Differential Equations Class Notes

Dan Wysocki

Spring 2013

Contents

1 Introduction 2

2 Classification of Differential Equations 6

2.1 Linear vs. Non-Linear . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

2.2 Seperable Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

2.3 Exact Differential Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

2.4 Integrating Factor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

2.5 Modeling with First Order Equations . . . . . . . . . . . . . . . . . . . . . . . . . . 21

2.6 Bernoulli’s Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

2.7 Homogenous Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

2.8 Second Order Linear Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

3 Transformations 33

1

3.1 Laplace Transform . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

3.2 Inverse Laplace Transform . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

3.3 Unit Step-function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42

3.4 Gamma Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47

4 Systems of Differential Equations 52

5 Final Examples 61

2

1 Introduction

Differential equations are equations that have one or more derivatives or differentials.

1

x

dy

dx=

1

yFirst order (1.1)

2x+ y y′′ = 5 Second order (1.2)

(1.3)

y′ =dy

dx

4d2y

dx2+ xy = sinx (1.4)

What is a solution to a Differential Equation?

xdy

dx= 2y (1.5)

Show that y(x) = x2 is a solution.

dy

dx= 2x

x (2x) = 2(x2)

2x2 = 2x2

For

(1 + xexy)dy

dx+ 1 + yexy = 0, (1.6)

show that x+ y + exy = 0 is a solution.

3

Implicit Solution

d

dx(x+ y + exy) =

d

dx(0)

1 +dy

dx+ exy

(y + x

dy

dx

)= 0

1 +dy

dx+ y exy + x exy

dy

dx= 0

dy

dx(1 + x exy) + 1 + y exy = 0

For

dy

dx= 2x, (1.7)

show that the following are solutions

1. y(x) = x2

d

dxy(x) =

d

dxx2

dy

dx= 2x

2. y(x) = x2 + 1

d

dxy(x) =

d

dxx2 + 1

dy

dx= 2x

4

3. y(x) = x2 + C

d

dxy(x) =

d

dxx2 + C

dy

dx= 2x

dp

dt=

1

2p− 450 (1.8)

Show that p(t) = 900 + C et/2 is a solution.

dp

dt= 0 + C e

t/2 1

2

=C

2et/2

LHS =C

2et/2

RHS =1

2

(900 + C e

t/2)− 450

=C

2et/2

What if p(0) = 850 ← (initial condition)

850 = 900 + C e0/2

850 = 900 + C ⇒ C = −50

We now have an Initial Value Problem (IVP).

dpdt

= 12p− 450

p(0) = 850

5

Solution is

p(t) = 900− 50 et/2

dy

dx= f(x, y),

dy

dxis the slope.

Example: Consider a mouse population that reproduces at a rate proportional to the current

population, with a rate constant equal to 1/2 mice/month assuming no owls present. When owls are

present, they eat the mice. Suppose that the owls eat on average 15 mice per day. Assuming there

are 30 days in a month, write a differential equation to describe the above.

p = mouse population, t = time in months

dp

dt∝ p

dp

dt= kp =

1

2p− 450, (450 is the mice eaten per month)

When p = 900,dp

dt= 0. This is the equilibrium solution.

When p > 900,dp

dt> 0

When p < 900,dp

dt< 0

6

Next time: Classifications of Diff Eqns & Seperable Diff Eqns.

2 Classification of Differential Equations

Ordinary vs. Partial Differential Equation (ODE vs. PDE)

→ based on number of independent variables

(dedpendent variable)→ y = f(x← (independent variable))

• If there is one independent variable → ODE

• If there is more than one independent variable → PDE

Example 2.1

dp

dt=

1

2p− 450→ p(t) one independant variable (2.1)

Example 2.2

∂2v(x, t)

∂x2=∂v(x, t)

∂t→ PDE (Heat equation) (2.2)

7

Example 2.3

∂2v(x, t)

∂x2=∂2v(x, t)

∂t2→ PDE (Wave equation) (2.3)

Based on number of unknowns

→ system of equations

dx

dt= 4x+ 3t

dy

dt= 5x− 2y

Order of a differential equation is the highest derivative that appears in the equation.

Equation 1 y′ + 3y = 0 1st order linearEquation 2 y′′ + 3y′ − 2t = 0 2nd order linear

Equation 3d4y

dt4− d2y

dt2+ 1 = e2t 4th order linear

Equation 4 uxx + ut = 0 2nd order linearEquation 5 uxx + uuyy = sin t 2nd order non-linearEquation 6 uxx + sin(u)uyy = cos t 2nd order non-linearEquation 7 uxx + uyy + sin(u) = 0 2nd order nonlinearEquation 8 uxx + uyy + sin(t) = 0 2nd order linearEquation 9 uxx + uyy + u = 0 2nd order linear

Table 1: Examples of different order and linear vs. non-linear differential equations.

2.1 Linear vs. Non-Linear

A linear ODE is of the form:

an(t)y(n) + an−1(t)y(n−1) + . . .+ a1(t)y′ + a0(t)y = g(t) (2.4)

This is similar to a polynomial, which take the form

anxn + an−1x

n−1 + . . .+ a1x+ a0 = f(x) (2.5)

8

But instead of it being the independent variables raised to decreasing powers, it’s the derivative of

the independent variables with decreasing orders of the derivatives.

2.2 Seperable Equations

Consider a first order ODE

dy

dx= f(x, y) . . . (2.6)

If we can express Equation 2.6 in the form of

M(x) dx+N(y) dy = 0 (2.7)

then Equation 2.6 is called a seperable equation.

Solvedy

dx=

x2

1− y2(2.8)

∫(1− y2) dy =

∫x2 dx∫

N(y) dy =

∫M(x) dx

y − y3

3=x3

3+ C (Implicit Solution)

Solvedy

dx=

3x2 − 4x+ 2

2(y − 1)(2.9)

9

∫2(y − 1) dy =

∫3x2 − 4x+ 2 dx

y2 − 2y = x3 − 2x2 + 2x+ C

Complete the square by adding 1 to both sides (note that C is still a constant, so the 1 disappears)

(y − 1)2 = x3 + 2x2 + 2x+ C

y − 1 = ±√x3 + 2x2 + 2x+ C

y = 1±√x3 + 2x2 + 2x+ C (Explicit Solution)

If, also, y(0) = −1, solve for y

−1 = 1±√

03 + 02 + 0 + C = 1±√C

−2 = ���−

±√C

C = (−2)2 = 4

y = 1±√x3 + 2x2 + 2x+ 4

−1 = 1±√

4 = 1���−

±2

y = 1−√x3 + 2x2 + 2x+ 4

How about y(0) = 3, C = 4?

3 = 1±√

4 = 1���+

±2

y = 1 +√x3 + 2x2 + 2x+ 4

10

Solve

y′ = 4x−x34+y3

y(0) = 1(2.10)

dy

dx=

4x− x3

4 + y3∫4 + y3 dy =

∫4x− x3 dx

4y +1

4y4 = 2x2 − 1

4x4 + C

4(1) +1

414 = 0− 0 + C

4 +1

4= C

C =17

4

4y +1

4y4 = 2x2 − 1

4x4 +

17

4

2.3 Exact Differential Equations

Assume we have a solution

Ψ(x, y) = C (2.11)

Then we will find what this solution is.

Theorem 2.4

M(x, y) +N(x, y)dy

dx= 0 is called exact if

∂M(x, y)

∂y=∂N(x, y)

∂x.

For Equation 2.11,

∂Ψ

∂x+∂Ψ

∂y

dy

dx= 0. (2.12)

∂Ψ

∂x= M(x, y),

∂Ψ

∂y= N(x, y) (2.13)

11

Example 2.5

2x+ y2 + 2xydy

dx= 0

2x+ y2 = M(x, y)

2xy = N(x, y)

∂M

∂y= 2y,

∂N

∂x= 2y∫

∂Ψ

∂x=

∫ (2x+ y2

)dx

= x2 + xy2 + c1∫∂Ψ

∂y=

∫2xy dy

= xy2 + c2

Ψ(x, y) = x2 + xy2 + h(y)

Ψ(x, y) = xy2 + h(x)

∂Ψ

∂y= 2xy + h′(y)

h′(y) = 0∫h′(y)dy =

∫0dy

h(y) = c3

x2 + xy2 + c3 = c4

x2 + xy2 = C

Example 2.6

2xy +(x2 + 1

) dydx

= 0

2xy = M(x, y)

x2 + 1 = N(x, y)

12

First we must see if they are exact

∂M

∂y= 2x

∂N

∂x= 2x

∂M

∂y=∂N

∂x

x2y + y = C

∂Ψ(x, y)

∂x= 2xy∫

∂Ψ(x, y)

∂xdx =

∫2xy dx

Ψ(x, y) = x2y + h(y)

Now take the derivative with respect to y and compare it to the known ∂Ψ∂y

to find h(y)

∂Ψ

∂y= x2 + h′(y)

��x2 + h′(y) =��x

2 + 1∫h′(y) dy =

∫1 dy

h(y) = y + c1

Ψ(x, y) = x2y + y + c1

x2y + y = C

Example 2.7

(x2 + y

)+ (x+ cos(y))

dy

dx= 0

M(x, y) = x2 + y

N(x, y) = x+ cos(y)

∂M

∂x= 2x

∂N

∂x=

13

Example 2.8

2xy + (x2 + 1)dy

dx= 0

∂M

∂y= 2x,

∂N

∂x= 2x

14

Example 2.9

x2 dy

dx= x3 − 2xy

y(1) = 3

M(x, y) = −x3 + 2xy

N(x, y) = x2

∂M

∂y= 2x

∂N

∂x= 2x

It is exact, so there must be a solution of the form:

Ψ(x, y) = C

∂Ψ

∂x+∂Ψ

∂y

dy

dx= 0

∂Ψ

∂x= −x3 + 2xy

∂Ψ

∂y= x2

Integrate:∫∂Ψ

∂y=

∫x2 dy

Ψ(x, y) = x2y + h(x)

∂Ψ

∂x= 2xy + h′(x)

−x3 + 2xy = 2xy + h′(x)

h′(x) = −x3

h(x) =

∫−x3 dx

= −x4

4+ C

Ψ(x, y) = x2y − x4

4+ C

x2y − x4

4= C

15

y(1) = 3

12 · 3− 14

4= C

3− 1

4= C

C = 11/4

x2y − x4

4= 11/4

Example 2.10

(x2 + y) + (x− sin(y))dy

dx= 0

M = (x2 + y), N = (x− sin(y))

∂M

∂y= 1

∂N

∂x= 1

It is exact

2.4 Integrating Factor

Example 2.11

x2 + y

x+x− sin(y)

x

dy

dx= 0

M =x2 + y

x, N =

x− sin(y)

x∂M

∂y= 1/x

∂N

∂x=

sin(y)

x2

It is not exact

Both examples have the same solution, but one is exact and the other is not. Example 2.11 is

simply Example 2.10 divided by x.

16

• “Key term” (x) has been divided out!

• Key term → Integrating factor.

y′ + p(x)y = g(x) (2.14)

The Integrating factor is

µ(x) = e∫p(x) dx

µ′(x) = e∫p(x) dx · d

dx

∫p(x) dx

µ′(x) = e∫p(x) dx · p(x)

µ′(x) = µ(x) p(x)

µ(x) y′ + µ(x) p(x) y = µ(x) g(x)

µ(x) y′ + µ′(x) y = µ(x) g(x)∫d

dx(µ(x) y) =

∫µ(x) g(x) dx

µ(x) y =

∫µ(x) g(x) dx

y =1

µ(x)

∫µ(x) g(x) dx

Example 2.12

Solve y′ + 2y = ex

p(x) = 2, g(x) = ex

µ(x) = e∫

2 dx = e2x

y =1

e2x

∫e2x ex dx

= e−2x

∫e3x dx

= e−2x

(1

3e3x + c

)=

1

3ex + c e−2x

17

Example 2.13

dy

dx= 4− 2y

xdy

dx+

2y

x= 4

p(x) =2

x, g(x) = 4

µ(x) = e∫p(x) dx

= e2 lnx

ln(ab) = b ln(a)

= elnx2

= x2

y =1

x2

∫4x2

= x−2

(4

3x3 + c

)y(x) =

4

3x+ c x−2

Example 2.14 dydx

+ 2xy = 1,

y(1) = 2

p(x) = 2x, g(x) = 1

µ(x) = e∫p(x) dx

= ex2

y =1

ex2

∫ex

2

dx

You can’t integrate ex2

18

Example 2.15

x2y3 + x(1 + y2)y′ = 0

My = 3x2y2, Nx = 1 + y2

My 6= Nx Not exact

µ(x, y) =1

xy3

x2y3

xy3+x(1 + y2)

xy3y′ = 0

x+

(1

y3+

1

y

)y′ = 0

My = 0, Nx = 0

My = Nx Exact!∫M dx =

∫x dx =

x2

2+ C∫

N dy =

∫ (y−3 + y−1

)dy

= − 1

2y2+ ln |y|+ C

x2

2− 1

2y2+ ln |y| = C

Theorem 2.16

IfNx −My

M= Q(y), then

M +Ny′ = 0

has an integrating factor µ(y) = e∫Q(y) dy

19

µM + µNy′ = 0

e∫Q(y) dyM + e

∫Q(y) dyNy′ = 0

My = e∫Q(y) dyM

Nx = e∫Q(y) dyN

My = Nx

My =∂

∂y

(e∫Q(y) dyM

)= e

∫Q(y) dy ·Q(y)M + e

∫Q(y) dy ·My

Nx =∂

∂x

(e∫Q(y) dyN

)= e

∫Q(y) dy ·Nx

Nx −My = MQ(y)

Nx = My +MQ(y)

Nx = e∫Q(y) dy (My +MQ(y))

= e∫Q(y) dy ·My +MQ(y)e

∫Q(y) dy = My

My = Nx, Exact!

20

Example 2.17

y +(2xy − e−2y

)y′ = 0

N(x, y) = 2xy − e−2y

M(x, y) = y

Nx = 2y, My = 1

Using Theorem 2.16:

Nx −My

M=

2y − 1

y= 2− 1

y

Q(y) = 2− 1

y

µ(y) = e∫

2− 1ydy

= e2y−ln |y|

= e2y · e− ln |y|

= e2y · eln | 1y|

=1

ye2y

1

ye2y[y +

(2xy − e−2y

)y′]

= 0

e2y +1

ye2y(2xy − e−2y

)y′ = 0

e2y +

(2xe2y − 1

y

)y′ = 0

M = e2y

N =

(2xe2y − 1

y

)My = 2e2y

Nx = 2e2y

My = Nx Exact!∫M dx = xe2y∫N dy = xe2y − ln |y|

xe2y − ln |y| = C

21

2.5 Modeling with First Order Equations

Example 2.18 Mixing Problem

At time t = 0, a tank contains Q0 lb of salt dissolved in 100 gal of water. Assume that water

containing 1/4 lb of salt/gal is entering the tank at a rate of r gal/min and that the well-stirred

mixture is draining from the tank at the same rate. Set up the initial value problem that describes

this flow process. Find the amount of salt Q(t) in the tank at any time, and also find the limiting

amount QL that is present after a very long time. If r = 3 and Q0 = 2QL, find the time T after

which the salt level is within 2% of QL. Also find the flow rate that is required if the value of T is

not to exceed 45 min.

Let Q(t) be the amount of salt at time t

dQ

dt= Rate in − Rate out

Rate in = (Amount)× (Rate)

Amount = 1/4 lb/gal

Rate = r gal/min

Rate out: Amount : Q [lb]

100 gal

Rate : r [gal/min]

22

dQ

dt=

(1

4

lb

gal

)(rgal

min

)−(Q

100

lb

gal

)(rgal

min

)=

1

4r − Q

100r

dQ

dt+

Q

100r =

1

4r

µ(t) = e∫

r100

dt

= ert100

Q(0) = Q0

Example 2.19 Example 3 from Page 57 in the textbook.

Let Q(t) be the amount of chemical in the pond at any time t.

dQ

dt= Rate in − Rate out

= (2 + sin(2t))g

gal

(5

gal

year

)− Q

10

g

gal

(5

gal

year

)= (10 + 5 sin(2t))

g

year− Q

2

g

year

= 10 + 5 sin(2t)− Q

2

Q(0) = 0

p(t) =1

2

g(t) = 10 + 5 sin(2t)

µ(t) = e∫p(t) dt = e

t/2

Q = e−t/2

∫et/2 · (10 + 5 sin(2t)) dt

= e−t/2

[10

∫et/2 dt+ 5

∫et/2 sin(2t) dt

]

23

Integration by parts:

f(t) =

∫et/2 sin(2t) dt

v = et/2 dv = sin(2t) dt

dv = 1/2 et/2 dt v = −1/2 cos(2t)

= −1/2 et/2 cos(2t) + 1/4

∫et/2 cos(2t) dt

g(t) = 1/4

∫et/2 cos(2t) dt

v = et/2 dv = cos(2t) dt

du = 1/2 et/2 v = 1/2 sin(2t)

= 1/4

[1/2 e

t/2 sin(2t)− 1/4

∫et/2 sin(2t) dt

]I(t) =

∫et/2 sin(2t) dt

v = et/2 dv = sin(2t) dt

du = 1/2 et/2 v = −1/2 cos(2t)

= −1/2 et/2 cos(2t) + 1/4

∫et/2 cos(2t) dt

= −1/2 et/2 cos(2t) + 1/8 e

t/2 sin(2t)− I

1617

16I = −1/2 e−

t/2 cos(2t) + 1/8 et/2 sin(2t) + e

Now we can substitute I(t) back into g(t) and then f(t) to get our final solution.

Example 2.20 Page 59, Problem 3

A tank originally contains 100 gal of fresh water. Then water containing 1/2 lb of salt per gallon is

poured into the tank at a rate of 2 gal/min, and the mixture is allowed to leave at the same rate.

After 10 min the process is stopped, and fresh water is poured into the tank at a rate of 2 gal/min,

with the mixture again leaving at the same rate. Find the amount of salt in the tank at the end of

an additional 10 min.

24

dQ

dt= Rate in − Rate out

Q(0) = 0

dQ

dt= 1

lb

min− Q

50

lb

mindQ

dt= 1− Q

50

1 =dQ

dt+Q

50

µ(t) = e∫

1/50 dt = et/50

Q(t) = e−t/50

∫et/50 · 1 dt

= e−t/50(50 e

t/50 + C)

= 50 + Ce−t/50

Q(0) = 0 = 50 + C

C = −50

Q(t) = 50− 50e−t/50

Q(10) = 50− 50e−10/50

≈ 9.06 lb

2.6 Bernoulli’s Equation

Bernoulli’s Equation is of the form

dy

dx+ p(x)y = g(x)yn (2.15)

This is a more general case of the form we’ve used for integrating factors, where n was equal to 0

dy

dx+ p(x)y = g(x)�

�71

y0

25

Let y = v1

1−n ; yn = vn

1−n

dy

dx=v

11−n−1

1− ndv

dx=

vn

1−n

1− ndv

dx

vn

1−n

1− ndv

dx+ p(x)v

11−n = g(x)v

n1−n

Now divide everything byn

1− n

{v

11−n

vn

1−n= v

11−n− n

1−n = v1−n1−n = v1 = v

dv

dx+ (1− n)p(x)v = g(x)(1− n)

dy

dx+ p(x)y = g(x)yn

Example 2.21

Solvedy

dx+ 3y = exy2

This is a Bernoulli with n = 2

y = v1

1−2 = v−1 ⇒ y =1

vdv

dx+ (1− 2)p(x)v = g(x)(1− 2)

dv

dx− p(x)v = −g(x)

dv

dx− 3v = −ex

µ(x) = e∫−3 dx = e−3x

v(x) = e3x

∫e−3x · (−1)ex dx

= −e3x

∫e−3xex dx

= −e3x

∫e−2x dx

= −e3x

(−e−2x

2+ c1

)=ex

2− c1e

3x

=ex

2+ Ce3x

26

but y =1

v= v−1

so y(x) =

(ex

2+ Ce3x

)−1

y = 0 is a solution as well

2.7 Homogenous Equations

wrote it in my notebook

2.8 Second Order Linear Equations

of the formd2y

dx2= f(x, y, y′) or p(x)y′′ + q(x)y′ + r(x)y = g(x)

Example 2.22 Solve y′′ − y = 0 or y′′ = y.

{y1(x) = ex, y2(x) = e−x

y(x) = c1y1(x) + c2y2(x)

= c1ex + c2e

−x

y′(x) = c1ex − c2e

−x

y′′(x) = c1ex + c2e

−x = y(x)

So y(x) = c1ex + c2e

−x is the general solution.

27

If we have ay′′+by′+cy = 0 where a, b, and c are constants, then y = emx is a solution if y′ = memx,

y′′ = m2emx

am2emx + bmemx + cemx = 0

emx[am2 + bm+ c

]= 0

So am2 + bm+ c = 0 (characteristic equation)

There are three cases for characteristic equations

1. Two distinct roots r1 6= r2

y(x) = c1er1x + c2e

r2x

2. Complex roots m = α± βi

y(x) = c1eαx cos(βx) + c2e

αx sin(βx)

3. Repeated roots r1 = r2 = r

y(x) = c1erx + c2xe

rx

A second order linear equation

y′′ + p(x)y′ + q(x)y = g(x)

is called homogenous if g(x) = 0 and inhomogenous if g(x) 6= 0.

If r1 and r2 are two distinct roots of a characteristic equation, then the general solution is given by

y(x) = c1er1x + c2e

r2x

28

Example 2.23 Solve y′′ + 3y′ − 10y = 0

emx[m2 + 3m− 10

]= 0

m2 + 3m− 10 = 0

(m+ 5)(m− 2) = 0

m = −5, m = 2

y(x) = c1e−5x + c2e

2x

For an IVP you would need y(x0) = y0 and y′(x0) = y0

Example 2.24

I.V.P.

y′′ + 4y′ + 3y = 0

y(0) = 1, y′(0) = 1

m2 + 4m+ 3 = 0

(x+ 1)(x+ 3) = 0

y(x) = c1e−x + c2e

−3x

y′(x) = −c1e−x − 3c2e

−3x

y(0) = 1 = c1 + c2

y′(0) = 1 = −c1 − 3c2 1 1 1

−1 −3 1

∼1 0 2

0 1 −1

y(x) = 2e−x − e−3x

29

Example 2.25 Solve x2 + 2x+ 2 = 0

x =−2±

√22 − 4(1)(2)

2(1)

= −1± i

Example 2.26 Solve y′′ + 2y′ + 2y = 0

m2 + 2m+ 2 = 0

m = −1± i

r1 = −1 + i

r2 = −1− i

y(x) = c1e(−1+i)x + c2e

(−1−i)x

= c1e−x+xi + c2e

−x−xi

= c1e−x [cos(x) + i sin(x)] + c2e

−x [cos(−x) + i sin(−x)]− sin(θ) = sin(θ)

cos(θ) = cos(−θ)

= c1e−x cos(x) + ic1e

−x sin(x) + c2e−x cos(x)− ic2e

−x sin(x)

= (c1 + c2)e−x cos(x) + (c1 − c2)ie−x sin(x)

A = c1 + c2

B = i(c1 − c2)

y(x) = Ae−x cos(x) +Be−x sin(x)

So, if α± iβ are the roots, the general solution is

y(x) = Aeαx cos(βx) +Beαx sin(βx)

Example 2.27 The roots are: m = −2± 7i

30

Then the general solution must be

y(x) = Ae−2x cos(7x) +Be−2x sin(7x) = e−2x [A cos(7x) +B sin(7x)]

Example 2.28 y′′ + 14y′ + 149y = 0

y(0) = −10, y′(0) = 8

Example 2.29 y′′ + 4y′ + 13y = 0

y(0) = 0, y′(0) = 7

For repeated roots, r1 = r2, so y(x) = er1x = er2x

Example 2.30 y′′ + 4y′ + 4y = 0

m2 + 4m+ 4 = 0

(m+ 2)2 = 0, m = −2

y(x) = v(x)e−2x

y′(x) = v′(x)e−2x − 2v(x)e−2x

= e−2x [v′(x)− 2v(x)]

y′′(x) = v′′(x)e−2x − 2v′(x)e−2x + 4v(x)e−2x − 2v′(x)e−2x

= e−2x [v′′(x)− 4v′(x) + 4v(x)]

e−2x [v′′(x)− 4v′(x) + 4v(x)] + 4e−2x [v′(x)− 2v(x)] + 4e−2xv(x) = 0

e−2x [v′′(x)− 4v′(x) + 4v(x) + 4v′(x)− 8v(x) + 4v(x)] = 0

31

e−2xv′′(x) = 0

v′ = c1

v =

∫c1 dx = c1x+ c2

v(x) = c1x+ c2

y(x) = (c1x+ c2) e−2x

y(x) = c2e−2x + c1xe

−2x

y′(x) = −2c2e−2x + c1e

−2x − 2c1xe−2x

xe−2x e−2x

e−2x − 2xe−2x −2e−2x

c1

c2

If r1 = r2 = r, the general solution is y(x) = c1e

rx + c2xerx

Example 2.31 16y′′ − 40y′ + 25y = 0

y(0) = 3, y′(0) = −9/4

16m2 − 40m+ 25 = 0

(4m− 5)2 = 0

m = 5/4

32

y(x) = c1e5x/4 + c2xe

5x/4

y′(x) = 5/4 c1e5x/4 + c2e

5x/4 + 5/4 c2xe5x/4

y(0) = 3 = c1 + 0

y′(0) = −9/4 = 5/4 c1 + c2 + 0

c1 = 3

c2 = −9/4− 5·3/4 = −6

y(x) = 3e5x/4 − 6xe5x/4

= 3e5x/4 (1− 2x)

Example 2.32 y′′ + 14y′ + 149y = 0

y(0) = −10, y′(0) = 8

m2 + 14m+ 149 = 0

m = −7± 10i

y(x) = c1e−7x cos(10x) + c2e

−7x sin(10x)

y′(x) = −7c1e−7x cos(10x)− 10c1e

−7x sin(10x)− 7c2e−7x sin(10x) + 10c2e

−7x cos(10x)

= (−7c1 + 10c2)e−7x cos(10x) + (−10c1 − 7c2)e−7x sin(10x)

Example 2.33 y′′ + 4y′ + 13y = 0

y(0) = 0, y′(0) = 7

33

m2 + 4m+ 13 = 0

m = −2± 3i

y(x) = c1e−2x cos(3x) + c2e

−2x sin(3x)

y′(x) = −2c1e−2x cos(3x)− 3c1e

−2x sin(3x)− 2c2e−2x sin(3x) + 3c2e

−2x cos(3x)

y(0) = 0 = c1

y′(0) = 7 = −2c1 + 3c2

1 0 1

−2 3 7

∼1 0 1

0 1 3

y(x) = e−2x cos(3x) + 3e−2x sin(3x)

MISSING NOTES GO HERE

3 Transformations

3.1 Laplace Transform

The Laplace transform of f(t) is denoted by

L{f(t)} =

∫ ∞0

f(t)e−st dt = F (s) (3.1)

34

Example 3.1 Compute L{eat}

F (s) =

∫ ∞0

eate−st dt

=

∫ ∞0

e(a−s)t dt

=

[1

a− se(a−s)t

]∞0

=1

a− s[e−∞ − e0

], s > a

= − 1

a− s=

1

s− a, s > a

Example 3.2 Compute L{sin(at)}

F (s) =

∫ ∞0

sin(at)e−st dt

IBP

u = e−st, du = −se−st

v = − cos(at)a

, dv = sin(at) dt

F (s) = −cos(at)e−st

a− s

a

∫ ∞0

cos(at)e−st dtu = e−st, du = −se−st dt

v = sin(at)a

, dv = cos(at) dt

F (s) = −cos(at)e−st

a− s

a

(sin(at)e−st

a+s

a

∫ ∞0

sin(at)e−st dt

)= −cos(at)e−st

a− s

a

(sin(at)e−st

a+s

aF (s)

)= −cos(at)e−st

a− s

a2sin(at)e−st − s2

a2F (s)

F (s)

(1 +

s2

a2

)= −e

−st

a

(cos(at) +

s

asin(at)

)F (s) =

[−e−st

a

(cos(at) +

s

asin(at)

) a2

a2 + s2

]∞0

= 0−(−1

a(1 + 0)

)a2

a2 + s2, s > 0

=a

a2 + s2, s > 0

35

Properties

1. L{f(t) + g(t)} = L{f(t)}+ L{g(t)}

2. L{c f(t)} = c L{f(t)}

Example 3.3 Compute L{3 sin(2t)}

L{3 sin(2t)} = 3L{sin(2t)}

= 32

s2 + 4, s > 0

=6

s2 + 4, s > 0

Example 3.4 Compute L{et − e2t}

L{et − e2t} = L{et} − L{e2t}

=1

s− 1− 1

s− 2, s > 2

= − 1

(s− 2)(s− 1), s > 2

3.2 Inverse Laplace Transform

L{f(t)} = F (s)

L−1 {L{f(t)}} = L−1 {F (s)} (3.2)

Properties

1. L−1{F (s) +G(s)} = L−1{F (s)}+ L−1{G(s)}

2. L−1{c F (s)} = c L−1{F (s)}

36

Example 3.5 Compute L−1

{3

s

}

L−1

{3

s

}= 3L−1

{1

s

}= 3(1) = 3

Example 3.6

L−1

{4

s+ 1

}= 4L−1

{1

s− (−1)

}= 4e−t

Example 3.7

L−1

{2

s2 + 3s+ 2

}= 2L−1

{1

(s+ 1)(s+ 2)

}A

s+ 1+

B

s+ 2=

2

(s+ 1)(s+ 2)

A(s+ 2) +B(s+ 1) = 2

Set s = −2

B = −2, A = 2

f(s) = 2L−1

{1

s− (−1)

}+ (−2)L−1

{1

s− (−2)

}= 2e−t − 2e−2t

Do for homework:

37

Example 3.8

f(t) = L−1

{2s+ 4

s2 + 2s+ 5

}s2 + 2s+ 5 = s2 + 2s+ 1 + 4

= (s+ 1)2 + 4

= (s+ 1)2 + 22

f(t) = L−1

{2s+ 4

(s+ 1)2 + 22

}From the table:

eat sin(bt) = L−1

{b

(s− a)2 + b2

}, s > a

eat cos(bt) = L−1

{s− a

(s− a)2 + b2

}, s > a

f(t) = L−1

{2(s+ 1) + 2

(s+ 1)2 + 22

}= 2L−1

{s+ 1

(s+ 1)2 + 22

}(a = −1, b = 2)

+ L−1

{2

(s+ 1)2 + 22

}(a = −1, b = 2)

= 2e−t cos(2t) + e−t sin(2t)

Theorem 3.9

L{f ′(t)} = sL{f(t)} − f(0)

38

Proof

L{f ′(t)} =

∫ ∞0

f ′(t)e−st dtu = e−st, du = −se−st dt

v = f(t), dv = f ′(t) dt

=[e−stf(t)

]∞0−∫ ∞

0

(−s)e−stf(t) dt

=

(���

����:0

e−s·∞f(∞)− e−s·0f(0)

)+ s

∫ ∞0

f(t)e−st dt

= 0− f(0) + 5L{f(t)}

Example 3.10

L{f ′′(t)} = L{

[f ′(t)]′}

= sL{f ′(t)} − f ′(0)

= s (sL{f(t)} − f(0))− f ′(0)

= s2 L{f(t)} − sf(0)− f ′(0)

Example 3.11

L{f ′′′(t)} = L{[f ′′(t)]′}

= sL{f ′′(t)} − f ′′(0)

= s[s2 L{f(t)} − sf(0)− f ′(0)

]− f ′′(0)

= s3 L{f(t)} − s2f(0)− sf ′(0)− f ′′(0)

1. L{f ′(t)} = sL{f(t)} − f(0)

2. L{f ′′(t)} = s2 L{f(t)} − sf(0)− f ′(0)

3. L{f ′′′(t)} = s3 L{f(t)} − s2f(0)− sf ′(0)− f ′′(0)

39

4. L{f (n)(t)} = sn L{f(t)} − sn−1f(0)− sn−2f ′(0)− . . .− f (n−1)(0)

= sn L{f(t) +∑n

i=1−sn−if i−1(0)

Example 3.12

Solve

y′ = y

y(0) = 1

or

dydx

= y

y(0) = 1

∫1

ydy =

∫dx

ln |y| = (x+ c1)

|y| = ex+c1

= ec1ex

y = ±ec1ex

= Cex

y = Cex

but y(0) = 1

1 = Ce0

C = 1

y = ex

40

Now let’s do it again using the Laplace transform

y′(x) = y(x)

L{y′(x)} = L{y(x)}

sL{y(x)} − y(0) = L{y(x)}

sL{y(x)} − L{y(x)} = y(0)

L{y(x)}(s− 1) = 1

L{y(x)} =1

s− 1

L−1{L{y(x)}} = L−1

{1

s− 1

}, a = 1

y(x) = ex

Example 3.13

Solve

y′′ + 2y′ + 5y = 0

y(0) = 2, y′(0) = 0

L{y′′}+ 2L{y′}+ 5L{y} = L{0}

s2 L{y} − sy(0)− y′(0) + 2 (sL{y} − y(0))5 L{y} = 0

L{y}(s2 + 2s+ 5

)= 2s+ 4

L{y} =2s+ 4

s2 + 2s+ 5

L−1 {L{y}} = L−1

{2s+ 4

s2 + 2s+ 5

}y(x) = 2e−x cos(2x) + e−x sin(2x)

Example 3.14

Solve: y′′ + y = sin(2t), y(0) = 2, y′(0) = 1

41

L{y′′} − L{y} = L{sin(2t)}

s2 L{y} − sy(0)− y′(0) + L{y} =2

s2 + 4, s > 0

s2 L{y} − 2s− 1 + L{y} =2

s2 + 4

L{y} (s2 + 1)− 2s− 1 =1

s2 + 4

L{y} (s2 + 1) =1

s2 + 4+ 2s+ 1

=2 + 2s3 + 8s+ s2 + 4

s2 + 4

L{y} =2s3 + s2 + 8s+ 6

(s2 + 4)(s2 + 1)

y = L−1

{2s3 + s2 + 8s+ 6

(s2 + 4)(s2 + 1)

}2s3 + s2 + 8s+ 6

(s2 + 4)(s2 + 1)=as+ b

s2 + 1+cs+ d

s2 + 4

2s3 + s2 + 8s+ 6 = (as+ b)(s2 + 4) + (cs+ d)(s2 + 1)

= as3 + bs2 + 4as+ 4b+ cs3 + ds2 + cs+ d

= s3(a+ c) + s2(b+ d) + s(4a+ c) + (4b+ d)

a+ c = 2

b+ d = 1

4a+ c = 8

4b+ d = 6

1 0 1 0 2

0 1 0 1 1

4 0 1 0 8

0 4 0 1 6

∼

1 0 0 0 2

0 1 0 0 5/3

0 0 1 0 0

0 0 0 1 −2/3

2s3 + s2 + 8s+ 6 = 2s3 + s2(5/3− 2/3) + 8s+ 20/3− 2/3

42

3.3 Unit Step-function

Definition The function u(t) is called the unit step-function.

u(t) =

0, t < 0

1, t > 0

(3.3)

uc(t) is a transformation of the unit step-function

uc(t) = u(t− c) =

0, t− c < 0

1, t− c > 0

(3.4)

43

Example 3.15 For a < b, find u(t− a)− u(t− b).

u(t− a)− u(t− b) =

0, (t < a) ∪ (t > b)

1, a < t < b

Example 3.16

Draw f(t) =

3, t < 2

1, 2 < t < 5

t, 5 < t < 8

t2/10, t > 8

Theorem 3.17 For C ≥ 0

L{uc(t)} = L{u(t− c)} =e−cs

s, s > 0

44

Proof

L{uc(t)} =

∫ ∞0

uc(t)e−st dt

Recall uc(t) =

0, t < c

1, t > c

=

∫ c

0���*

0uc(t)e

−st dt+

∫ ∞c���*1

uc(t)e−st dt

= 0 +

∫ ∞c

e−st dt

=

[−e−st

s

]∞c

c = −1

s

[0− e−sc

]=e−sc

s, s > 0

Example 3.18

L{uc(t)f(t− c)} = e−csF (s), s > 0

=

∫ ∞0

uc(t)f(t− c)e−st dt

=

∫ c

0���*

0uc(t)f(t− c)e−st dt+

∫ ∞c���*

1uc(t)f(t− c)e−st dt

=

∫ ∞c

f(t− c)e−st dt

Let v = t− c

=

∫ ∞0

f(v)e−s(v+c) dv

=

∫ ∞0

f(v)e−sve−sc dv

= e−sc∫ ∞

0

f(v)e−sv dv

= e−sc L{f(v)}

= e−scF (s)

45

Example 3.19 For homework:

L{t2u(t− 1)

}= L

{t2u1(t)

}

MISSING NOTES HERE

Example 3.20

F (s) = L{∫ t

0

e−(t−τ) sin τ dτ

}= L

{e−t ∗ sin t

}= L

{e−t}· L {sin t}

=

(1

s+ 1

)(1

s2 + 1

)=

1

(s+ 1)(s2 + 1), s > 0

Example 3.21

F (s) = L{∫ t

0

sin(t− τ) cos τ dτ

}= L{sin t ∗ cos t}

=

(1

s2 + 1

)(s

s2 + 1

)=

s

(s2 + 1)2, s > 0

46

Example 3.22

f(t) = L−1

{1

s2(s2 + 1)

}L−1

{1

s2

}= t

L−1

{1

s2 + 1

}= sin t

L{t} =1

s2

L{sin t} =1

s2 + 1

L{t}L {sin t} =

(1

s2

)(1

s2 + 1

)L{t ∗ sin t} =

1

s2(s2 + 1)

t ∗ sin t = L−1

{1

s2(s2 + 1)

}=

∫ t

0

(t− τ) sin τ dτu = t− τ, u′ = −dτ

v = − cos τ, v′ = sin τ

= [− cos τ(t− τ)]t0 −∫ t

0

cos τ dτ

= t− [sin τ ]t0 = t− sin t

Example 3.23

L−1

{G(s)

s2 + 1

}F (s) =

1

s2 + 1

f(t) = sin t

sin t ∗ g(t) = L−1

{G(s)

s2 + 1

}=

∫ t

0

sin(t− τ)g(τ) dτ

47

Example 3.24

y′′ + 4y = g(t); y(0) = 3, y′(0) = −1

L{y′′}+ L{4y} = L{g(t)}

s2 L{y} − 3s+ 1 + 4L{y} = G(s)

L{y} (s2 + 4)− 3s+ 1 = G(s)

L{y} =G(s) + 3s− 1

s2 + 4

Let F (s) =1

s2 + 4

f(t) =1

2sin(2t)

L−1 {G(s)F (s)} =1

2sin(2t) ∗ g(t)

y =1

2sin(2t) ∗ g(t) + L−1

{3s

s2 + 4− 1

s2 + 4

}=

1

2sin(2t) ∗ g(t) + 3 cos(2t)− 1

2sin(2t)

=1

2

∫ t

0

g(t− τ) sin(2τ) dτ + 3 cos(2t)− 1

2sin(2t)

3.4 Gamma Function

Γ(a+ 1) =

∫ ∞0

tae−t dt, a > −1 (3.5)

Theorem 3.25

Γ(a+ 1) = aΓ(a)

48

Example 3.26

Γ(4) = 3Γ(3)

= 3 · 2Γ(2)

= 3 · 2 · 1Γ(1)

= 3 · 2 · 1 · 1 = 6 = 3!

So for positive integers, Γ(a+ 1) = a!

Proof

Γ(a+ 1) =

∫ ∞0

tae−t dtu = ta, u′ = a ta−1

v = −e−t, v′ = e−t

=������:

0[−tae−t

]∞0

+ a

∫ ∞0

ta−1e−t dt

= 0 + aΓ(a)

49

Example 3.27

Γ(1) = Γ(0 + 1)⇒ a = 0

=

∫ ∞0

t0e−t dt

=

∫ ∞0

e−t dt

=[−e−t

]∞0

= 1

Γ(3) = Γ(2 + 1)

= 2Γ(2)

= 2 · 1 = 2

Γ(4) = Γ(3 + 1)

= 3Γ(2 + 1)

= 3 · 2Γ(1 + 1)

= 3!

Γ(n+ 1) = n!, n ∈ Z+

50

Proof

Γ

(1

2

)= Γ

(−1

2+ 1

)=

∫ ∞0

t−1/2e−t dt[

Γ

(1

2

)]2

=

(∫ ∞0

t−1/2e−t dt

)2

Let u2 = t, 2u du = dt[Γ

(1

2

)]2

=

(∫ ∞0

u−1e−u2 · 2u du

)2

=

(2

∫ ∞0

e−u2

du

)2

=

(∫ ∞−∞

e−u2

du

)2

=

(∫ ∞−∞

e−x2

dx

)(∫ ∞−∞

e−y2

dy

)=

∫ ∞−∞

∫ ∞−∞

e−(x2+y2) dx dy

In cylindrical coordinates: x2 + y2 = r2

=

∫ 2π

0

∫ ∞0

e−r2

r dr dθ

(Let w = r2, dw = 2r dr)

=1

2

∫ 2π

0

∫ ∞0

e−w dw dθ

=1

2

∫ 2π

0

1 dθ[Γ

(1

2

)]2

=1

2(2π) = π

Γ

(1

2

)=√π

51

Example 3.28

Γ

(3

2

)= Γ

(1

2+ 1

)=

1

2Γ

(1

2

)=

√π

2

Γ

(7

2

)= Γ

(5

2+ 1

)=

5

2Γ

(3

2+ 1

)=

5

2· 3

2Γ

(1

2+ 1

)=

15

4· 1

2Γ

(1

2

)=

15√π

8

Example 3.29

Γ

(−5

3

)= Γ

(−8

3+ 1

)= −8

3Γ

(−11

3+ 1

)=

88

9Γ

(−14

3+ 1

)

Notice that it’s just getting more and more complicated. So we need to do the opposite of what we’ve

52

been doing

−5

3Γ

(−5

3

)= Γ

(−5

3+ 1

)= Γ

(−2

3

)Γ

(−5

3

)= −3

5Γ

(−2

3

)−2

3Γ

(−2

3

)= Γ

(−2

3+ 1

)= Γ

(1

3

)Γ

(−2

3

)= −3

2Γ

(1

3

)Γ

(−5

3

)=

9

10Γ

(1

3

)

Example 3.30

Γ

(−13

7

)= − 7

13Γ

(−6

7

)=

49

78Γ

(1

7

)

4 Systems of Differential Equations

Example 4.1

dxdt

= 3x+ 5y

dydt

= 3x+ y

x(0) = 1, y(0) = 2

53

L{x′(t)} = 3L{x}+ 5L{y}

L {y′(t)} = 3L{x}+ L{y}sL{x} −��

�* 1x(0) = 3L{x}+ 5L{y}

sL{y} −���*2

y(0) = 3L{x}+ L{y}(s− 3)L{x} − 5L{y} = 1

−3L{x}+ (s− 1)L{y} = 2(s− 3) −5

−3 (s− 1)

L{x}L {y}

=

1

2

L{x}L {y}

=

(s− 3) −5

−3 (s− 1)

−1 1

2

=

1

s2 − 4s− 12

(s− 1) 5

3 (s− 3)

1

2

=

1

s2 − 4s− 12

(s+ 9)

(2s− 3)

L{x} =

s+ 9

s2 − 4s− 12

L{y} =2s− 3

s2 − 4s− 12

x =1

8e−2t

(15e8t − 7

)y =

1

8e−2t

(9e8t + 7

)

Example 4.2 x′(t) = 3x− 5y

y′(t) = x+ 5y

x(0) = −3, y(0) = 4

54

L{x′(t)} = 3L{x} − 5L{y}

L {y′(t)} = L{x}+ 5L{y}sL{x} −��

�*−3x(0) = 3L{x} − 5L{y}

sL{y} −���*4

y(0) = L{x}+ 5L{y}(s− 3)L{x}+ 5L{y} = −3

−L{x}+ (s− 5)L{y} = 4(s− 3) 5

−1 (s− 5)

L{x}L {y}

=

−3

4

L{x}L {y}

=1

s2 − 8s+ 20

(s− 5) −5

1 (s− 3)

−3

4

=

1

s2 − 8s+ 20

−3s− 5

4s− 15

L{x} =

−3s− 5

s2 − 8s+ 20

L{y} =4s− 15

s2 − 8s+ 20

x = −1

2e4t(17 sin(2t) + 6 cos(2t))

y =1

2e4t(sin(2t) + 8 cos(2t))

Example 4.3 ∫ex

2

dx

We can’t integrate this using the techniques we learned in Calculus, but we can approximate it as a

power series

f(x) =∞∑n=0

an(x− a)n, an =f (n)(a)

n!

So we can write f(x) as

f(x) = a0 + a1(x− a) + a2(x− a)2 + a3(x− a)3 + . . .+ an(x− a)n

55

Example 4.4

y′′ − xy = 0, y(0) = 1, y′(0) = 2

y(x) =∞∑n=0

anxn

= y(0) +y′(0)

1!x+

y′′(0)

2!x2 +

y′′′(0)

3!x3 + . . .

Now, we know y(0) and y′(0), but nothing higher, so first, let’s plug in our initial conditions to the

original equation to find y′′(0)

y′′(x)− xy(x) = 0

y′′(0)− 0 ·���*1

y(0) = 0

y′′(0) = 0

Now let’s take the derivative of the equation, and plug in our initial conditions to find y′′′(0)

y′′′(x)− y(x)− xy′(x) = 0

y′′′(0)−���*1

y(0)− 0 ·���*2

y′(0) = 0

y′′′(0) = 1

Now let’s do it for the fourth derivative

y(4)(x)− y′(x)− y′(x)− xy′′(x) = 0

y(4)(x)− 2y′(x)− xy′′(x) = 0

y(4)(0)− 2���*2

y′(0)− 0 ·����*1

y′′(0) = 0

y(4)(0) = 4

56

We can keep going, but we’ll stop here with our approximation, giving us

y(x) = 1 + 2x+1

6x3 +

4

24x4 + . . .

Example 4.5

y′′ − x2x = 0, y(0) = 1, y′(0) = 0

y′(x) =∞∑n=1

n anxn−1

y′′(x) =∞∑n=2

n(n− 1)anxn−2

y′′(x)− x2y =∞∑n=2

n(n− 1)anxn−2 − x2

∞∑n=0

anxn

=∞∑n=2

n(n− 1)anxn−2 −

∞∑n=0

anxn+2 = 0

=∞∑n=0

(n+ 2)(n+ 1)an+2xn −

∞∑n=2

an−2xn = 0

= (2)(1)a2 + (3)(2)a3x+∞∑n=2

(n+ 2)(n+ 1)an+2xn −

∞∑n=2

an−2xn = 0

= 2a2 + 6a3x+∞∑n=2

xn[(n+ 2)(n+ 1)an+2 − an−2] = 0

2a2 = 0, 6a3 = 0, (n+ 2)(n+ 1)an+2 − an−2 = 0, n ≥ 2

an+2 =an−2

(n+ 2)(n+ 1), n ≥ 2

Example 4.6

Solve (−4x2 − 1)y′′ + 5xy′ = 0; y(0) = −3, y′(0) = −8

Recall∞∑n=0

f (n)(a)(x− a)n

n!=∞∑n=0

an(x− a)n

We already know y(0) and y′(0), so it would be simplest to let a = 0, as we need to know y(a) and

y′(a). We’re also going to need further derivatives, so lets find y′′(0) by evaluating the equation with

x = 0.

(−1)y′′(0) + 0 = 0⇒ y′′(0) = 0

57

Now we need the third derivative

−8xy′′ + (−4x2 − 1)y′′′ + 5y′ + 5xy′′ = 0

and evaluate at x = 0

0 + (−1)y′′′(0) + 5(−8) + 0 = 0⇒ y′′′(0) = −40

Now let’s use the power series

y(x) =y(0)

0!+y′(0)

1!x′ +

y′′(0)

2!x2 + . . .

= −3− 8x− 20

3x3 + . . .

Example 4.7

Solve (−3x+ 6)y′′ + 8y′ + (2x− 5)y = 0; y(0) = 5, y′(0) = 0

First, evaluate at x = 0 to find y′′(0)

6y′′(0) + 0− 5���*5

y′(0) + (−5)���*0

y(0) = 0⇒ y′′(0) =25

6

Now take the derivative and evaluate at x = 0 to find y′′′(0)

−3y′′(x) + (−3x+ 6)y′′′(x) + 8y′′(x) + 2y(x) + (2x− 5)y′(x) = 0

−3����*

25/6

y′′(0) + 6y′′′(0) + 8����*

25/6

y′′(0) + 2���*5

y(x)− 5���*0

y′(x) = 0

−75

6+ 6y′′′(0) +

100

3+ 10 = 0

y′′′(0) = −185

36

58

We can go further, but if we stop here, we have

y(x) = y(0) +y′(0)

1!x+

y′′(0)

2!x2 +

y′′′(0)

3!x3 + . . .

y(x) = 5 +25

12x2 − 185

216x3 + . . .

Example 4.8

Solve xy′′ + y′ + y = 0; y(0) = −1, y′(0) = 1

Radius of Convergence

Definition Let p(x), q(x), and r(x) be analytic functions. A point x0 is a singular point for the

differential equation

p(x)y′′(x) + q(x)y′(x) + r(x)y(x) = 0 (4.1)

ifq(x0)

p(x0)or

r(x0)

p(x0)is undefined. Otherwise, x0 is called a regular point.

Try solving the following: xy′′ + y′ + y = 0

y(0) = −1, y′(0) = 1

If we try evaluating at x = 0 to find y′′(0), we end up multiplying the y′′(0) by zero, so we can’t

get it by that method. We could rewrite this problem by dividing everyting by x

y′′ +1

xy′ +

1

xy = 0

Now we see that x = 0 is a singular point.

Find the singular points for the following equations:

x2y′′ + xy′ = 0⇒ x0 = 0

2y′′ +y

x− 1= 0⇒ x0 = 1

59

Theorem 4.9 Suppose

y(x) =∞∑n=0

an(x− x0)n

is a series solution do Equation 4.1. Then, the radius of convergence for y(x) is at least as large as

the distance from x0 (regular point) to the nearest singular point of the equation.

Example 4.10 Find the radius of convergence about x0

(x+ 1)y′′ + y′ + (x+ 1)y = 0, about x0 = 0

So first we divide by (x+ 1)

y′′ +y′

x+ 1+ y = 0

x = −1 is a singular point. So the radius of convergence is at least 1.

Example 4.11 Find the radius of convergence about x0

(x2 − 2x− 3)y′′ + (x+ 1)y′ + 3y = 0, about x0 = 0

First divide by (x+1), to get singular point x = −1. Then divide the original equation by (x2−2x−3)

to get

y′′ +���

�(x+ 1)y′

(x− 3)����(x+ 1)

+3y

(x− 3)(x+ 1)= 0

which has singular points x = −1 and x = 3. So the radius of convergence is at least 1.

Example 4.12

(x3 + x2 + x+ 1)y′′ + 2xy′ + (x− 1)y = 0, about x0 = 0

Factor the big polynomial first

x3 + x2 + x+ 1 = x2(x+ 1) + 1(x+ 1) = (x2 + 1)(x+ 1)

60

Now divide that out to get

y′′ +2xy′

(x2 + 1)(x+ 1)+

(x− 1)y

(x2 + 1)(x+ 1)= 0

x = −1, x = ± ι. Radius of convergence is at least√

2

61

5 Final Examples

Example 5.1

L{t cosh(t)}

cosh(t) =et + e−t

2

F (s) =1

2L{tet + te−t

}=

1

2L{tet}

+1

2L{te−t

}L{tneat

}, n ∈ Z+ =

n!

(s− a)n+1, s > a

F (s) =1

2

1!

(s− 1)1+1+

1

2

1!

(s+ 1)1+1, s > 1

=1

2

(1

(s− 1)2+

1

(s+ 1)2

), s > 1

Example 5.2

L{f(t)} , where f(t) = 3 cos(2t) +

∫ t

0

e2(t−τ)τ sin(3τ) dτ

L{f(t)} = 3L{cos(2t)}+ L{∫ t

0

e2(t−τ)τ sin(3τ) dτ

}= 3

s

s2 + 4+ L

{e2t ∗ t sin(3t)

}= 3

s

s2 + 4+ L

{e2t}L{t sin(3t)}

= 3s

s2 + 4+

1

s− 2

6s

(s2 + 9)2

= 3s

s2 + 4+

6s

(s− 2)(s2 + 9)2

62

Example 5.3

L−1

{2s+ 9

s2 + 8s+ 17

}

f(t) = L−1

{2s+ 9

s2 + 8s+ 16 + 1

}= L−1

{2s+ 9

(s+ 4)2 + 12

}= 2L−1

{s

(s+ 4)2 + 12

}+ 9L−1

{1

(s+ 4)2 + 12

}= 2L−1

{s+ 4− 4

(s+ 4)2 + 12

}+ 9L−1

{1

(s+ 4)2 + 12

}= 2L−1

{s+ 4

(s+ 4)2 + 12

}+ L−1

{1

(s+ 4)2 + 12

}= 2e−4t cos(t) + e−4t sin(t) = e−4t (2 cos(t) + sin(t))

Example 5.4

y′′ + 6y′ + 8y = δ(t), y(0) = 0, y′(0) = 0

s2 L{y}+ 6sL{y}+ 8L{y} = e−0s

L{y} (s2 + 6s+ 8) = 1

L{y} =1

s2 + 6s+ 8

=1

(s+ 2)(s+ 4)

1

(s+ 2)(s+ 4)=

A

s+ 2+

B

s+ 4

1 = A(s+ 4) +B(s+ 2)

A = 1/2, B = −1/2

L{y} =1

2

1

s+ 2− 1

2

1

s+ 4

y =1

2

(e−2t − e−4t

)

63

Example 5.5

y′′ + 2xy′ + 4y = 0, y(0) = 1, y′(0) = 2

Just use:∞∑n=0

y(n)(0)

n!xn

The solution is:

y(x) = 1 + 2x− 2x2 − 2x3 +4

3x4 + . . .

Example 5.6 Find the radius of convergence for the series solution about x0 = 0 to the equation:

(x2 + 2x+ 10)(x2 + 6x+ 8)y′′ + 2xy′ − 4y = 0

y′′ +2xy′

(x+ 2)(x+ 4)(x2 + 2x+ 10)− 4y

(x+ 2)(x+ 4)(x2 + 2x+ 10)= 0

You can actually just use the quadratic formula. The roots will be the singular points, so the singular

points are x = −1± 3i and x = −4, −2. The closest point is −1− 3i, and it is 2 away from 0, so

the radius of convergence is at least 2.

Example 5.7

y′′ + 3y′ + 2y =1

1 + ex= g

λ2 + 3λ+ 2 = 0

λ = −2, λ = −1

yh = c1e−2x + c2e

−x = y1 + y2

W =

∣∣∣∣∣∣∣y1 y2

y′1 y′2

∣∣∣∣∣∣∣yp = −y1

∫y2 · gW

+ y2

∫y1 · gW

64