Linear Differential Equations

SMS 2308 - Mathematical Methods

Lecture Notes

Samsun Baharin Haji Mohamad

March 30, 2012

1

Contents

1 Introduction to Differential Equations 31.1 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.2 Solution of Differential Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.3 Linearly Independent Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

2 First Order and First Degree ODEs 52.1 Separation of Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52.2 Homogeneous Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

2.2.1 Equations Reducible to Homogeneous Form . . . . . . . . . . . . . . . . . . 92.3 Linear First-order Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

2.3.1 Bernoullis Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142.4 Exact Differential Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

2.4.1 Reduction to Exact Form, Integrating Factor . . . . . . . . . . . . . . . . . 16

3 Second Order Linear Differential Equation 183.1 Homogeneous Linear ODEs with Constant Coefficients . . . . . . . . . . . . . . . . 183.2 Non-homogeneous ODEs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

3.2.1 Method of Undetermined Coefficients . . . . . . . . . . . . . . . . . . . . . 213.2.2 Method of Variation of Parameters . . . . . . . . . . . . . . . . . . . . . . . 23

4 Other Types of ODE 244.1 ODE with y is missing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 244.2 ODE with x is missing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

4.3 Equation of the type d2ydx2 = f(y) . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

4.4 Euler-Cauchy Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 274.4.1 Method 1: Using substitution x = ez . . . . . . . . . . . . . . . . . . . . . . 274.4.2 Method 2: Using y = xm . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

4.5 One Solution is Known . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

5 Higher Order Linear ODEs 335.1 Homogeneous Linear ODEs with Constant Coefficients . . . . . . . . . . . . . . . . 335.2 Non-Homogeneous Linear ODEs . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

5.2.1 Method of Variation of Parameters . . . . . . . . . . . . . . . . . . . . . . . 355.2.2 Annihilator Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36

2

1 Introduction to Differential Equations

1.1 Definitions

A differential equation is an equation which involve differential coefficients or differentials. Forexample, the equations below are all differential equations.

1. exdx+ eydy = 0

2. d2xdt2 + n

2x = 0

3. y = x dydx + xdxdy

4.

[1 +

(dydx

)2]3/2/ d

2ydx2 = c

5. dxdy wy = a cos(pt), dydt + wx = a sin(pt)

6. xux + yuy = 2u

7. 2yt2 = c

2 2yx2

An ordinary differential equations (ODE) is differential equations where all the differentialcoefficients have reference to a single independent variable. Item 1 to 5 in the list are all ODEs.

A partial differential equation (PDE) is differential equations which there are two or moreindependent variables and partial differential coefficients with respect to any of them. Item 6 and7 are PDEs.

The ORDER of a differential equation is the order of the highest derivative appearing in it.The DEGREE of a differential equation is the degree of the highest derivative occuring in it,

after the equation has been expressed in a form free from any radicals or fractions as far as thederivatives are concerned.

From the list above, we can see that

1. Item 1 is of the first order and first degree.

2. Item 2 is of the second order and first degree.

3. Item 3 written as y dydx = x(dydx

)2+ x is clearly of the first order but of second degree.

4. Item 4 written as

[1 +

(dydx

)2]3= c2

(d2ydx2

)2is of the second order and second degree.

1.2 Solution of Differential Equations

A solution of a differential equation is a relation between the variables which satisfies the givendifferential equation.

y = A cos(nx+ )

is a solution of

d2y

dx2+ ny2 = 0

The general (or complete) solution of a differential equation is that in which the number ofarbitrary constants are equal to the order of the differential equation. Thus y = A cos(nx + ) is

3

a solution for d2ydx2 +ny

2 = 0 as the number of arbitrary constants (A,) are the same as the order

of d2ydx2 + ny

2 = 0.A particular solution is a solution that can be obtained from general solution by giving

particular values to the arbitrary constants. For example

y = A cos(nx+ pi/4)

is the particular solution of the equation d2ydx2 + ny

2 = 0 as it can be derived from the generalsolution y = A cos(nx+ ) by putting = pi/4.

A differential equation may sometimes have an additional solution which cannot be obtainedfrom the general solution by assigning a particular value to the arbitrary constant. Such a solutionis called a singular solution.

1.3 Linearly Independent Solution

Two solutions y1(x) and y2(x) of the differential equation

d2y

dx2+ P (x)

dy

dx+Q(x)y = 0

are said to be linearly independent if c1y1(x) + c2y2(x) = 0 such that c1 = 0 and c2 = 0.If c1 and c2 are not both zero, then the two solutions y1(x) and y2(x) are said to be linearly

dependent.

If y1(x) and y2(x) any two solutions ofd2ydx2 +P (x)

dydx +Q(x)y = 0, then their linear combination

c1y1(x) + c2y2(x) where c1 and c2 are constants, is also a solution ofd2ydx2 + P (x)

dydx +Q(x)y = 0.

4

2 First Order and First Degree ODEs

Some special methods to find solutions for these ODEs will be discussed here,

1. Separation of Variables

2. Homogeneous equations

3. Linear equations

4. Exact equations.

2.1 Separation of Variables

If in an equation, it is possible to collect all function of x and dx on one side and all the functionsof y and dy on the other side, then the variables are said to be separable. Thus the general formof such equation is

f(x)dx = g(y)dy

Integrating both sides, we get f(x)dx =

g(y)dy + c

as its solution.

Example 1

Solvedy

dx=x(2 log x+ 1)

sin y + y cos y

We separate them into(sin y + y cos y)dy = (x(2 log x+ 1))dx

Integrating both side, we getsin ydy +

y cos ydy = 2

log x xdx+ 2

xdx

cos y +[y sin y

sin y 1dy + c

]= 2

[(log x x

2

2

1

x x

2

2dx

)+x2

2

] cos y + y sin y + cos y + c = 2x2 log x x

2

2+x2

2

Hence the solution is2x2 log x y sin y = c

Example 2

Solvedy

dx= e3x2y + x2e2y

We separate them intoe2ydy = (e3x + x2)dx

5

Integrating both side, we get e2ydy =

(e3x + x2)dx+ c

e2y

2=e3x

3+x3

3+ c

Hence the solution is3e2y = 2(e3x + x3) + 6c

Example 3

Solvedy

dx= sin(x+ y) + cos(x+ y)

Setting x+ y = t so that dy/dx = dt/dx 1. The given equation becomesdt

dx 1 = sin t+ cos t

dx =dt

1 + sin t+ cos t

Integrating both side, we getdx =

dt

1 + sin t+ cos t+ c

x =

2d

1 + sin 2 + cos 2+ c (setting t = 2)

=

2d

2 cos2 + 2 sin cos c

=

sec2

1 + tan d + c

= log(1 + tan ) + c

Hence the solution is

x = log

[1 + tan

1

2(x+ y)

]+ c

Example 4

Solvey

x

dy

dx+

x2 + y2 12(x2 + y2) + 1

= 0

Setting x2 + y2 = t, we get

2x+ 2ydy

dx=dt

dx

ory

x

dy

dx=

1

2x

dt

dx 1

6

The given equation becomes

1

2x

dt

dx 1 + t 1

2t+ 1= 0

1

2x

dt

dx= 1 t 1

2t+ 1=

t+ 2

2t+ 1

2xdx =2t+ 1

t+ 2dt

2xdx =

(2 3

t+ 2

)dt

Integrating both side, we get 2xdx =

(2 3

t+ 2

)dt+ c

x2 = 2t 3 log(t+ 2) + c

Hence the solution isx2 + 2y2 3 log(x2 + y2 + 2) + c = 0

2.2 Homogeneous Equations

Homogeneous equations are of the form

dy

dx=f(x, y)

g(x, y)

where f(x, y) and g(x, y) are homogeneous functions of the same degree in x and y.To solve a homogeneous equation,

1. Set y = ux, then dydx = u+ xdudx

2. Separate the variables u and x, and integrate.

Example 1

Solve(x2 y2)dx xydy = 0

Rearrange the equation, we getdy

dx=x2 y2xy

Setting y = ux, then dydx = u+ xdudx , the equation becomes,

u+ xdu

dx=

1 u2u

xdu

dx=

1 2u2u

Separating the variables, we getu

1 2u2 du =dx

x

7

Integrating both sides, we get u

1 2u2 du =dx

x+ c

14

4u1 2u2 du =

dx

x+ c

14

log(1 2u2) = log x+ c4 log x+ log(1 2u2) = 4c

log x4(1 2u2) = 4c

x4(

1 2y2

x2

)= e4c = c (Set u = y/x)

Hence the solution isx2(x2 2y2) = c

Example 2

Solve (x tan

y

x y sec2 y

x

)dx+

(x sec2

y

x

)dy = 0

The equation can be rearrange into homogeneous equation

dy

dx=(yx

sec2y

x tan y

x

)cos2

y

x

Setting y = ux, then dydx = u+ xdudx , the equation becomes,

u+ xdu

dx= (u sec2 u tanu) cos2 u

xdu

dx= u tanu cos2 u u

Separating the variables, we getsec2 u

tanudu = dx

x

Integrating both side, we get sec2 u

tanudu =

dx

x

log tanu = log x+ log cx tanu = c

Hence the solution isx tan

y

x= c

Example 3

Solve

(1 + ex/y)dx+ ex/y(

1 xy

)dy = 0

8

Rewrite the equation into its homogeneous form, we get

dx

dy=

ex/y(

1 xy)

(1 + ex/y)

Setting x = uy, then dxdy = u+ ydudy , the equation becomes

u+ ydu

dy= e

u(1 u)(1 + eu)

ydu

dy= e

u(1 u)(1 + eu)

u = u+ eu

(1 + eu)

Separating the variables, we get

dyy

=1 + eu

u+ eudu =

d(u+ eu)

u+ eu

Integrating both side, we get

dy

y=

d(u+ eu)

u+ eu

log y = log(u+ eu) + cy(u+ eu) = ec = c

Hence the solution isx+ yex/y = c

2.2.1 Equations Reducible to Homogeneous Form

The equation of the formdy

dx=

ax+ by + c

ax+ by + c

can be reduced to the homogeneous form as follows:

Case I. When aa 6= bbSetting

x = X + h, y = Y + k, (h, k are constants)

anddx = dX, dy = dY

the equation becomesdY

dX=

aX + bY + (ah+ bk + c)

aX + bY + (ah+ bk + c)

we choose h and k so that dYdX become homogeneous equation.Set ah+ bk + c = 0 and ah+ bk + c = 0 so that

h

bc bc =k

ca ca =1

ab baor

h =bc bcab ba, and k =

ca caab ba

9

Thus when ab ba 6= 0, dYdX becomesdY

dX=

aX + bY

aX + bY

which is homogeneous in X, Y and be solved by setting Y = uX.

Example

Solvedy

dx=y + x 2y x 4

Settingx = X + h, y = Y + k, (h, k are constants)

anddx = dX, dy = dY

the equation becomesdy

dx=y + x+ (k + h 2)y x+ (k h 4)

Letting k + h 2 = 0 and k h 4 = 0, then h = 1 and k = 3 then the equation will bedYdX =

Y+XYX which is homogeneous in X and Y .

Setting Y = uX, then dYdX = u+XdudX , the equation becomes,

u+Xdu

dX=u+ 1

u 1Xdu

dX=

1 + 2u u2u 1

u 11 + 2u u2 du =

dX

X

Integrating both sides, we get

12

2 2u

1 + 2u u2 du =dX

X

12

log(1 + 2u u2) = logX + c

log

(1 +

2Y

X2 Y

2

X2

)+ logX2 = 2c

log(X2 + 2XY Y 2) = 2cX2 + 2XY Y 2 = e2c = c

Setting X = x h = x+ 1 and Y = y k = y 3, the previous equation becomes

(x+ 1)2 + 2(x+ 1)(y 3) (y 3)2 = cx2 + 2xy y2 4x+ 8y 14 = c

which is the required solution.

Case II. When aa =bb

When ab ba = 0, the above method fails as h and k become infinite or undetermined.

10

Seta

a=

b

b=

1

mor

a = am, b = bm

then dydx becomesdy

dx=

(ax+ by) + c

m(ax+ by) + c

Setting ax+ by = t, so that a+ b dydx =dtdx or

dydx =

1b

(dtdx a

), then the previous equation becomes

1

b

(dt

dx a)

=t+ c

mt+ c

dt

dx= a+

bt+ bc

mt+ c

=(am+ b)t+ ac + bc

mt+ c

so the variables are separable. In the solution, setting t = ax+ by, we get the required solution ofdydx

Example

Solve(3y + 2x+ 4)dx (4x+ 6y + 5)dy = 0

The equation can be rearrange as

dy

dx=

(2x+ 3y) + 4

2(2x+ 3y) + 5

Setting 2x+ 3y = t, then 2 + 3 dydx =dtdx The equation becomes,

1

3

(dt

dx 2)

=t+ 4

2t+ 5

dt

dx=

7t+ 22

2t+ 52t+ 5

7t+ 22dt = dx

Integrating both sides, 2t+ 5

7t+ 22dt =

dx (

2

7 9

7 1

7t+ 22

)= x+ c

2

7t 9

49log(7t+ 22) = x+ c

Setting t = 2x+ 3y, we have

14(2x+ 3y) 9 log(14x+ 21y + 22) = 49x+ 49c21x 42y + 9 log(14x+ 21y + 22) = c


11

2.3 Linear First-order Equations

A differential equation is said to be linear if the dependent variable and its differential coefficientsoccur only in the first degree and not multiplied together.

Thus the standard form of a linear equation of the first order is,

dy

dx+ Py = Q

where P Q are the functions of x.To solve the equation, we multiply both sides by e

Pdx, so we get

dy

dx ePdx + y

(ePdxP

)= Qe

Pdx

d

dx

(yePdx)

= QePdx

Integrating both side, we get

yePdx =

QePdx + c

as the required solution.

Example 1

Solve

(x+ 1)dy

dx y = e3x(x+ 1)2

Divide the equation with (x+ 1), we get

dy

dx yx+ 1

= e3x(x+ 1)

Here P = 1x+1 and Pdx =

dx

x+ 1= log(x+ 1) = log(x+ 1)1

The integrating factor is

ePdx = elog(x+1)

1=

1

x+ 1

Thus the solution for the equation is

y

x+ 1=

[e3x(x+ 1)]

1

x+ 1dx+ c

=

e3xdx+ c

=1

3e3x + c

So the solution is

y =

(1

3e3x + c

)(x+ 1)

12

Example 2

Solve (e2x

x

)dx

dy= 1

Rewrite the equation intody

dx+

yx

=e2x

x

where P = 1x

and Pdx =

1x

= 2x

then the integrating factor becomes ePdx = e2

x

The solution for the equations is

y e2x =

e2x

x e2xdx+ c

=

1xdx+ c

So the solution isy e2

x = 2

x+ c

Example 3

Solve

3x(1 x2)y2 dydx

+ (2x2 1)y3 = ax3

Setting z = y3 and 3y2 dydx =dzdx , the equation becomes

x(1 x2)dzdx

= (2x2 1)z = ax3

dz

dx+

2x2 1x x3 z =

ax3

x x3

Here P = 2x21

xx3 and Pdx =

2x2 1x x3 dx

=

( 1x 1

2

1

1 + x+

1

2

1

1 x)dx

= log x 12

log(1 + x) 12

log(1 x)

= log[x

1 x2]

The integrating factor is

exp

(2x2 1x x3 dx

)= e log[x

1x2] =

[x

1 x2]1

13

So the solution is

z[x

1 x2] = a

x3

x(1 x2) 1

x

1 x2 dx+ c

= a

x(1 x2)3/2dx+ c

= a2

(2x)(1 x2)3/2dx+ c

z[x

1 x2] = a(1 x2)1/2 + cInserting z = y3, we get

y3 = ax+ cx

1 x2

2.3.1 Bernoullis Equation

The equationdy

dx+ Py = Qyn

where P, Q are function of x is reducible to linear equation of the first order. This is a Bernoullisequation.

To solve it, we divide both sides with yn so that

yndy

dx+ Py1n = Q

Setting y1n = z so that

(1 n)yn dydx

=dz

dxThen, the equation becomes

1

1 ndz

dx+ Pz = Q

dz

dx+ P (1 n)z = Q(1 n)

which is a linear equation of the first order and can be solved easily.

Example

Solve

xdy

dx+ y = x3y6

.Divide both sides with xy6, we get

y6dy

dx+y5

x= x2

Setting y5 = z and

5y6 dydx

=dz

dxThe equation becomes

15

dz

dx+z

x= x2

dz

dx 5xz = 5x2

14

which is the same as dydx + Py = Q pattern. So we can use the integrating factor.Integrating factor is

ePdx = e

(5/x)dx

= e5 log x

= elog x5

= x5

The solution is

z x5 =

(5x2) x5dx+ c

y5x5 = 5 x2

2 + c

Divide both side with y5x5, we get

1 = (2.5 + cx2)x3y5


2.4 Exact Differential Equations

A differential equation of the form

M(x, y)dx+N(x, y)dy = 0

is said to be exact if its left hand side member is the exact differential of some function u(x, y)which is

du Mdx+Ndy = 0The solution is therefore

u(x, y) = c

From calculus, we see that the partial derivatives of u(x, y) is

du =u

xdx+

u

ydy

Comparing with du Mdx+Ndy, then we see thatu

x= M

u

y= N

If we differentiate M with respect to y and N with respect to x, we get

M

y=

2u

yx

N

x=

2u

xy

orM

y=N

x

We integrate ux = M with respect to x, we get

u =

Mdx+ k(y) k(y) is a constant

15

To determine k(y), we derive uy from this equation, and useuy = N to get dk/dy, and integrate

dk/dy to get k.

Example

Solvecos(x+ y)dx+ (3x2 + 2y + cos(x+ y))dy = 0

Test for exactness

M = cos(x+ y)

N = (3x2 + 2y + cos(x+ y))

Thus

M

y= sin(x+ y)

N

x= sin(x+ y)

So the equation is exact.Implicit general solution

u =

Mdx+ k(y)

=

cos(x+ y)dx+ k(y)

= sin(x+ y) + k(y)

To find k(y), we differentiate u = sin(x+ y) + k(y) with respect to y, we get

u

y= cos(x+ y) +

dk

dy

N = 3y2 + 2y + cos(x+ y)

So, dkdy is

dk

dy= 3y2 + 2y

By integration k(y) isk(y) = y3 + y2 + c

Inserting k(y) into u = sin(x+ y) + k(y), we obtain the answer

u(x, y) = sin(x+ y) + y3 + y2 = c

2.4.1 Reduction to Exact Form, Integrating Factor

We can transform a non-exact equation,

P (x, y)dx+Q(x, y)dy = 0

by multiplying it with a function F (x, y) producing an exact equation

FPdx+ FQdy = 0

16

which can be solve by using the method from the previous section. F (x, y) is called IntegratingFactor.

How to find Integrating factor

For Mdx + Ndy = 0, the exactness condition is M/y = N/x. For F (x) and FPdx +FQdy = 0, the exactness condition is

(FP )/y = (FQ)/x

By product rule, we getFPy = F

Q+ FQx

Dividing by FQ and rearrange, we get

1

F

dF

dx= R where R =

1

Q

(P

y Qx

)If we integrate the equation with respect to x and taking the exponent on both side, we get

the integrating factor which is

F (x) = exp

R(x)dx

and if our initial is F (y), we get

F (y) = exp

R(y)dy where R(y) =

1

P

(Q

x Py

)

Example

Find an integrating factor and solve the initial value problem of

(ex+y + yey)dx+ (xey 1)dy = 0, y(0) = 1

Test for exactness

P

y= ex+y + ey + yey

Q

x= ey

Since Py 6= Qx , so we have to find the integrating factor to make it an exact equation.R(x) fail, since it on both x and y, so we have to use R(y), which is

R(y) =1

P

(Q

x Py

)=

1

ex+y + yey(ey (ex+y + ey + yey))

= 1

Hence the integrating factor is F (y) = exp 1dy = e1. Multiplying the integrating factor to

the original equation, we get(ex + y)dx+ (x ey)dy = 0

17

which is an exact equation and be solve using the previous method.

u =

(ex + y)

= ex + xy + k(y)

u

y= x+

dk

dy= N = x ey

dk

dy= ey

k = ey + cHence, the general solution is

u(x, y) = ex + xy + ey = c

The particular solution is

u(0, 1) = e0 + 0 1 + e1 = 3.72ex + xy + ey = 3.72

3 Second Order Linear Differential Equation

3.1 Homogeneous Linear ODEs with Constant Coefficients

A second-order homogeneous linear ODEs with constant coefficient is

y

+ ay+ by = 0

The solution is

y = ex

y

= ex

y

= 2ex

If we put into the equation and rearrange, we will get

(2 + a+ b)ex = 0

Hence is a solution of the important characteristic equation

2 + a+ b = 0

The roots for this quadratic equation is

1 =1

2(a+

a2 4b) 2 = 1

2(a

a2 4b)

then the functiony1 = e

1x y2 = e2x

are solutions for second-order homogeneous linear ODEs with constant coefficient.From algebra, we know that the quadratic equation might have three different kind of roots,

depending on the sign of the discriminant a2 4b, which are1. Two real roots if a2 4b > 0.2. A real double root if a2 4b = 0.

18

3. Complex conjugate roots if a2 4b < 0

Case I: Two Distinct Real Root 1 and 2

In this case, the basic solution is

y1 = e1x and y2 = e

2x

and the general solution isy = c1e

1x + c2e2x

Example

Solve the initial value problem

y

+ y 2y = 0, y(0) = 4, y(0) = 5

The characteristics equation is2 + 2 = 0

Its roots are,1 = 1 and 2 = 2

So the general equation isy = c1e

x + c2e2x

For finding particular solution, we differentiate once

y

= c1ex 2c2e2x

and using the initial condition, we get

y = ex + 3e2x

Case II: Real Double Root = a/2

When a2 4b = 0 we only get one root, = 1 = 2 = a/2, hence the only solution is

y1 = e(a/2)x

To obtain the second independent solution y2, we set y2 = uy1. Substitute this and its derivativey2 and y

2 , we get

(uy1 + 2uy + uy1 ) + (a(u

y1 + uy1) + buy1 = 0

Collecting termuy1 + u(2y

1 + ay1) + u(y

1 + ay

1 + by1) = 0

The second and third expression is zero, so we are left with

uy1 = 0

By two integrations, we getu = c1x+ c2

19

we choose c1 = 1 and c2 = 0 then u = x, so y2 = xy1, so the general solution is

y = (c1 + c2x)e(a/2)x

Example


y + y + 0.25y = 0 y(0) = 3.0, y(0) = 3.5

The characteristic equation is

2 + + 0.25 = (+ 0.5)2 = 0

It has double root = 0.5, this give the general solution as

y = (c1 + c2x)e0.5x

Taking the derivative and find c1 and c2, so we get the particular solution as

y = (3 2x)e0.5x

Case III: Complex Roots 12

+ i and 12 i

The discriminant is a2 4b < 0, two complex roots, or we can obtain a real solution by using

y1 = eax/2 cosx and y2 = eax/2 sinx

where = b 14a2. The real general solution for Case III is

y = eax/2(A cosx+B sinx)

where A, B are arbitrary constants.

Example


y + 0.4y + 9.04y = 0, y(0) = 0, y(0) = 3

The characteristic equation is2 + 0.4+ 9.04 = 0

so the roots are 1 = 0.2 + 3i and 1 = 0.2 3i. Hence = 3. The general solution is

y = e0.2x(A cos 3x+B sin 3x)

Differentiate the general solution and using the initial value, we get the particular solution as

y = e0.2x sin 3x

20

3.2 Non-homogeneous ODEs

A general solution for a non-homogeneous ODEs

y + p(x)y + q(x)y = r(x), r(x) 6= 0on open interval I is a solution of the form

y(x) = yh(x) + yp(x)

where, yh = c1y1 + c2y2 is a general solution from homogeneous ODE and yp is any solution ofnon-homogenous ODE containing no arbitrary constant.

3.2.1 Method of Undetermined Coefficients

Method of undetermined coefficients is suitable for linear ODEs with constant coefficients a and b,

y + ay + by = r(x)

when r(x) is an exponential function, a power of x, a cosine or sine, or sums or products of suchfunctions. These functions have derivative similar to r(x) itself. So, we choose a form for yp similarto r(x), but with unknown coefficients to be determined by substituting yp and its derivatives intothe ODE. Table below shows the choice for yp for important form of r(x).

Term in r(x) Choice for ypkex Cex

kxn (n = 0, 1, ) Knxn +Kn1xn1 + +K1x+K0k cosx K cosx+M sinxk sinx K cosx+M sinxkex cosx ex(K cosx+M sinx)kex sinx ex(K cosx+M sinx)

And the choice of rules for the method of undetermined coefficients are

1. Basic Rule: If r(x) is one of the functions in the first column from the table, choose ypfrom the same line and determine the coefficients by substituting yp and its derivatives intothe equation

2. Modification rule: If a term in our choice for yp happens to be a solution of the ho-mogeneous ODE corresponding to the equation, we multiply the choice of yp by x or x

2

etc.

3. Sum Rule: If r(x) is a sum of functions in the first column of the table; we choose yp bysumming up the functions in the corresponding line.

Example 1: Rule 1


y + y = 0.001x2, y(0) = 0, y(0) = 1.5

The general solution for homogeneous ODE y + y = 0 is

yh(x) = A cosx+B sinx

Solution for non-homogeneous ODE, setting yp = Kx2, then y

p = 2K. Substitute into the

equation, we get2K +Kx2 = 0.001x2

21

Comparing both sides, we see that

2K = 0, K = 0.001

which the value of K contradict each other, so yp = Kx2 is not the solution for this equation.

Looking from the table, the choice of yp for the term r(x) = Kxn is yp = K2x

2 + K1x + K0,then

yp + yp = 2K2 + 2K2x

2 +K1x+K0 = 0.001x2

Comparing both sides, we get

K2 = 0.001, K1 = 0, K0 = 0.002This gives

yp = 0.001x2 0.002

and the general solution becomes,

y = yh + yp = A cosx+B sinx+ 0.001x2 0.002

To find the value for A and B, we use the initial condition

y(0) = A cos 0 +B sin 0 + 0.001(0)2 0.002= A 0.002 = 0

So, A = 0.002. To find B, we diffentiate the general solution once and put the initial condition

y = yh + y

p

= A sinx+B cosx+ 0.002xy(0) = A sin 0 +B cos 0 + 0.002(0)

= B = 1.5

So, B = 1.5. This gives the answer as

y = 0.002 cosx+ 1.5 sinx+ 0.001x2 0.002

Example 2: Rule 2


y + 3y + 2.25y = 10e1.5x, y(0) = 1, y(0) = 0The general solution for the homogeneous ODE,

y + 3y + 2.25y = 0

where 2 + 3+ 2.25 = (+ 1.5)2 = 0 is

yh = (c1 + c2x)e1.5x

The solution of non-homogeneous ODE cannot be taken as yp = Ce1.5x since this a solution

for homogeneous ODE. So, by using rule number 2, we modify our choice function by x2. That is,we use

yp = Cx2e1.5x

yp = C(2x 1.5x2)e1.5x

yp = C(2 3x 3x+ 2.25x2)e1.5x

22

Substitute these equations into the main equation, by ignoring the e1.5x, we get

C(2 6x+ 2.25x2) + 3C(2x 1.5x2) + 2.25Cx2e1.5x = 10

Comparing both sides, we get 2C = 10, so C = 5 and the solution is

yp = 5x2e1.5x

Hence the general equation becomes

y = yh + yp = (c1 + c2x)e1.5x 5x2e1.5x

Using the initial condition to find c1 and c2, we get

c1 = 1, and c2 = 1.5

This give the answer which is

y = (1 + 1.5x)e1.5x 5x2e1.5x

Example 3: Rule 3


y + 2y + 5y = e0.5x + 40 cos 10x 190 sin 10x, y(0) = 0.16, y(0) = 40.08

The general solution for homogeneous ODE y+ 2y+ 5y = 0 from the characteristics equation

2 + 25 = (+ 1 + 2i)(+ 1 2i) = 0

isyh = e

x(A cos 2x+B sin 2x)

The solution for non-homogeneous ODE, we choose our yp as summation of two function fromthe table, that is

yp = Ce0.5x +K cos 10x+M sin 10x

Substitute into the equation and comparing both side, we get

C = 0.16, K = 0, M = 2

Hence the solution is

y = yh + yp = ex(A cos 2x+B sin 2x) + 0.16e0.5x + 2 sin 10x

From the initial condition, we found that A = 0 and B = 10, hence the complete solution is

y = 10ex sin 2x+ 0.16e0.5x + 2 sin 10x

3.2.2 Method of Variation of Parameters

A particular solution yp for non-homogeneous linear ODE where r(x) is not an exponential functionex, a power of x, a cosine or sine, or sums or products of such functions can be determine by usingthe following formula

yp(x) = y1

y2r(x)

Wdx+ y2

y1r(x)

Wdx

23

where y1 and y2 are solutions for the corresponding homogeneous ODE,

y + P (x)y +Q(x)y = 0

and W is the Wronskian of y1 , y2,

W = y1y2 y2y

1

Example

Solve the non-homogeneous ODE

y + y = secx

From the homogeneous solution we get y1 = cosx and y2 = sinx. Using these information, wecan get the Wronskian as

W = cosx cosx sinx ( sinx) = 1By using the formula, yp is

yp = cosx

sinx secx1

dx+ sinx

cosx secx

1dx

= cosx ln | cosx|+ x sinx

The complete solution will be

y = yh + yp

= c1 cosx+ c2 sinx+ cosx ln | cosx|+ x sinxy = (c1 + ln | cosx|) cosx+ (c2 + x) sinx

4 Other Types of ODE

4.1 ODE with y is missing

The equation which do not contain y directly can be written as

f

(d2y

dx2,dy

dx, x

)= 0

on substituting

dy

dx= p

d2y

dx2=dp

dx

we get

f

(dp

dx, p, x

)= 0

24

Example

Find the solution for this ODE

(1 x2)D2 xD = 2

By setting dydx = p andd2ydx2 =

dpdx , we get

(1 x2)dpdx xp = 2

dp

dx x

1 x2 p =2

1 x2Integrating factor is

I.F = e ( x

1x2)

= e1/2log(1x2)

= elog1x2

=

1 x2

Hence, the solution is

p

1 x2 = 2

1 x21 x2 dx

=

2

1 x2 dx

= 2 sin1 x+ c1

or p =2 sin1 x

1 x2 +c1

1 x2

ordy

dx=

2 sin1 x1 x2 +

c11 x2

On integrating with respect to x, we get

y = (sin1 x)2 + c1 sin1 x+ c2

4.2 ODE with x is missing

The equations that do not contain x directly are of the form

f

(d2y

dx2,dy

dx, y

)= 0

On substituting

dy

dx= p

d2y

dx2=dp

dx=dp

dy dydx

d2y

dx2=dp

dy p

25

we get

f

(dp

dx, p, y

)= 0

Example

Solve the following equation

yd2y

dx2+

(dy

dx

)2=dy

dx

Put dydx = p andd2ydx2 =

dpdy p into the equation, we get

ydp

dy p+ (p)2 = p

or ydp

dy= 1 p

ordp

1 p =dy

y

or log(1 p) = log y + log c1or

1

1 p = c1y

or p = 1 1c1y

ordy

dx=c1y 1c1y

dydx

= p

orc1y

c1y 1dy = dx

or

(1 +

1

c1y 1)dy = dx

y + 1c1

log(c1y 1) = x+ c2

4.3 Equation of the type d2y

dx2= f(y)

Multiplying with 2 dydx , we get

2dy

dx d

2y

dx2= 2

dy

dx f(y)

Integrating with respect to x, we get(dy

dx

)2= 2

f(y)dy + c1

= (y), (y) = 2f(y)dy + c1

dy

dx=(y)

dy(y)

= dx

So the solution is dy(y)

= x+ c2

26

Example


d2y

dx2=y, y(0) = 1, y(0) =

23

Multiplying with 2 dydx , we get

2dy

dx d

2y

dx2= 2

dy

dx y

Integrating with respect to x,we get(dy

dx

)2=

4

3y3/2 + c1

=4

3y3/2, c1 = 0 using IVP

dy

dx=

23y3/4

y3/4dy =23dx

Integrating both side and using IVP values, we get

y1/4 =23x+ c2

=23x+ 1, c2 = 1 using IVP

4.4 Euler-Cauchy Equation

An equation of the form

a2x2 d

2y

dx2+ a1x

dy

dx+ a0y = r(x)

where a0, a1anda2 are constants is called Euler-Cauchy Equation. We will discuss two methodwhich can be use to solve this equation.

4.4.1 Method 1: Using substitution x = ez

An Euler-Cauchy equation can be reduced to a linear equation with constant coefficients by chang-ing the independent variable from x to z where

x = ez, z = log x,d

dz= D

For x dydx

dy

dx=dy

dz dzdx

=1

x dydz

xdy

dx=dy

dz

xdy

dx= Dy

27

And for x2 d2ydx2 , we get

d2y

dx2=

d

dx

(dy

dx

)=

d

dx

(1

x dydz

)= 1

x2dy

dz+

1

x

d2y

dz2dz

dx

= 1x2dy

dz+

1

x

d2y

dz21

x

=1

x2

(d2y

dz2 dydz

)=

1

x2(D2 D)y

x2d2y

dx2= (D2 D)y

Substitution of x dydx = Dy and x2 d

2ydx2 = (D

2 D)y into the first equation will reduces equationinto a differential equation with constant coefficients.

Example


x2d2y

dx2 2xdy

dx 4y = x4

Set

x = ez, z = log x,d

dz= D, x

dy

dx= Dy, x2

d2y

dx2= (D2 D)y

The equation becomes

D(D 1)y 2Dy 4y = e4z(D2 3D 4)y = 0

Since m1 = 1 and m2 = 4, then the homogeneous solution isyh(z) = c1e

z + c2e4z

And by using method of undetermined coefficient where we set yp = C xe4z, the non-homogeneoussolution is

(D2 3D 4)y = 8Ce4z + 16Cze4z 3(Ce4z + 4Cze4z) 4Cze4z= 5Ce4z

e4z = 5Ce4z

Comparing both side, we get C = 15 , thus particular solution is

yp(z) =1

5ze4z

So the solution is

y(z) = yh(z) + yp(z)

= c1ez + c2e4z +

ze4z

5

28

But since our original independent variable is x, we substitute back x = ez and z = lnx into theequation to get our final solution,

y(x) =c1x

+ c2x4 +

1

5x4 lnx

4.4.2 Method 2: Using y = xm

We write the Euler-Cauchy equation in the form

x2y + axy + by = 0

Substituting

y = xm

y = mxm1

y = m(m 1)xm2

into Euler-Cauchy equation, we get

x2m(m 1)xm2 + axmxm1 + bxm = 0 orm(m 1) + am+ b = 0m2 + (a 1)m+ b = 0

The roots are calculated using this formula

m1 =1

2(1 a) +

1

4(1 a)2 b

m2 =1

2(1 a)

1

4(1 a)2 b

Case I: Real roots, m1 6= m2If the roots m1 and m2 are real and different, the solutions are

y1 = xm1 , y2 = x

m2

and the general solution will be

y = c1xm1 + c2x

m2

Example

Solve the following Euler-Cauchy equation

x2y + 1.5xy 0.5y = 0the auxiliary equation is

m2 + (1.5 1)m 0.5 = 0The roots are m1 = 0.5 and m2 = 1, so the general solution is

29

y = c1x+

c2x

Case II: Real double roots, m1 = m2

The first solution is (from m1 =12 (1 a))

y1 = x(1a)/2

To find a second linearly independent solution, we use the reduction of order method by settingy2 = u y1 where we get

u =

1

y21exp

(p dx

)dx

where p = ax , which after some manipulation, we get u as

u = lnx

Therefore, the second linearly independent solution is

y2 = u y1= lnx x(1a)/2

Then, the general solution will be

y = (c1 + c2 lnx)x(1a)/2

Example

Solve the Euler-Cauchy equation

x2y 5xy + 9y = 0

Answer

The auxiliary equation is

m2 6m+ 9 = 0which give double real roots m1 = m2 = 3, so the general solution is

y = (c1 + c2 lnx)x3

Case III: Complex Roots, m1,2 = i

We show one example

Example

Solve the Euler-Cauchy equation

x2y + 0.6xy + 16.04y = 0

30

the auxilirary equation is

m2 0.4m+ 16.04 = 0which give complex roots

m1 = 0.2 + 4i

m2 = 0.2 4i

The basis solution

y1 = xm1 = x0.2+4i = x0.2(eln x)4i = x0.2e(4 ln x)i

y2 = xm2 = x0.24i = x0.2(eln x)4i = x0.2e(4 ln x)i

Using Eulers Formula and t = 4 lnx, we get

y1 = xm1 = x0.2[cos(4 lnx) + i sin(4 lnx)]

y2 = xm2 = x0.2[cos(4 lnx) i sin(4 lnx)]

After some manipulation we get (try to prove yourself)

y1 = x0.2 cos(4 lnx)

y2 = x0.2 sin(4 lnx)

Hence the general solution is

y = x0.2[A cos(4 lnx) +B sin(4 lnx)

4.5 One Solution is Known

If y = u is a given solution belonging to the homogeneous equation of the differential equation,then put y = u v is other solution for differential equation. Let

d2y

dx2+ P (x)

dy

dx+Q(x)y = R(x) (1)

Since u is a solution of the differential equation, we can write

d2u

dx2+ P (x)

du

dx+Q(x)u = 0

For y = u v

y = u vdy

dx= v

du

dx+ u

dv

dxd2y

dx2= v

d2u

dx2+ 2

dv

dx

du

dx+ u

d2v

dx2

31

Substituting y, dydxandd2ydx2 into (1), we get

vd2u

dx2+ 2

dv

dx

du

dx+ u

d2v

dx2+ P (x)

(vdu

dx+ u

dv

dx

)+Q(x)uv = R(x)

Rearrange, we get

v

[d2u

dx2+ P (x)

du

dx+Q(x)u

]+ u

[d2v

dx2+ P (x)

dv

dx

]+ 2

du

dx

dv

dx= R(x)

The first bracket is equal to 0, so the remaining equation divided by u, we get

d2v

dx2+ P (x)

dv

dx+

2

u

du

dx

dv

dx=R(x)

ud2v

dx2+

[P (x) +

2

u

du

dx

]dv

dx=R(x)

u

Thus, the formula to find v is

d2v

dx2+

[P (x) +

2

u

du

dx

]dv

dx=R(x)

u(2)

Example

Solve the following equation, given that y = u = x is a homogeneous solution for this equation

d2v

dx2+

[P (x) +

2

u

du

dx

]dv

dx=R(x)

u

d2v

dx2+

[2x(1 + x)x2

+2

x(1)

]dv

dx=x

x

d2v

dx2 2dv

dx= 1

ordz

dx 2z = 1, z = dv

dx

The last equation is linear differential equation, its solution is

ze2x =

1 e2xdx+ c1

=e2x

2 + c1

z =1

2 + c1e2x

dv

dx=

1

2 + c1e2x

dv = (1

2 + c1e2x)dx

v =x2

+c12e2x + c2

Then, the complete solution y = u v is

y = u v = x(x

2+c12e2x + c2

)

32

5 Higher Order Linear ODEs

An Higher Order ODE which can be written as below

yn + pn1(x)yn1 + . . .+ p1(x)y + p0(x)y = r(x)

is called Higher Order Linear ODE. The general solution can be represent as

y(x) = c1y1(x) + . . .+ cnyn(x).

5.1 Homogeneous Linear ODEs with Constant Coefficients

We consider nth-order homogeneous linear ODEs with constant coefficient which can write in form

yn + an1yn1 + . . .+ a1y + a0y = 0

Substituting y = ex, we obtain the characteristic equation,

n + an1n1 + . . .+ a1+ a0 = 0

By factoring the equation or using other methods we can determine the roots for this equation.

Distinct Real Roots

If all the roots we get are real and different, (1 6= 2 6= . . . 6= n), then the n solutions,

y1 = e1x, . . . , yn = e

nx

cosntitute the basic solution for

y = c1e1x + c2e

2x + . . .+ cn1en1x + cnenx

Example

Solve the ODE

y 2y y + 2y = 0

Answer


3 22 + 2 = 0The roots are 1 = 1, 2 = 1 and 3 = 2. Therefore, the general solution is

y = c1ex + c2ex + c3e2x

Simple Complex Roots

If complex roots occur, they exist in conjugate pairs. Thus if = + i is a complex rootexist then it conjugate = i will also exist. The two corresponding linearly independentsolutions are

y1 = ex cos(x), y2 = e

x sin(x)

33

Example

Solve the ODE

y y + 100y 100y = 0

Answer


3 2 + 100 100 = 0The roots are 1 = 1, 2 = 10i and 2 = 10i which has a complex roots. The general solution

will be

y = c1ex + c2e

0x cos 10x+ c3e0x sin 10x= c1e

x + c2 cos 10x+ c3 sin 10x

y = c1ex +A cos 10x+B sin 10x

Multiple Real Roots

If among the nth roots we have a real double root occur, for example 1 = 2, we take y1 andxy1 as the corresponding linearly independent solutions.

In general, if we have as a real root of order m, the m corresponding linearly independentsolution are

ex, xex, x2ex, . . . , xm1ex.

Example

Solve the ODE

y 3y + 3y y = 0

Answer


5 34 + 33 2 = 0The roots are 1 = 2 = 0 and 3 = 4 = 5 = 1. So the general solution will be

y = c1 + c2x+ (c3 + c4x+ c5x2)ex

Multiple Complex Roots

The general solution will be the same as Simple Complex Roots except if = + i is acomplex double root with corresponding linearly independent solutions are

34

ex cos(x), ex sin(x), xex cos(x), xex sin(x)

and the corresponding general solution are

y = ex [(A1 +A2x) cosx+ (B1 +B2x) sinx]

5.2 Non-Homogeneous Linear ODEs

We consider nth-order non-homogeneous linear ODEs with constant coefficient which can write inform

yn + an1yn1 + . . .+ a1y + a0y = r(x)

The general solution is in the form

y(x) = yh(x) + yp(x)

where, yh(x) is the general solution for the corresponding homogeneous ODE. yp(x) is theparticular solution and can be found by using one of these methods.

1. Method of Undetermined Coefficients (same as for Second Order ODE)

2. Method of Variation of Parameters

3. Annihilator Method (guessing yp pattern)

5.2.1 Method of Variation of Parameters

The particular solution for the non-homogeneous higher order ODEs can be determined using thefollowing formula

yp(x) =

nk=1

yk(x)

Wk(x)

W (x)r(x) dx

= y1(x)

W1(x)

W (x)r(x) dx+ . . .+ yn(x)

Wn(x)

W (x)r(x) dx

Example

Solve the non-homogeneous Euler-Cauchy equation

x3y 3x2y + 6xy 6y = x4 lnx, (x > 0)

Answer

For general homogeneous solution, we use y = xm which give

m(m 1)(m 2) 3m(m 1) + 6m 6 = 0The roots are 1, 2, 3 and give as a basis

y1 = x, y2 = x2, y3 = x

3.

Hence the corresponding general solution of the homogeneous ODE is

35

yh = c1x+ c2x2 + c3x

3.

For particular non-homogeneous solution, we start by finding the Determinants

W =

x x2 x3

1 2x 3x2

0 2 6x

= 2x3W1 =

0 x2 x3

0 2x 3x2

1 2 6x

= x4W2 =

x 0 x3

1 0 3x2

0 1 6x

= 2x3W3 =

x x2 01 2x 00 2 1

= x2Putting all the result into the standard equation, we get

yp = y1(x)

W1(x)

W (x)r(x) dx+ y2(x)

W2(x)

W (x)r(x) dx+ y3(x)

W3(x)

W (x)r(x) dx

= x

x4

2x3x lnx dx+ x2

2x32x3

x lnx dx+ x3

x2

2x3x lnx dx

= x

x

2x lnx dx x2

x lnx dx+ x3

1

2xx lnx dx

=x

2

(x3

3lnx x

3

9

) x2

(x2

2lnx x

2

4

)+x3

2(x lnx x)

yp =1

6x4(

lnx 116

)Hence the complete solution is

y = yh + yp

= c1x+ c2x2 + c3x

3 +1

6x4(

lnx 116

)5.2.2 Annihilator Method

Here we try to make r(x) becomes 0 by multiplying the non-homogeneous equation with fixedDifferential operator.

1. for r(x) = erx, we mulitply the non-homogeneous equation with (Dr) to get (Dr)r(x) = 0Example

Derx = rerx

(D r)erx = 0

2. for r(x) = xkerx we multiply the non-homogeneous equation with (D r)k+1 to get (D r)k+1r(x) = 0

36

Example

D(erxu) = erxDu+ rerxu = erx(D + r)u

0 = D(erxu) erxDu rerxuD2(erxu) = erxD2u+ 2rerxDu+ r2erxu = erx(D + r)2u

0 = D2(erxu) erxD2u 2rerxDu r2erxu

3. for r(x) = cosx or sinx, we multiply the non-homogeneous equation with (D2 + 2) toget (D2 + 2)r(x) = 0Example

D(cosx) = sinxD2(cosx) = 2 cosx

0 = D2(cosx) + 2 cosx

4. for r(x) = xkex cosx or xkex sinx, we multiply the non-homogeneous equation with[(D )2 + 2]k+1 to get [(D )2 + 2]k+1 r(x) = 0

Example

Find a general solution for the following non-homogeneous ODEs using Annihilator method,

y 3y + 4y = xe2x

Answer

For homogeneous solution we use

(D3 3D2 + 4)y(x) = (D + 1)(D 2)2y(x) = 0.To annihilate r(x) = xe2x we multiply both side with (D 2)2, we get

(D 2)2(D3 3D2 + 4)y(x) = (D 2)2(xe2x)(D 2)2(D + 1)(D 2)2y(x) = 0

(D + 1)(D 2)4y(x) = 0The general solution is

y(x) = c1ex + c2e2x + c3xe2x + c4x2e2x + c5x3e2x

which comprise of

yh = c1ex + c2e2x + c3xe2x

yp = c4x2e2x + c5x

3e2x

Using method of undetermined coefficients, we get c4 = 1/18 and c5 = 1/18, so the generalsolution is

y(x) = c1ex + c2e2x + c3xe2x 1

18x2e2x +

1

18x3e2x

37

Introduction to Differential EquationsDefinitionsSolution of Differential EquationsLinearly Independent Solution

First Order and First Degree ODEsSeparation of VariablesHomogeneous EquationsEquations Reducible to Homogeneous Form

Linear First-order EquationsBernoulli's Equation

Exact Differential EquationsReduction to Exact Form, Integrating Factor

Second Order Linear Differential EquationHomogeneous Linear ODEs with Constant CoefficientsNon-homogeneous ODEsMethod of Undetermined CoefficientsMethod of Variation of Parameters

Other Types of ODEODE with y is missingODE with x is missingEquation of the type d2ydx2 = f(y)Euler-Cauchy EquationMethod 1: Using substitution x = ezMethod 2: Using y = xm

One Solution is Known

Higher Order Linear ODEsHomogeneous Linear ODEs with Constant CoefficientsNon-Homogeneous Linear ODEsMethod of Variation of ParametersAnnihilator Method

Linear Differential Equations

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