

2 2 2 2 '

'

'

2 2 3 2

x y dx

y xF xy xy f y

2

F3) x y 3xy 2xy f y

y

F4) N ; x y 3xy 2xy x y 3xy 2xy f y

y

f y 0

5) f y 0 sin ce f y 0

6) F x y 2xy 2xy C general solution

Definition :

Linear Differential Equation is one in which the dependent var iable and its derivatives

appear to the first deg ree only and the coefficients are either a cons tan t or function only of the

indepen

LINEAR DIFFERENTIAL EQUATION

dent var iable.

The differential equation

dyyP x Q x

dx

is a linear first order differential equation sin ce only y and its derivative appear int o the equation

and they are only to the first power. As indicated in the equation above, P and Q ar

e functions of

x alone.

To solve the general solution of a linear differential equation, the first thing to do is to

reduce the given equation in the form

dyyP x Q x

dx

which is the general form of a first order differential equation linear in y

P x dx P x dx

G y dy G y dy

,

or in the form

dxxG y H y

dy

which is the general form of a first order differential equation linear in x.

dy1) yP x Q x ye Q x e dx C

dx

dx2) xG y H y xe H y e dy C

dy

General Solution of a Linear Differential Equation

3

3

3

P x dx P x dx

dx dx

3x x

Example 1. Solve the general solution of a given linear differential equation.

dy yx 3

dx x

Solution :

dy yx 3 linear in y

dx x

1P and Q x 3

x

therefore the general solution is

ye Qe dx C

ye x 3 e dx

ln x 3 ln x

3

4

5 2

5 2

C

ye x 3 e dx C

yx x 3 xdx C

xy x 3x dx C

x 3xxy C

5 2

10xy 2x 15x C general solution

Example 2. Solve the general solution of a given linear differential equation

dx 3x2y

dy y

Solution :

dx 3x2y linear in x

dy y

G

G y dy G y dy

3dy 3dy

y y

3ln y 3ln y

3 3

3 4

3and H 2y

y

therefore the general solution is,

xe He dy C

the general solution is,

xe 2ye dy C

xe 2ye dy C

xy 2y y dy C

xy 2y dy C

53

3 5

P x dx P x dx

2yxy C

5

5xy 2y C general solution

Example 3.Solve the general solution of a given linear differential equation

dy 2yx 1

dx x

Solution :

dy 2yx 1 linear in y

dx x

2P and Q x 1

x


ye Qe dx C

ye

2dx 2dx

x x

2ln x 2ln x

2 2

2

2

2

2

3

x 1 e dx C

ye x 1 e dx C

yx x 1 x dx C

y 1x dx C

x x

y 1ln x C

x x

y x ln x x C general solution

Example 4. Solve the general solution of a given linear differential solution.

dyx y x 3x

dx

2

2

2

2

P x dx P x dx

2x

Solution :

Reducing the given equation int o linear form gives,

dy yx 3x 2

dx x

dy yor x 3x 2 linear in y

dx x

1P x and Q x x 3x 2

x


ye Qe dx C

PHYSICAL APPLICATION OF FIRST ORDER

FIRST DEGREE DIFFERENTIAL EQUATIONS

It has been found experimentally that radioactive substance decompose at a rate propor -

tional to the quantity of substance present.

If we let Q(t) represent the quantity of substance at time t, then the statement above

Radioactive Decay

may

be exp ressed mathematically by the differential equation

dQkQ

dt

where k is the cons tan t of proportionality. Re arranging the equation gives

dQkdt

Q

int egrating both sides of the equation

dQkdt

Q

ln Q kt C working equation

Ex

ample1. Radium decomposes at a rate proportional to the amount present. If of 100 grams set

aside now there will be left 96 grams ten years hence. Find how much will be left after 20 years.

What is the half life of the radium?

Solution :

Q 100 wh

en t 0

Q 96 when t 10

Q ? when t 20

Q 50 when t ?

Usin g the working equation

ln Q kt C

when Q 100, t 0

ln100 k(0) C; C ln100

when Q 96, t 10

ln 96 k(10) ln100

ln 96 ln100 10k

196ln 10k

100

1 96k ln 0.00408

10 100

a) Q ? when t 20

ln Q ( 0.00408)(20) ln100

ln Q 4.524

Q ln (4.524) 92.16 grams

b) Q 50 when t ?

ln 50 ( 0.00408)(t) ln100

ln 50 ln100 0.00408 t

50ln 0.00408 t

100

50ln

100t 169.890.00408

0

0

0

0

0

0

years

Example 2. If 5% of the radioactive subs tan ce decompose in 5 years, what percentage will be

present at the end of 500 years?1000 years?

Solution :

Q Q when t 0

Q 0.95Q when t 50

Q xQ when t 500

Q xQ when t 1000

when Q Q , t 0

ln Q k

0

0

0 0

0 0

3

0

3

0 0

0 0

0

0

1

(0) C; C ln Q

when Q 0.95Q , t 50

ln 0.95Q k(50) ln Q

ln 0.95Q ln Q 50k

ln 0.95 50k

ln 0.95k 1.03 x 10

50

a) Q xQ when t 500

ln xQ ( 1.03 x 10 )(500) ln Q

ln xQ ln Q 0.515

xQln 0.515

Q

ln x 0.515

x ln ( 0.515) 0.59

75 or 59.75%

03

0 0

0 0

0

0

1

b) Q xQ when t 1000

ln xQ ( 1.03 x 10 )(1000) ln Q

ln xQ ln Q 1.03

xQln 1.03

Q

ln x 1.03

x ln ( 1.03) 0.3570 or 35.70%

Example 3. If the half life of a radioactive subs tan ce is1800 years, what percentage is present

at the en

0

0

0

0

0

0 0

0

0 0

0

d of 100 years? In how many years does only10% of the subs tan ce remain ?

Solution :

Q 0.5Q when t 1800

Q Q when t 0

Q xQ when t 100

Q 0.10Q when t ?

when Q Q , t 0

ln Q k(0) C; C ln Q

when Q 0.5Q , t 1800

ln 0.5Q k(1800) ln Q

ln 05Q l

0

0

0

4

0

4

0 0

0 0

0

0

1

0

4

0

n Q 1800k

0.5Qln 1800k

Q

ln 0.5k 3.85 x 10

1800

a) Q xQ , t 100

ln xQ ( 3.85 x 10 )(100) ln Q

ln xQ ln Q 0.0385

xQln 0.0385

Q

ln x 0.0385

x ln ( 0.0385) 0.9622 or 96.22%

b) Q 0.10Q , t ?

ln 0.10Q ( 3.85 x 10 )(t)

04

0 0

40

0

4

ln Q

ln 0.10Q ln Q ( 3.85 x 10 ) t

0.10Qln 3.85 x 10 t

Q

ln 0.10t 5980.74 years

3.85 x 10

00

0

0

0 0

Example 4. A certain radioactive subs tan ce has a half life of 38 hrs. Find how long it will take

for 90% of the radioactivity to be dissipated.

Solution :

Q Q when t 0

Q 0.5Q when t 38

Q 0.10Q when t ?

when Q Q , t 0

ln Q k(0) C; C ln Q

wh

0

0 0

0 0

0

0

0

0 0

0 0

0

0

en Q 0.5Q , t 38

ln 0.5Q k(38) ln Q

ln 0.5Q ln Q 38k

0.5Qln 38k

Q

ln 0.5k 0.01824

38

therefore, when Q 0.10Q ; t ?

ln 0.10Q ( 0.01824)(t) ln Q

ln 0.10Q ln Q 0.01824 t

0.10Qln 0.01824 t

Q

ln 0.10t 126.24 hr

0.01824

s.

Example1. A bacterial population P is known to have a rate of growth proportional to P itself .

If between noon and 2 pm, the population tripples, at what time, no control being exerted,

should P becomes100 times it was at

Population Growth

o

noon ?

Solution :

P initial population

P population at any time t

dPrate of increase

dt

sin ce rate of increase is proportional to P itself then,

dPkP

dt

dPkdt

P

00 0

0

0 0

0 0

0

0

0

0

int egrating both sides of the equation gives,

dPk dt

P

ln P kt C working equation

when P P , t 0

ln P k(0) C; C ln P

when P 3P , t 2

ln 3P k(2) ln P

ln 3P ln P 2k

3Pln 2k

P

ln 3k 0.5493

2

when P 100P , t ?

ln100P (0.5493)(t)

0

0 0

0

0

ln P

ln100P ln P 0.5493 t

100Pln 0.5493 t

P

ln100t 8.38 pm

0.5493

Example 2. If the population of the city doubled in the past 25 years and the present population

is100000, when will the city have a population of 500000?

Solution :

P

50000 when t 0

P 100000 when t 25

P 500000 when t ?

Solution :

when P 50000, t 0

ln 50000 k(0) C; C ln 50000

when P 100000, t 25

ln100000 k(25) ln 50000

ln100000 ln 50000 25k

100000ln 25k

50000

ln 2k 0.02773

25

therefore; when P 500000, t ?

ln 500000 (0.02773)(t) ln 50000

ln 500000 ln 50000 0.02773 t

500000ln 0.02773 t

50000

ln10t 83 years

0.02773

then

t 83 25 58 years from now

Example 3. Express the following proposition as a different

0

ial equation : The population of the

city increases at a rate which is proportional to the current population and the difference bet.

200000 and the current population.

Solution :

P initial population

P population at any time t

dPrat

dt

e of increase

the rate of increase is jo int ly proportional to both P and (200000 P), then

dPkP(200000 P) differential equation

dt

where k is the cons tan t of proportionality

Example 4. The initial population of the city is 100000 and aft

er 20 years, the population is

50000. What will be the population after 35 years, following the rate of increase given in ex. 3.

Solution :

from ex. 3,

dPkP(200000 P)

dt

dPkdt

P(200000 P)

int egrating both sides of the equation gives,

dP

P(

k dt200000 P)

1 A B

P(200000 P) P 200000 P

1 A(200000 P) BP

solving for A and B,

1A B

200000

therefore;

dP dPk dt

200000P 200000(200000 P)

1 1ln P ln(200000 P) kt C

200000 200000

1 Pln kt C working equation

200000 200000 P

when P 10000, t 0

1 10000ln k(0) C

200000 200000 10000

1

20000

5

5

5

6 5

7

7

ln 0.0526 C0

C 1.4722 x 10

when P 50000, t 20

1 50000ln k(20) 1.4722 x 10

200000 200000 50000

1ln 0.3333 20k 1.4722 x 10

200000

5.493 x 10 1.4722 x 10 20k

k 4.6144 x 10

when t 35, P ?

1 Pln 4.6144 x 10 (3

200000 200000 P

5

7 5

5) 1.4722 x 10

Pln 200000[4.6144 x 10 (35) 1.4722 x 10 ]

200000 P

Pln 0.28568

200000 P

P 114190 after 35 years

Newtons Law of Cooling

Experiments has shown that under certain conditions, a good approximation to the tempe -

rature of an object can be obtained by using Newton's Law of Cooling.

Newton 's Law of Cooling stated that "the temperature of the body changes at a rate that

is proportional to the difference in temperature between the outside medium and the body itself ".

We shall assume that the cons tan t of proportionality is the same whether the temperature

is increa sin g or

bb m

b

m

b

b m

b

decrea sin g.

Expressin g the above statement int o mathematical equation give,

dTk T T

dt

where :

T temperature of the body

T temperature of the outside medium

re arranging the equation and int egrating gives,

dTkdt

T T

ln T

m

o o

o

T kt C working equation

Example1. A thermometer reading 18 C is brought int o a room where the temperature is 70 C;

1 min ute later the thermometer reading is 31 C. Find the temperature reading 5 min s. after the

thermometer is first brou

b m

b

b

b m

b m

b m

ght int o the room.

Solution :

T 18 when t 0; T 70

T 31 when t 1

T ? when t 5

when T 18, t 0, T 70

ln T T kt C

ln 18 70 k(0) C; C ln( 52)

when T 31, t 1, T 70

ln 31 70 k(1) ln( 52)

ln( 39) k ln( 52)

ln( 39) ln( 52) k

39k ln

0.2877

52

b m

b

5

b

5

b

b

b

when T ?, t 5, T 70

ln T 70 ( 0.2877)(5) ln( 52)

39ln T 70 ln 52

52

39T 70 52

52

T 12.34 70

T 57.66 C

Example 2. A pie ois removed from a 350 C oven and placed in the kitchen with 70 C

sorrounding tempera

b m

b

b

b m

b m

ture. In half an hour, the pie has a temperature of 150 C. How soon

will it be at 100 C and thus ready to eat ?

Solution :

T 350 when t 0; T 70

T 150 when t 30

T 100 when t ?

when T 350, t 0, T 70

ln T T kt C

ln 350 70 k(0) C; C ln 280

whe

b m

b m

n T 150, t 30, T 70

ln 150 70 k(30) ln 280

ln80 30k ln 280

ln80 ln 280 30k

80ln 30k

280

80ln

280k 0.0417630

when T 100, t ?, T 70

ln 100 70 0.04176 t ln 280

ln 30 ln 280 0.04176t

30ln 0.04176t

280

2.2336t 53.4

0.04176

9 min

Example 3. At 9 : 00 AM, a thermometer reading 70 C is taken outdoor where the temperature

is15 C. At 9 : 05 AM, the thermometer reading is 45 C. At 9 :10 AM, the thermometer is taken

indoors where the temperature is fixed at 70 C. Find the readi

1

2

1

m m

b b

b b

b

b

b

ng at 9 : 20 AM.

Solution :

outside T 15 inside T 70

T 70, t 0 T ?, t 0

T 45, t 5 T ?, t 10

T ?, t 10 ref . time : 9 :10 AM

OUTSIDE

when T 70, t 0

ln 70 15 k(0) C; C ln 55

when T 45, t 5

ln 45 15 5k ln 55

ln 30 ln 55 5k

30ln 5k

55

3ln

k

1

1

1

1

1

1

2

2

b

b

2

b

2

b

b

b

b

b

0

55

5

when T ?, t 10

30ln

55ln T 15 10 ln 555

30ln T 15 ln 55

55

30T 15 55

55

T 31.36 C reading at 9 :10 AM

INSIDE (ref . time is 9 :10 AM)

when T 31.36, t 0

ln 31.36 70 k(0) C

ln 38.64 C

when T ?, t 10

ln T

30

ln5570 10 ln 36.845

2

2

2

2

b

2

b

b

30ln T 70 ln 36.84

55

30T 70 36.84

55

T 58.5 C reading at 9 : 20 AM

Example 4. If the temperature of the air is 300 K and the subs tan ce cools from 370 K to 340 K

in 15 min utes, find whe the temperature will be 310 K.

S

b m

b

b

b

b

b

olution :

T 370 when t 0; T 300

T 340 when t 15

T 310 when t ?

when T 370, t 0

ln 370 300 k(0) C; C ln 70

when T 340, t 15

ln 340 300 15k ln 70

ln 40 ln 70 15k

40ln

70k 0.037315

when T 310, t ?

ln 310 300 0.0373t ln 70

ln10 ln

70 0.0373t

10ln 0.0373t

70

10ln

70t0.0373

t 52.15 min utes

0Suppose that at time t = 0, a quantity Q of a subs tan ce is present in a container. Assume

that at time t 0, a fluid containing a concentration C of a subs tan ce is allowed to enter the

container at a cons tan t rate and that the m

Mixture Problems

ixture is kept at a uniform concentration throughout

by a mixing device. Also assume that at t 0, the mixture in the container with concentration C

is allowed to escape at a cons tan t rate .

The problem is to det er min e the amt. Q of the subs ta

n ce in the container at any time t.

dQThe rate of change of the amount of the subs tan ce in the container equals the rate at

dt

which a fluid enters the container times the concentration of the subs tan ce in the entering fluid

min us the rate at which a fluid leaves the container times the concentration of the subs tan ce in

the container.

entering leaving

rate rate

C concentration C concentration

dQentering leaving

dt

dQC C working equation

dt

Example1. Pure wat

er is poured at the rate of 3 gal / min int o a tan k containing 300 kg of salt

dissolved in 100 gallons of water and the solution, kept well stirred, pours out at 2 gal / min . Find

the amount of salt at the end of 100 min utes.

Given :

Required : Q when t 1

00 min

Solution :

rate of filling 3 2 1gpm

number of gal lons added at any time t

1gpm t t gallons

volume at any time t

100 t

then, the concentration of the subs tan ce in the container at any time t is,

QC

100 t

Usin g the derived equation gives,

dQ QC C 3(0) 2

dt 100 t

dQ Q2

dt 100 t

dQ 2Q0

dt 100 t

separating the var iables,

dQ 2dt0

Q 100 t

int egrating

2

2

2 2

2 2

2

2

term by term,

dQ dt2 0

Q 100 t

ln Q 2ln 100 t C

ln Q 100 t C

Q 100 t C

when t 0,Q 300

300 100 0 C; C 300 100

therefore,

Q 100 100 300 100

300 100Q 75 kg of salt

200

Example 2. A tan k initially holds 100 gallons of brine solution c

ontaining 1 kg of salt. At t 0,

another brine solution containing 1 kg of salt per gallon is poured int o the tan k at the rate of

3 gpm, while the well stirred mixture leaves the tan k at the same rate.

Find the time at which the mixture contains 2 kg

of salt.

Given :

3 3dt dt

100 100

3 3t t

100 100

3 3t t

100 100

3t

100

Required : t when Q 2 kg

Solution :

QC

100

dQ Q(3)(1) 3

dt 100

dQ Q3 3

dt 100

dQ Q3 3 linear differential equation

dt 100

Qe 3e dt C

Qe 3e dt C

Qe 100e C

Q 100 Ce

when t 0, Q 1

1 100 C; C

3t

100

3t

100

99

therefore,

2 100 99e

98e

99

3 98t ln

100 99

t 0.338 min

Example 3. A tan k contains 80 gallons of pure water. A brine solution with 2 kg / gal of salt

enters at 2 gpm, and the well stirred mixture leaves at the same rate.

Find t

he time at which the brine leaving will contain 1 kg / gal of salt.

Given :

dt dt

40 40

t t

40 40

t t

40 40

t

40

t

40

Required : t so that C 1 kg / gal

Solution :

dQ Q Q(2)(2) 2 4

dt 80 40

dQ Q4 linear

dt 40

Qe 4e dt C

Qe 4e C

Qe 160e C

Q 160 Ce

when t 0, Q 0

0 160 C; C 160

therefore,

Q 160 160e

Qwhen C 1 ; Q 8

80

t

40

t

40

0

80 160 160e

e 0.5

tln(0.5)

40

t 27.73 min

In this topic, we use the notation t, s, v, a, m, and F for time, dis tan ce, velocity, accele

ration, mass, and force respectively. From calculus, we have

ds dv dvv and a v

dt dt ds

If a particle of mass m moves in a strai

Motion in a Straight Line

2 2 2

ght line under the inf luence of one or more forces

having resul tan t F, then, in accordance with Newton 's law of motion, we have

dF mv

dt

assu min g that m is cons tan t, then

dv a W dvF m ma W

dt g g dt

where :

g 9.8 m / s 98 cm / s 32.2 ft / s

Example1. A boat with its load weighs 322 lbs. If the force exerted upon the boat by the motor

in the direction of the motion is equivalent to a cons tan t force of 15 lbs, if the resis tan ce (in lb)

to motion is equal numerically to twice the speed (

in ft / s), that is, 2v lb and if the boat starts from

rest, find the speed after 10 seconds.

Solution :

W dvF

g dt

322 dv15 2v

32.2 dt

dv15 2v 10

dt

15 2v dt 10dv

10dv 5dvdt

15 2v 7.5 v

int egrating

dvdt 5

7.5 v

t 5ln 7.5 v

C

0.0745

5

0.0745

5

when t 0, v 0

0 5ln 7.5 0 C

C 5ln 7.5

therefore,

10 5ln 7.5 v 5ln 7.5

10 5ln 7.5 5ln 7.5 v

0.0745 5ln 7.5 v

0.0745ln 7.5 v

5

7.5 v e

v 7.5 e

v 6.48 ft / s

Example 2. An iceboat with load weighs 322 lbs. It is prope

o

o

lled by a force of 2 v v lb when

moving at the rate of v ft / s in a v ft / s tail wind. There is a cons tan t resis tan ce to motion of

10 lbs. (a) Find the speed v at time t sec from rest in a 40 ft / s wind. (b) Find its speed after 10 s

from rest.

Solution :

(a) F

o

o

o

o

o

o

W dv

g dt

322 dv2 v v 10

32.2 dt

dvv v 5 5

dt

5dvdt

v v 5

int egrating

5dvdt

v v 5

t 5ln v v 5 C

when t 0, v 0 and v 40

0 5ln 40 0 5 C

C 5ln 35

therefore,

t 5ln 40 v 5 5ln 35

t 5ln 35 v 5ln 35

tln35

5

tln35

5

10ln35

5

ln35 2

5ln 35 t 5ln 35 v

tln 35 v ln 35

5

35 v e

v 35 e

(b) when t 10

v 35 e

v 35 e

v 30.26 ft / s

Example 3. A boat is being towed at the rate of 20 kph. At the ins tan t (t 0) that the towing

line is cast off , a man in the b

oat begins to row in the direction of motion exerting a force of

90 N. If the combined mass of the man and the boat is 225 kg and the resis tan ce is equal to

26.25v, find the speed of the boat after 1/ 2 min ute.[Ans. 3.5 m / s]

Example1. The rate of change of air pressure with altitude (distance above the earth) is propor -

tional to the air pressure. If the air pressure on the ground is101 KPa and if at an altitude of

3050 m it is 70 KPa, find the air

Other Rate Problems

pressure at an altitude of 4575 m.

Solution :

h 0 P 101 KPa

h 3050 P 70 KPa

h 4575 P ?

dPkP

dh

dPkdh

P

int egrating

dPkdh

P

ln P kh C working equation

when h 0, P 101

ln101 k(0) C; C ln101

when h 3050, P 70

ln 70 3050k ln101

ln 70 ln1

4

4

4.065

01 3050k

70ln

101k 1.2 x 103050

when h 4575

ln P 1.2 x 10 4575 ln101

ln P 4.065

P e

P 58.28 Kpa

Example 2. Water leaks from a cylinder through a small orifice in its base at a rate proportional

to the square root of the volume rem

aining at any time. If the cylinder contains 64 gallons ini

tially and 15 gallons leaks out the first day, when will 25 gallons remain ? How much will remain

at the end of four days?

Solution :

v 64 when t 0

v 49 when t 1

12

1

2

t ? when v 25

t 4 when v ?

water leaks at the rate proportional to the square root of the volume remaining at any

time, then

dvk v

dt

separating the var iables and int egrating,

dvkdt

v

v dv kdt

2v kt C

2 v kt C working equation

when

t 0, v 64

2 64 k(0) C; C 16

when t 1, v 64 15 49

2 49 k(1) 16

k 2(7) 16 2

when t ?, v 25

2 25 2t 16

2t 16 2(5)

t 3 days

when t 4, v ?

2 v 2(4) 16

2 v 8

v 16 gallons

Differential Equations Complete Manual

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