Leo Creedon CE3 Predegree Mathematics These notes are in ... · Mathematics Leo Creedon CE3...

Mathematics

Leo Creedon

CE3 Predegree Mathematics

These notes are in draft format and do notexactly reflect the notes delivered in class

February 24, 2009

Contents

1 Mathematics Review Predegree Maths 21.1 Partial Fractions . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

2 First Order Differential Equations Predegree Maths 222.1 Separable Differential Equations . . . . . . . . . . . . . . . . . 232.2 First Order Linear Differential Equations (The Integrating Fac-

tor Method) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 302.3 Homogeneous Differential Equations . . . . . . . . . . . . . . 39

3 Laplace Transforms Predegree Maths 453.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 463.2 Proof by Induction . . . . . . . . . . . . . . . . . . . . . . . . . 473.3 Inverse Laplace Transforms . . . . . . . . . . . . . . . . . . . . 533.4 Solving Differential Equations using Laplace Transforms . . . 61

4 Second Order Differential Equations 714.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72

1

Chapter 1

Mathematics Review PredegreeMaths

2

CHAPTER 1. MATHEMATICS REVIEW PREDEGREE MATHS 3

Exercise 1.1 12 + 1

6 =

Exercise 1.2 12 −

13 =

Exercise 1.3 12.

16 =

Exercise 1.4 12 ÷

16 =

Exercise 1.5 ab + c

d =

Exercise 1.6 ab −

cd =

Exercise 1.7 ab .

cd =

Exercise 1.8 ab ÷

cd =

Exercise 1.9 (ab)c =

Exercise 1.10 (23)2 =

Exercise 1.11 2−3 =

Exercise 1.12 ln(ab) =



Exercise 1.15 ln(e) =

Exercise 1.16 ln(1) =


Exercise 1.17 eln x =

Exercise 1.18 ln(ex) =

Exercise 1.19 e0 =

Exercise 1.20 Draw a rough graph of y = ex:

Exercise 1.21 Draw a rough graph of y = ln(x):

Exercise 1.22 Draw a rough graph of y = x2:

Exercise 1.23 Draw a rough graph of y = x2 + 1:


Exercise 1.24 Draw a rough graph of y = (x + 1)2:

Exercise 1.25 ddxx

2 =

Exercise 1.26 ddx(3x6 − 10x5 + x− 1) =

Exercise 1.27 ddx(√

x) =

Exercise 1.28 ddx(5x4 + 3x2 + 10x + 2 +

√x + 3x

32) =

Exercise 1.29 ddx sin x =

Exercise 1.30 ddt(t

−12) =

Exercise 1.31 ddx(ln(x)) =


Exercise 1.32 ddx(x cos x) =

Exercise 1.33 ddx(sin x cos x) =

Exercise 1.34 ddx[(3x2 + 5x + 2)(cos x)] =

Exercise 1.35 ddx

(cos x2x

)=

Exercise 1.36 ddx

3x2−7x+25x2+1

=

Exercise 1.37 ddx sin(2x) =


Exercise 1.38 ddx

√3x =

Exercise 1.39 ddx[(cos x)100] =

Exercise 1.40 ddx[(3x2 + 5x + 2)−1] =

Exercise 1.41 ddx cos(3x2 + 2x− 10) =

Exercise 1.42 ddx2 tan(

√x) =

Exercise 1.43 ddx

√cos(2x) =

Exercise 1.44 Use implicit differentiation to find dydx, where

xy + sin y = 0:


Exercise 1.45 Use implicit differentiation to find dydx, where

y +√

y = x. Hence find the slope of the tangent line to the

curve when y = 4 (i.e. find dydx when y = 4).

Exercise 1.46 ddx ln(5x2 + 2) = ...

Exercise 1.47 Find dydx, where y = (10x2+1)100(2x−1)9

(x3+1)11

(i) by direct differentiation

(ii) by taking ln of both sides and then differentiating (log-

arathmic differentiation)


Review of Integration:∫ b

a f (x)dx = the area under the curve f (x), between a

and b. Also, if∫

f (x)dx = g(x) + C, (a function) thenddxg(x) = f (x). In fact

∫ b

a f (x)dx = [g(x)]ba = g(b)−g(a),

a number. This is called the Fundamental Theorem

of Calculus.

Exercise 1.48∫

[f (x) + g(x)]dx =

Exercise 1.49∫

kf (x)dx =

Exercise 1.50∫ a

b f (x)dx =

Exercise 1.51∫ a

a f (x) =

Exercise 1.52∫

x2dx =

Exercise 1.53∫ 1

0 x2dx =

Exercise 1.54∫

(3x5 − 6x3 + 2x2 − x + 5)dx =

Exercise 1.55∫ π

20 sin xdx =

Exercise 1.56∫

(cos x + tan x)dx =


Exercise 1.57∫ √

sds =

Exercise 1.58∫ e

11xdx =

Exercise 1.59∫

( 1√x

+ 6x2 + 3x4 − x−3 + 2x− 10)dx =

Exercise 1.60 Find I =∫

cos(3x)3dx using the substitu-

tion u = 3x.

Exercise 1.61 Find I =∫ √

3x5 + x(15x4 + 1)dx, using

the substitution u = 3x5 + x.


(6x2 + sin x)32(12x + cos x)dx

using the substitution u =



sin(3x2 − x)(6x− 1)dx.

Exercise 1.64 Find I =∫ 4

0 ex22xdx.


sin(2x)dx, by letting u =

Exercise 1.66 Use integration by parts to find I =∫

x sin xdx.


x cos xdx.


Exercise 1.68 Find I =∫ e

1 xlnxdx.


lnxx dx.



lnxdx.

Exercise 1.71 Find I =∫ 3

22x−1x2−x

dx.


xexdx.


1.1 Partial Fractions

A rational function is the quotient of two polynomi-

als. e.g. f (x) = 3x2+2x−15x4+x+7

and g(x) = 1x are rational

functions.

Given a rational function, we can often rewrite it as

the sum of a few easier fractions (the partial fractions

decomposition of a rational function).

Roots of Quadratic Functions

Given a quadratic function f (x) = ax2 + bx + c, its

roots are x = −b±√

b2−4ac2a .

If b2 − 4ac > 0 then we get two real roots of f (x) =

(x− r1)(x− r2).

If b2 − 4ac = 0 then we get one (repeated) real root

of f (x) = (x− r)2.


If b2 − 4ac < 0 then we get no real roots of f (x) (we

get two complex roots).

In this case f (x) has no factorisation and we say that

f (x) is an irreducible quadratic.

Next, Let f (x) be a rational function(

f1(x)f2(x)

). If we

factorise the denominator f2(x) into the product of

distinct linear terms, repeated linear terms and irre-

ducible quatratic terms then we can decompose f (x)

as follows:


Case 1: For each distinct linear term (x − r) in the

denominator, the decomposition contains a term Ax−r

where A is an unknown constant.

Case 2: For each repeated linear term (x − r)2 in the

denominator, the decomposition contains terms Ax−r+

B(x−r)2

where A and B are unknown constants.

For each repeated linear term (x− r)3 in the denomi-

nator, the decomposition contains terms Ax−r + B

(x−r)2+

C(x−r)3

, where A, B and C are unknown constants.

Similarly for (x− r)4, (x− r)5, etc.

Case 3: For each irreducible quadratic term ax2+bx+

c in the denominator, the decomposition contains the

term Ax+Bax2+bx+c

where A and B are unknown constants.


Example 1.73 The partial fractions decomposition of 2x+3(x−1)(x−2)

is Ax−1 + B

x−2, by Case 1.

Example 1.74 The partial fractions decomposition of x2+3x−1(x−6)(x−3)2

is Ax−6 + B

x−3 + C(x−3)2

, by Case 1 and Case 2.

Example 1.75 The partial fractions decomposition of x2+2x+1(x−4)(x2+1)

is Ax−4 + Bx+C

x2+1, by Case 1 and Case 3.

Example 1.76 The partial fractions decomposition of5x3+3x+1

(x−3)(x+1)3(x−6)2(x2+2)(x2+3)is

Ax−3 +

[B

x+1 + C(x+1)2

+ D(x+1)3

]+

[E

x−6 + F(x−6)2

]+ Gx+H

x2+2+

Ix+Jx2+3

.

(Note that x2 + 2 and x2 + 3 are irreducible quadratics).

Example 1.77 In Example 1.1 we saw that 2x+3(x−1(x−2) =

Ax−1 + B

x−2. Next, we should find the constants A and B.

Multiply both sides by (x− 1)(x− 2) to get

2x + 3 = Ax−1(x− 1)(x− 2) + B

x−2(x− 1)(x− 2)

⇒ 2x + 3 = A(x− 2) + B(x− 1). ∗


Equation ∗ is true for all values of x, so substitute in some

clever values of x.

x = 2 : 2(2) + 3 = A(0) + B(2− 1)

⇒ 7 = B.

x = 1 : 2(1) + 3 = A(1− 2) + B(0)

⇒ 5 = −A ⇒ A = −5.

∴ the partial fractions decomposition is 2x+3(x−1)(x−2) = −5

x−1 +7

x−2.

Example 1.78 In Example 1.2 we had that x2+3x−1(x−6)(x−3)2

=A

x−6 + Bx−3 + C

(x−3)2. Find A, B and C.

Multiply by (x− 6)(x− 3)2 to get

x2 +3x− 1 = A(x− 3)2 +B(x− 6)(x− 3)+C(x− 6) ∗x = 3 : 9 + 9− 1 = A(0) + B(0) + C(−3)

⇒ 17 = −3C ⇒ C = −173 .

x = 6 : 36 + 18− 1 = A(32) + B(0) + C(0)

⇒ 53 = 9A ⇒ A = 539 .

x = 0 : 0 + 0− 1 = A(−3)2 + B(−6)(−3) + C(−6)

⇒ −1 = 9A + 18B − 6C ⇒ −1 = 9539 + 18B − 6

(−173

)⇒ −1 = 53 + 18B + 34 ⇒ −1 = 87 + 18B

⇒ −88 = 18B ⇒ B = −8818 ⇒ B = −44

9 .

∴ the partial fractions decomposition is


x2+3x−1(x−6)(x−3)2

=539

x−6 +−449

x−3 +−173

(x−3)2.

Exercise 1.79 In Example 1.3 we had that x2+2x+1(x−4)(x2+1)

=A

x−4 + Bx+Cx2+1

. Find A, B and C.

x2 + 2x + 1 = A(x2 + 1) + (Bx + C)(x− 4) ∗.

x = 4 : 16 + 8 + 1 = A(16 + 1) + 0 ⇒ A = 2517.

x = 0 : 0 + 0 + 1 = A(1) + (B(0) + C)(−4) ⇒


∴ we have that x2+2x+1(x−4)(x2+1)

=2517

x−4 +−817 x+ 2

17x2+1

.

Chapter 2

First Order Differential EquationsPredegree Maths

22

CHAPTER 2. FIRST ORDER DIFFERENTIAL EQUATIONS PREDEGREE MATHS23

2.1 Separable Differential Equations

A differential equation is an equation containing deriv-

atives (we often write D.E. for differential equation).

Example 2.1 dydx = 2xy is a D.E.

We can solve this D.E. (i.e. we can find y in terms of x).

Proceed as follows:1y

dydx = 2x ⇒ 1

ydy = 2xdx ⇒∫

1ydy = 2

∫xdx

⇒ ln|y| = 2x2

2 + C ⇒ ln|y| = x2 + C ⇒ |y| = ex2+C

⇒ y = ±ex2+C .

If a function y = f (x) satisfies a D.E. then it is called

a solution of the D.E.

In Example 2.1, y = ±ex2+C is a solution of the D.E.dydx = 2xy.

To show this for y = ex2+C , we substitute y = ex2+C

into the D.E. to get:ddx(ex2+C) = ex2+C(2x+0) = 2xex2+C = 2xy as required,

so y is a solution of the D.E.

(you can check that y = −ex2+C is also a solution).

Example 2.2 Show that y = x4

16 is a solution of the D.E.dydx = xy

12 .


Substituting in we get that the L.H.S. (left hand side) of

the D.E. is ddx

(x4

16

)= 1

16ddxx

4 = 1164x

3 = 14x

3.

The R.H.S. of the D.E. becomes x(

x4

16

)12

= x(x4)12

(16)12

= x√

x4√16

=

xx2

4 = x3

4 = L.H.S. as required, so y = x4

16 is a solution of

the D.E.

The order of a D.E. is the order of the highest deriv-

ative in the D.E.

Example 2.3 d2ydx2 + 5

(dydx

)3

− 4y = ex has order 2, and(dydx

)6

+ 3d2ydx2 +

√d3ydx3 = 0 has order 3.

A first order D.E. of the form dydx = g(x)h(y) is called

a separable D.E.

(where g(x) is a function of x and h(y) is a function of

y).

This D.E. can be solved as follows:1

h(y)dydx = g(x) ⇒

∫1

h(y)dy =∫

g(x)dx.

Then integrate and solve for y if possible.

Example 2.4 Solve the separable D.E. dydx = 3x2

sin y .

sin ydy = 3x2dx ⇒∫

sin ydy = 3∫

x2dx

⇒ − cos y = x3 + C ⇒ cos y = −x3 − C


⇒ cos−1(cos y) = cos−1(−x3−C) ⇒ y = cos−1(−x3−C).

Exercise 2.5 Solve the separable D.E. dydx = −e−3x.

Exercise 2.6 Solve the separable D.E. dydx = (x+1)2, given

that when x = 0, y = 0.


Exercise 2.7 Solve the separable D.E. dydx = (x + 4)50,

given that when x = 0, y = 451

51 .

Exercise 2.8 Solve the separable D.E. dydx = 4y

x .


Exercise 2.9 Solve the separable D.E. dydx = cos x+2

sin y .

Exercise 2.10 Solve the separable D.E. dydx = 2x

y+1.


Exercise 2.11 Solve the separable D.E. dydx = 1+x

1+y .


Exercise 2.12 Solve the separable D.E. dydx = y2+xy2

x2y−x2 .

Exercise 2.13 Solve xy dydx = x2+1

y+1 .


Exercise 2.14 Solve dydx = x2y + x2 + 3y + 3.

2.2 First Order Linear Differential Equations (The

Integrating Factor Method)

Definition 2.15 A D.E. of the form dydx + P (x)y = Q(x)

is called a first order linear D.E.

We can find the general solution of this D.E. as fol-

lows:

1.) Find the integrating factor e∫

P (x)dx.

2.) Multiply the D.E. by the integrating factor (I.F.) to

get:


e∫

P (x)dx(

dydx + P (x)y

)= e

∫P (x)dxQ(x) ∗

3.) Note (or remember) that the L.H.S. of ∗ equalsddx(ye

∫P (x)dx)

(= ddx(y.I.F.)).

∴ ∗ becomes d(ye∫

P (x)dx) = e∫

P (x)dxQ(x)dx.

4.) Integrate both sides and solve for y.

This algorithm is called the integrating factor method.

Example 2.16 Solve the first order linear differential equa-

tion dydx − 3y = 0 using the integrating factor method.

1.) I.F. = e∫

P (x)dx = e∫−3dx = e−3x

(note that we do not include the arbitrary constant C).

2.) Multiply the D.E. by the I.F. to get

e−3x(dydx − 3y) = 0e−3x ∗

3.) Recall that the L.H.S. of ∗ equals ddx(y.I.F.).

∴ ddx(ye−3x) = 0 ⇒ d(ye−3x) = 0dx.

4.) Integrate and solve for y:∫d(ye−3x) =

∫0dx

⇒ ye−3x = 0 + C

⇒ y = Ce3x.

Exercise 2.17 Use the I.F. Method to solve dydx + 5y = 4.


1.) I.F. = e∫

5dx = e5x.

2.) Multiply by the I.F. to get

e5x(dydx + 5y) = 4e5x. ∗

3.) Recall that the L.H.S. of ∗ equals ddx(y.I.F.),

so ddx(ye5x) = 4e5x, so d(ye5x) = 4e5xdx.

4.) Integrate to get:∫d(ye5x) = 4

∫e5xdx

⇒ ye5x = 45e

5x + C

⇒ y = 45 + C

e5x .

Note that we could have done this question using

separable variables:dydx = 4− 5y ⇒ dy = (4− 5y)dx ⇒

∫1

4−5ydy =∫

dx

⇒ −15 ln |4− 5y| = x + C

(check: let u = 4− 5y).

Now solve for y.


Exercise 2.18 Solve xdydx − 4y = x6ex.

This is not in the correct form for the I.F. Method, so we

divide across by x to get:dydx −

4xy = x5ex. Now we can use the I.F. Method on this

D.E.:


Exercise 2.19 Solve dydx + 2y = 3.

Note that dydx + 2y = 3 is a separable D.E. so it can be

solved as follows:dydx = 3− 2y ⇒ 1

3−2ydy = dx ⇒ ...


Exercise 2.20 Solve dydx + 3y = 2e8x.


Exercise 2.21 Solve dydx − 3y = 2e5x.


Exercise 2.22 Use differentiation to show that y = xex +

Cex is a solution of dydx = y + ex.


2.3 Homogeneous Differential Equations

The polynomial anxn + . . .+a2x

2 +a1x+a0 has degree

n (if an 6= 0).

A D.E. is homogeneous of degree n if each term in

the D.E. is a polynomial of degree n (in the variables

x and y).

Example 2.23 x2dy + y2dx = 0 is a homogeneous D.E.

of degree 2.

(x2 + 6y2)dx + (5x2 − xy)dy = 0 is a homogeneous D.E.

of degree 2.

(x2 + 6y2)dx + (x3 + 8xy2)dy = 0 is not a homogeneous

D.E. (since 2 6= 3).

(x2 + 3y2)dx + (x2 − 2xy)dy = x2 is not a homogeneous

D.E.

Note: To solve a homogeneous D.E., just let y = vx

(where v is a new variable) and you will get a sepa-

rable D.E. which should be easy to solve.


Example 2.24 Solve the homogeneous D.E. (x − y)dx +

xdy = 0.

Let y = vx, so (x− vx)dx + xd(vx) = 0

⇒ (1− v)dx + d(vx) = 0 (dividing by x)

⇒ (1− v)dx + vdx + xdv = 0 (product rule)

⇒ dx− vdx + vdx + xdv = 0 ⇒ dx + xdv = 0

⇒ dx = −xdv ⇒ −1x dx = dv (variables are separated)

⇒∫ −1

x dx =∫

dv ⇒ (−1) ln |x| + C = v.

But y = vx, so v = yx.

∴ − ln |x| + C = yx ⇒ y = −x ln |x| + Cx.

Exercise 2.25 Solve the homogeneous D.E.

ydx + (x + y)dy = 0.

Let y = vx. ∴ vxdx + (x + vx)d(vx) = 0

⇒ vdx + (1 + v)d(vx) = 0

⇒ vdx + (1 + v)(vdx + xdv) = 0 (product rule)

⇒ vdx + vdx + xdv + v2dx + vxdv = 0

⇒ 2vdx + v2dx + xdv + vxdv = 0

⇒ 2vdx + v2dx = −xdv − vxdv

⇒ (2v + v2)dx = (−x− vx)dv

⇒ (2v + v2)dx = −x(1 + v)dv

⇒ −1x dx = 1+v

v2+2vdv (separated)


⇒ ...

Exercise 2.26 Solve the homogeneous D.E. (x + y)dx −xdy = 0


Exercise 2.27 Solve the D.E. −ydx + (x +√

xy)dy = 0.


Exercise 2.28 Solve the D.E. (x + y)dx + xdy = 0


Exercise 2.29 Solve the D.E. ydx = 2(x + y)dy.

Chapter 3

Laplace Transforms PredegreeMaths

45

CHAPTER 3. LAPLACE TRANSFORMS PREDEGREE MATHS 46

3.1 Introduction

We will use Laplace transforms to solve differential

equations.

Definition 3.1 Given a function F (t), its Laplace trans-

form is

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

∫ ∞

0

F (t)e−stdt.

Example 3.2 Let y = f (x) = ex. As x goes to∞, ex goes

to ∞. (i.e. limx→∞ ex = ∞). As x goes to −∞, ex goes to

0. (i.e. limx→−∞ ex = 0).

Example 3.3 Find L(1).

By the definition L(1) =∫∞

0 1e−stdt = 1−s[e

−st]∞0 = −1s (e−s∞−

e−s0) = −1s (0− 1) (if s > 0) = 1

s .

∴ L(1) = 1s (where s > 0).

Example 3.4 Find L(t).

L(t) =∫∞

0 te−stdt. Use integration by parts: L.A.T.E.

Let u = t and dv = e−stdt, so du = dt and v = −1s e−st.

∴ L(t) = [uv −∫

vdu]∞0 = [t−1s e−st −

∫ −1s e−stdt]∞0


= −1s [∞e−s∞−0e−s0]+1

s

∫∞0 e−stdt = −1

s [0−0]+1s

1−s[e

−st]∞0

= 0− 1s2 [e

−s∞ − e−s0] = − 1s2 [0− 1] = 1

s2 .

∴ L(t) = 1s2 where s > 0.

Exercise 3.5 Find L(t2).

L(t2) =∫∞

0 t2e−stdt. Integration by parts. L.A.T.E.

Let u = t2 and dv = e−stdt. ∴ du = 2tdt and v = −1s e−st.

Thus L(t2) = [t2−1s e−st −

∫ −1s e−st2tdt]∞0

= [−t2

s e−st + 2s

∫te−stdt]∞0

= [−∞2

s e−s∞ − 02

s e−s0] + 2s

∫∞0 te−stdt

= [0−0]+2sL(t) (by the definition of the Laplace transform)

= 2s

1s2 (by the previous example)

= 2s3 .

∴ L(t2) = 2s3 (where s > 0).

3.2 Proof by Induction

If you want to prove that a statement is true for all

the positive whole numbers, then it is sufficient to:

i) Prove that the statement is true for the whole num-

ber n = 1.

ii) Prove that if the statement is true for any partic-

ular positive whole number n, then the statement is


true for the whole number n + 1.

This technique is called proof by induction.

Example 3.6 Use proof by induction to prove that∑ni=1 i = n(n+1)

2 , for n = 1, 2, 3, . . .

Proof:

i) The statement is true for n = 1, since this becomes∑1i=1 i = 1(1+1)

2 , i.e. 1 = 1.22 .

ii) Assume that the statement is true for a particular posi-

tive whole number n (i.e.∑n

i=1 i = n(n+1)2 ).

Now we will use this to prove that the statement is true

for n + 1 (i.e.∑n+1

i=1 i = (n+1)(n+2)2 ).∑n+1

i=1 i = (∑n

i=1 i) + (n + 1) = n(n+1)2 + (n + 1)

(by the assumption)

= n(n+1)2 + 2(n+1)

2 = n(n+1)+2(n+1)2 = (n+1)(n+2)

2 .

∴∑n+1

i=1 i = (n+1)(n+2)2 , completing the induction and com-

pleting the proof.

Example 3.7 L(tn) = n!sn+1 , for n = 1, 2, 3, . . .

Proof: Use induction:

i) The statement is true for n = 1

i.e. L(t1) = 1!s1+1 , i.e. L(t) = 1

s2 . This was proved in a

previous example.


ii) Assume that the statement is true for a particular posi-

tive whole number n (i.e. L(tn) = n!sn+1).

Next we will prove that the statement is true for n + 1

(i.e. L(tn+1) = (n+1)!sn+2 ).

L(tn+1) =∫∞

0 tn+1e−stdt. Use integration by parts: L.A.T.E.

Let u = tn+1 and dv = e−stdt. ∴ du = (n + 1)tndt and

v = −1s e−st.

∴ L(tn+1) = [uv −∫

vdu]∞0

= [tn+1−1s e−st −

∫ −1s e−st(n + 1)tndt]∞0

= [−tn+1

s e−st + n+1s

∫tne−stdt]∞0

= [−∞n+1

s e−s∞ − −0n+1

s e−s0] + n+1s

∫∞0 tne−stdt

= [0− 0] + n+1s L(tn)

(by the definition of the Laplace transform)

∴ L(tn+1) = n+1s L(tn) = n+1

sn!

sn+1

(by the inductive hypothesis) = (n+1)!sn+2 , as required.

This completes the induction and the proof.

Exercise 3.8 Find L(eat), where a is a constant.

L(eat) =∫∞

0 eate−stdt =∫∞

0 eat−stdt =∫∞

0 e(a−s)tdt

= 1a−s[e

(a−s)t]∞0 = 1a−s[e

(a−s)∞ − e(a−s)0]

= 1a−s[0− 1] (if a− s < 0)

= −1a−s = 1

s−a = L(eat) (where a < s).


Example 3.9 Find L(sin(wt)), where w is a constant.

L(sin(wt)) =∫∞

0 sin(wt)e−stdt.

Use integration by parts: L.A.T.E.

Let u = sin(wt) and dv = e−stdt.

∴ L(sin(wt)) = [uv −∫

vdu]∞0

= [sin(wt)−1s e−st −

∫ −1s e−st cos(wt)wdt]∞0

= [−e−st

s sin(wt) + ws

∫cos(wt)e−stdt]∞0

= [−e−st

s sin(wt)]∞0 + ws

∫∞0 cos(wt)e−stdt

= [−e−∞

s sin(∞)− e0

s sin 0] + ws L(cos(wt))

= [0− 0] + ws L(cos(wt)).

∴ L(sin(wt)) = ws L(cos(wt)) = w

s

∫∞0 cos(wt)e−stdt.

Integration by parts: L.A.T.E. Let u = cos(wt) and dv =

e−stdt, so du = −w sin(wt)dt and v = −1s e−st.

∴ L(sin(wt)) = ws [uv −

∫vdu]∞0

= ws [cos(wt)−1

s e−st −∫ −1

s e−st(−w) sin(wt)dt]∞0

= ws

{[−e−st

s cos(wt)]∞0 − ws

∫∞0 sin(wt)e−stdt

}= w

s

{[−e−∞

s cos(∞)− −e0

s cos 0]− ws L(sin(wt))

}ws

{[0 + 1

s ]−ws L(sin(wt))

}= w

s2 − w2

s2 L(sin(wt)).

∴ L(sin(wt)) = ws2 − w2

s2 L(sin(wt)).

Now solve for L(sin(wt)):

L(sin(wt)) + w2

s2 L(sin(wt)) = ws2


⇒ L(sin(wt))[1 + w2

s2 ] = ws2

⇒ L(sin(wt)) =ws2

1 + w2

s2

=ws2

s2+w2

s2

=w

s2

s2

s2 + w2=

w

s2 + w2.

This is Formula 5.

Linear Operations

Let θ be an operation sending functions to functions.

Example 3.10 ddxf (x) = f ′(x), so differentiation is an

operation sending functions to functions.∫f (x)dx is a function, so integration is an operation.

L(F (t)) = f (s), so the Laplace transform is an operation.

An operation θ is said to be a linear operation if

θ(kf (x)) = kθ(f (x)) and

θ(f (x) + g(x)) = θ(f (x)) + θ(g(x)),

where f (x) and g(x) are functions and k is a constant.

Example 3.11 ddx is a linear operation since d

dx(kf (x)) =

k ddx(f (x)) and d

dx(f (x) + g(x)) = ddx(f (x)) + d

dx(g(x)).

Integration is a linear operation since∫kf (x)dx = k

∫f (x)dx and

CHAPTER 3. LAPLACE TRANSFORMS PREDEGREE MATHS 52∫[f (x) + g(x)]dx =

∫f (x)dx +

∫g(x)dx.

Also, the Laplace transform is a linear operation since

L(kF (t)) =∫∞

0 kF (t)e−stdt = k∫∞

0 F (t)e−stdt = kL(F (t)).

Also, L(F (t) + G(t)) =∫∞

0 (F (t) + G(t))e−stdt

=∫∞

0 (F (t)e−st + G(t)e−st)dt

=∫∞

0 F (t)e−stdt +∫∞

0 G(t)e−stdt = L(F (t)) + L(G(t))

as required.

Example 3.12 Find L(2t + 1). By linearity of L we get

L(2t + 1) = L(2t) + L(1) = 2L(t) + L(1) = 2 1s2 + 1

s

(by Formulae 2 and 1).

∴ L(2t + 1) = 2s2 + 1

s .

Exercise 3.13 Find L(2t2 + 3t + 4).

Exercise 3.14 L(6t4 + 5t3 − 7t2 + t− 10 + 8 sin(2t))


Exercise 3.15 L(3t5+cos(4t)+sin t+e4t+te2t+et cos(3t))

3.3 Inverse Laplace Transforms

Let L(F (t)) = f (s). We say that the Laplace trans-

form of F (t) is f (s). We can also write that the in-

verse Laplace transform of f (s) is F (t) and we write

L−1(f (s)) = F (t).

∴ L−1(L(F (t))) = F (t) and L(L−1(f (s))) = f (s).

We can now use the table of Laplace transforms (back-

wards) to find inverse Laplace transforms.

Example 3.16 L−1(

1s

)= 1 (Formula 1)


L−1(

1s2

)= t

L−1(

2!s2+1

)= t2

L−1(

n!sn+1

)= tn

L−1(

ws2+w2

)= sin(wt)

Note: L−1 is a linear operation.

i.e. L−1(f (s) + g(s)) = L−1(f (s)) + L−1(g(s)) and

L−1(kf (s)) = kL−1(f (s)).

Example 3.17 Find L−1( 1s4).

We know that L−1( 3!s3+1) = t3 (by Formula 3).

∴ L−1( 6s4) = t3 ⇒ 6L−1( 1

s4) = t3

(since L−1 is a linear operation) ⇒ L−1( 1s4) = t3

6 .

Exercise 3.18 Find L−1( 1s6).

Exercise 3.19 Find L−1( 1s2+9

).


Exercise 3.20 Find L−1( 2ss2+9

).

Exercise 3.21 Find L−1( 3s+6s2+16

).


).


).


= 43 cos(2t) + 1

2 sin(2t).

Check: L(43 cos(2t) + 1

2 sin(2t)) =

Theorem 3.24 (Formula 15) L(tF (t)) = − ddsL(F (t)).

Proof: Omit.

Example 3.25 L(t sin t) = − ddsL(sin t) (Formula 15)

= − dds(

1s2+1

) = − dds[(s

2 + 1)−1]

= −(−1)(s2 + 1)−2 dds(s

2 + 1) = (s2 + 1)−2(2s) = 2s(s2+1)2

.

(Note: You could also have used Formula 10).

Exercise 3.26 Find L(t cos(2t)) using Formula 15.


L(t cos(2t)) = − ddsL(cos(2t)) (Formula 15)

= − dds(

ss2+4

) (Formula 6) = −[(s2+4)−s2s(s2+4)2

] (Quotient Rule)

= −[s2+4−2s2

(s2+4)2] = −[ −s2+4

(s2+4)2] = s2−4

(s2+4)2.

(Note: You could also have used Formula 11).

Exercise 3.27 Use Formula 15 to prove Formula 9.

L(teat) =

Theorem 3.28 (First Shifting Theorem (Formula 14))

L(eatF (t)) = f (s− a) = [L(F (t))]s→s−a

(i.e. it is L(F (t)), with s replaced by s− a).

Proof: L(eatF (t)) =∫∞

0 e−steatF (t)dt =∫∞

0 e−(s−a)tF (t)dt.

Recall that L(F (t)) =∫∞

0 e−stF (t)dt, so replacing s with

s− a gives us L(eatF (t)) as required.

Example 3.29 Use Formula 14 to find L(e4tt3).

L(e4tt3) = [L(t3)]s→s−4 (Formula 14)

= [ 3!s3+1 ]s→s−4 (Formula 3) = [ 6

s4 ]s→s−4 = 6(s−4)4

.

Exercise 3.30 L(e6t cos(4t)) =


Exercise 3.31 L(e7tt9) =

Example 3.32 Previously we proved Formula 9

(L(teat) = 1(s−a)2

) using Formula 15. Here we prove For-

mula 9 again, this time using Formula 14:

L(teat) = L(eatt) = [L(t)]s→s−a (Formula 14)

= [ 1s2 ]s→s−a (Formula 2) = 1

(s−a)2.

Exercise 3.33 L(e8tt cos(2t)) =

Definition 3.34 Intuitively, a function is continuous if

it can be drawn without removing the pen from the page.


Theorem 3.35 (Formula 12) Let F (t) and F ′(t) be con-

tinuous functions for t > 0.

Then L(F ′(t)) = sL(F (t))− F (0).

Proof: L(F ′(t)) =∫∞

0 e−stF ′(t)dt.

Use integration by parts: let u = e−st and dv = F ′(t)dt.

∴ du = −se−stdt and v =∫

F ′(t)dt = F (t).

∴ L(F ′(t)) = [uv−∫

vdu]∞0 = [e−stF (t)−∫

F (t)(−s)e−stdt]∞0

= [e−stF (t)]∞0 + s[∫

F (t)e−stdt]∞0

= [e−s∞F (∞)− e0F (0)] + s∫∞

0 e−stF (t)dt

= [0 − F (0)] + sL(F (t)) (by the definition of the Laplace

transform)

= sL(F (t))− F (0).

Example 3.36 Find L( ddt(sin(t)).

By Formula 12, this equals sL(sin(t)− sin 0

(here F (t) = sin t)

= s 1s2+12 − 0 (Formula 5)

= ss2+1

.

Alternativley, L( ddt(sin t)) = L(cos t) = s

s2+1(Formula 6).

Exercise 3.37 Use Formula 12 to find L( ddt(t sin t)).


Theorem 3.38 (Formula 13) Let F (t), F ′(t) and F ′′(t)

be continuous functions for t > 0.

Then L(F ′′(t)) = s2L(F (t))− sF (0)− F ′(0).

Proof: L(F ′′(t)) =∫∞

0 e−stF ′′(t)dt. Use integration by

parts: let u = e−st and dv = F ′′(t)dt.

∴ du = −se−stdt and v =∫

F ′′(t)dt = F ′(t).

∴ L(F ′′(t)) = [uv −∫

vdu]∞0

= [e−stF ′(t)−∫

F ′(t)(−s)e−stdt]∞0

= [e−stF ′(t)]∞0 + s∫∞

0 e−stF ′(t)dt

= [e−s∞F ′(∞)− e0F ′(0)] + sL(F ′(t))

= [0− F ′(0)] + s[sL(F (t))− F (0)] (Formula 12)

= −F ′(0) + s2L(F (t))− sF (0)

= s2L(F (t))− sF (0)− F ′(0) as required.

Example 3.39 Find L( d2

dt2(tet)) using Formula 13.

L( d2

dt2(tet)) = s2L(tet)− s(0e0)− [ d

dt(tet)]0


= s2 1(s−1)2

− 0− [tet + et1]0 (by Formula 9)

= s2

(s−1)2− [0e0 + e0] = s2

(s−1)2− 1 = s2

(s−1)2− (s−1)2

(s−1)2

= s2−(s−1)2

(s−1)2= s2−(s2−2s+1)

(s−1)2= 2s−1

(s−1)2.

Alternativley, L( d2

dt2(tet)) = L( d

dt(tet + et))

= L(tet + et + et) = L(tet) + 2L(et) = 1(s−1)2

+ 2 1s−1

(Formulae 9 and 4)

= 1(s−1)2

+ 2 s−1(s−1)2

= 1+2(s−1)(s−1)2

= 2s−1(s−1)2

, as before.

Exercise 3.40 Use Formula 13 to find L(F ′′(t)), where

F (t) = cos t.

L(F ′′(t)) = s2L(F (t))− sF (0)− F ′(0) =

3.4 Solving Differential Equations using Laplace

Transforms

To solve a differential equation using the Laplace trans-

form, we:

1: Take the Laplace transform of both sides of the


D.E. (using Formulae 12 and 13).

2: Solve this new equation to find L(y) in terms of s.

3: Find y by taking the inverse Laplace transform of

both sides (usually we need partial fractions for this).

Example 3.41 Solve the D.E. dydt−2y = 6 cos(2t)−6 sin(2t)

using Laplace transforms, given that when t = 0, y = 0

(i.e. y(0) = 0).

1: L(dydt )− 2L(y) = 6L(cos(2t))− 6L(sin(2t)).

Now use Formulae 12, 6 and 5 to take Laplace transforms

to get:

[sL(y)− y(0)]− 2L(y) = 6 ss2+22 − 6 2

s2+22

2: Solve to find L(y):

sL(y)− 0− 2L(y) = 6(s−2)s2+22

⇒ (s− 2)L(y) = 6(s−2)s2+22 ⇒ L(y) = 6

s2+22

3: Taking the inverse Laplace of both sides we get:

L−1(L(y)) = L−1( 6s2+22) ⇒ y = 3L−1( 2

s2+22)

⇒ y = 3 sin(2t) (by Formula 5).

Exercise 3.42 Use Laplace transforms to solve the D.E.dydt + 4y = e−4t, given that y(0) = 2 (i.e. when t = 0,

y = 2).


Exercise 3.43 Use Laplace transforms to solve the D.E.dydt − y = 1, given that y(0) = 0.


∴ y = −1 + et.

Alternativley, we could solve this D.E. using separable

variables:dydt − y = 1 ⇒

Lastly, we could solve the D.E. using the Integrating Fac-

tor Method:dydt − y = 1 ⇒


Exercise 3.44 Use Laplace transforms to solve the D.E.

3dydt + 2y = e6t, given that y(0) = 0.


⇒ y = 120e

6t − 120e

−23t.

Example 3.45 Use Laplace transforms to solve the D.E.d2ydt2

= y, given that y(0) = 1 and y′(0) = 2.

1: L(d2ydt2

) = L(y) ⇒ [s2L(y) − sy(0) − y′(0)] = L(y)

(Formula 13)

⇒ s2L(y)− s1− 2 = L(y).

2: ∴ s2L(y)− L(y) = s + 2 ⇒ (s2 − 1)L(y) = s + 2

⇒ L(y) = s+2s2−1

.

3: L(y) = s+2(s+1)(s−1) = A

s+1 + Bs−1.

∴ s + 2 = A(s− 1) + B(s + 1).

s = 1 : 1 + 2 = A0 + B2 ⇒ 3 = 2B ⇒ B = 32.

s = −1 : −1+2 = A(−2)+B0 ⇒ 1 = −2A ⇒ A = −12

∴ L(y) =−1

2s+1 +

32

s−1

⇒ y = L−1(L(y)) = −12L

−1( 1s+1) + 3

2L−1( 1

s−1)

⇒ y = −12e−1t + 3

2e1t (Formula 4).

Exercise 3.46 Use Laplace transforms to solve d2ydt2

= t,

given that y(0) = 1 and y′(0) = 2.


⇒ y = t3

6 + 2t + 1.

Alternativley, we could solve this D.E. by integrating twice:d2ydt2

= t ⇒

Exercise 3.47 Solve the D.E. d2ydt2

+ 3dydt + 7y = 2et, given


that y(0) = 1 and y′(0) = −1.

1:

⇒ L(y) = s2+s(s−1)(s2+3s+7)

= As−1 + Bs+C

s2+3s+7


∴ L(y) =211

s−1 +911s+14

11s2+3s+7

⇒ y = 211e

1t + 111L

−1( 9s+14(s+3

2)2+(−(32)2+7)

)

= 2et

11 + 911

{L−1

(s+3

2

(s−(−32))2+(

√192 )2

)+ L−1

(2818−

2718

(s−(−32))2+(

√192 )2

)}= 2et

11 + 911

{e−

32t cos(

√192 t) + 1

181

(√

192 )

L−1[

√192

(s−(−32 ))2+(

√192 )2

]

}= 2et

11 + 911

{e−

32t cos(

√192 t) + 1

182√19

e−32t sin(

√192 t)

}= 2et

11 + 911e

−32t

{cos(

√192 t) + 1

9√

19sin(

√192 t)

}= y.


Summary of First Order Differential Equations:

To solve a first order D.E. we try the following four

methods (in this order):

1: If it is separable then separate, integrate and solve

for y.

2: If it is homogeneous, substitute y = vx. This will

lead to a separable D.E.

3: If it is linear (i.e. it is of the form dydx + P (x)y =

Q(x)) then use the Integrating Factor Method (mul-

tiply the D.E. by I.F. = e∫

P (x)dx and recall that this

equals ddx(yI.F.). This leads to a separable D.E.).

4: Otherwise use the Laplace transform (find L of

both sides using the Formula sheet, solve for L(y),

decompose using partial fractions and then find y =

L−1(L(y)) ).

Chapter 4

Second Order DifferentialEquations

71

CHAPTER 4. SECOND ORDER DIFFERENTIAL EQUATIONS 72

4.1 Introduction

Consider a differential equation of the form

ad2ydx2 + bdy

dx + cy = 0, where a, b and c are constants.

From this D.E. we define the auxilliary equation

am2 + bm + c = 0.

We can solve this using the quadratic formula

m = −b±√

b2−4ac2a .

Case 1: b2 − 4ac > 0

Here we have two real roots of the auxilliary equa-

tion: m1 = −b−√

b2−4ac2a and m2 = −b+

√b2−4ac

2a .

This gives us the following general solution of the

D.E.: y = Aem1x+Bem2x, where A and B are unknown

constants.

Proof: We will show that y = Aem1x + Bem2x is a solu-

tion of the D.E. ad2ydx2 + bdy

dx + cy = 0.dydx = d

dx(Aem1x + Bem2x) = Am1em1x + Bm2e

m2x.

∴ d2ydx2 = Am2

1em1x + Bm2

2em2x.

Thus the L.H.S. of the D.E. becomes

a[Am21e

m1x+Bm22e

m2x]+b[Am1em1x+Bm2e

m2x]+c[Aem1x+

Bem2x]


= Aem1x(am21 + bm1 + c) + Bem2x(am2

2 + bm2 + c)

= Aem1x0 + Bem2x0 (since m1 and m2 are roots of the

quadratic am2 + bm + c = 0)

= 0 + 0 = 0.

∴ L.H.S. =0 as required, so y = Aem1x + Bem2x is a so-

lution of the D.E. ad2ydx2 + bdy

dx + cy = 0.

Example 4.1 Solve the D.E. d2ydx2 + 5dy

dx + 6y = 0.

The auxilliary equation here is m2 + 5m + 6 = 0.

∴ the roots are m =−5±√

52−4(1)(6)

2(1) = −5±√

25−242 = −5±1

2 .

Thus we have two real roots m1 = −5+12 = −2 and m2 =

−5−12 = −3. Thus Case 1 gives us that the general solution

of the D.E. is y = Aem1x + Bem2x = Ae−2x + Be−3x.

Exercise 4.2 Solve the D.E. d2ydx2 − 7dy

dx + 12y = 0.

The auxilliary equation is m2 − 7m + 12 = 0.

This has roots m =−(−7)±

√(−7)2−4(12)

2(1) = 7±√

49−482 =

7±12 ⇒ m1 = 7+1

2 = 4 and m2 = 7−12 = 3.

Thus we have two real roots, so Case 1 applies, so y =

Aem1x + Bem2x ⇒ y = Ae4x + Be3x.

Exercise 4.3 Solve the D.E. d2ydx2 + 2dy

dx − 3y = 0.


Case 2: b2 − 4ac = 0 in the auxilliary equation.

This gives one repeated real root m = −b±√

02a = −b

2a .

This gives us the following solution of the D.E.:

y = Aemx + Bxemx.

Proof: Omit.


dx + 9y = 0.

Auxilliary equation: m2+6m+9 = 0 ⇒ m =−6±√

62−4(9)

2 =−6±

√36−362 = −6±0

2 = −3 = m. Thus we have a repeated

real root m = −3, so Case 2 applies.

∴ y = Aemx + Bxemx = Ae−3x + Bxe−3x.


dx + 25y = 0.



dx + 16y = 0.

Recall: i is a number (called an imaginary number)

such that i2 = −1, so i =√−1. A number of the form

α+ iβ is called a complex number. α is called the real

part of α + iβ and β is called the imaginary part of

α + iβ. We can draw an Argand diagram of the com-

plex number 2 + 3i.

Case 3: b2 − 4ac < 0 in the auxilliary equation.

This gives us two complex roots: α± iβ.

In this case the general solution of the D.E. is

y = eαx[A cos(βx) + B sin(βx)].

Proof: Omit.



dx + 9y = 0.

m2 + 4m + 9 = 0 ⇒ m =−4±√

42−4(9)

2 = −4±√

16−362 =

−4±√−20

2 = −4±√

4√−5

2 = −4±2√−5

2 = −2 ±√−5 = −2 ±

√−1√

5 = −2± i√

5. This equals α± iβ. ∴ α = −2 and

β =√

5, so by Case 3, y = eαx[A cos(βx) + B sin(βx)] =

e−2x[A cos(√

5x) + B sin(√

5x)].


dx + 10y = 0.

Exercise 4.9 Solve the D.E. d2ydx2 + 16y = 0.



dx + 36y = 0.


dx − 3y = 0, given

that y(0) = 1 and y′(0) = 2.

m2 + 2m − 3 = 0 ⇒ m =−2±√

4−4(−3)

2 = −2±√

162 =

−2±42 = −1 ± 2 ⇒ we have two real roots m1 = 1 and

m2 = −3. ∴ y = Ae1x + Be−3x.

Next we can find the (arbitrary) constants A and B using

the initial conditions y(0) = 1 and y′(0) = 2.

y(0) = 1 and y = Ae1x + Be−3x imply 1 = Ae0 + Be−3(0)

⇒ 1 = A + B.

Also, dydx = Aex − 3Be−3x and y′(0) = 2


imply 2 = Ae0 − 3Be−3(0) ⇒ 2 = A− 3B. Next we solve

the simultaneous equations

A + B = 1

A− 3B = 2

⇒ 0 + 4B = −1 (subtracting) ⇒ B = −14.

∴ A + B = 1 ⇒ A− 14 = 1 ⇒ A = 5

4.

Thus the solution of the D.E. is

y = Aex + Be−3x = 54e

x − 14e−3x.

Exercise 4.12 Solve the D.E. 2d2ydx2 + 6dy

dx + 5y = 0, given

that y(0) = 1 and y′(0) = −1.


Exercise 4.13 Solve the D.E. 4d2ydx2 − 12dy

dx + 9y = 0, given

that y(0) = 1 and y′(0) = 2.


Summary:

To solve the D.E. ad2ydx2 + bdy

dx + cy = 0, find the roots of

the auxilliary equation am2 + bm + c = 0. If you get:

1: Two different real roots m1 and m2 then the gen-

eral solution is y = Aem1x + Bem2x.

2: One repeated different real root m then the gen-

eral solution is y = Aemx + Bxemx.

3: Two complex roots α ± iβ then the general solu-

tion is y = eαx[A cos(βx) + B sin(βx)].

Next we learn how to solve a D.E. of the form

ad2ydx2 + bdy

dx + cy = f (x) ∗First solve the D.E. ad2y

dx2 + bdydx + cy = 0 to get the com-

plementary function (= yc)

Then find a particular solution (= yp) of ∗.

Then the genaral solution of ∗ is of the from

y = yc + yp.


R.H.S. yp

a A1

ax + b A1x + B1

ax2 + bx + c A1x2 + B1x + C1

aebx A1ebx

a sin(bx) A1 sin(bx) + B1 cos(bx)

a cos(bx) A1 sin(bx) + B1 cos(bx)

Example 4.14 Solve d2ydx2 − 5dy

dx + 6y = 24 ∗.

Auxilliary equation: m2 − 5m + 6 = 0 ⇒ m1 = 2 and

m2 = 3. ∴ C.F. = yc = Ae2x + Be3x.

Next we find yp.

R.H.S. of ∗ = 24 ⇒ yp = C.

Now yp = C is a particular solution of the D.E. ∗, sod2

dx2C−5 ddxC+6C = 24 = 0−0+6C = 24 ⇒ C = 4 = yp.

Thus the general solution of ∗ is

y = yc + yp = Ae2x + Be3x + 4.

Exercise 4.15 Solve d2ydx2 + 14dy

dx + 49y = 4e5x ∗m2 + 14m + 49 = 0 ⇒ (m + 7)(m + 7) = 0


⇒ repeated real root m = −7. Thus by Case 2,

yc = Aemx + Bxemx = Ae−7x + Bxe−7x.

yp = Ce5x is a particular solution of ∗ (by the table).

Next find C by substituting yp = Ce5x into ∗:d2

dx2(Ce5x) + 14 ddx(Ce5x) + 49(Ce5x) = 4e5x

⇒ 25Ce5x + 14C5e5x + 49Ce5x = 4e5x

⇒ 25C + 70C + 49C = 4 ⇒ 144C = 4 ⇒ C = 136.

∴ yp = 136e

5x.

Thus the general solution of ∗ is

y = yp + yc = 136e

5x + Ae−7x + Bxe−7x.

Example 4.16 If the R.H.S. of ∗ is 3 cos(x), write down

yp: yp = A1 sin x + B1 cos x

If R.H.S. = 2e7x then yp = Ce7x.

If R.H.S. = 3 sinh x = 312(e

x− e−x then yp = Cex +De−x.

If R.H.S. = 2x2 − 7 then yp = A1x2 + B1x + C1.

If R.H.S. = x + 2ex then yp = A1x + B1 + C1ex.

Example 4.17 Solve d2ydx2 − 5dy

dx + 6y = 2 sin x ∗ ∗Note that the L.H.S. of ∗∗ =L.H.S. of ∗ from a previous

example, so they have the same yc.

∴ yc = Ae2x + Be3x (again).


R.H.S.= 2 sin x, so by the table, yp = C cos x + D sin x.

This is a solution of ∗∗, sod2

dx2(C cos x+D sin x)−5 ddx(C cos x+D sin x)+6(C cos x+

D sin x) = 2 sin x

⇒ ddx(−C sin x+D cos x)−5(−C sin x+D cos x)+6(C cos x+

D sin x) = 2 sin x

⇒ −C cos x−D sin x + 5C sin x− 5D cos x + 6C cos x +

6D sin x = 2 sin x

⇒ cos x(−C−5D+6C)+sin x(−D+5C +6D) = 2 sin x

⇒ (5C − 5D) cos x + (5C + 5D) sin x = 2 sin x

⇒ 5C − 5D = 0 and 5C + 5D = 2

(by comparing coefficients).

∴ 5C − 5D = 0

5C + 5D = 2

⇒ 10C + 0D = 2 (by adding)

⇒ C = 15. Also, 5C − 5D = 0, so 51

5 − 5D = 0 ⇒1− 5D = 0 ⇒ D = 1

5

∴ yp = 15 cos x + 1

5 sin x.

Also, yc = Ae2x + Be3x.

Thus the general solution is

y = yp + yc = 15 cos x + 1

5 sin x + Ae2x + Be3x.

Exercise 4.18 Solve d2ydx2 + 6dy

dx + 10y = 2 sin(2x) ∗


To find yp, we see that 2 sin(2x) is on our table.

∴ yp = C cos(2x) + D sin(2x).


Thus the general solution of the D.E. is y = yp + yc =−215 cos(2x) + 1

15 sin(2x) + e−3x(A cos x + B sin x).

Exercise 4.19 Solve the initial value problem (I.V.P.)d2ydx2 + 3dy

dx − 11y = 0, given that y(0) = 1 and y′(0) = 2.


Exercise 4.20 Solve the I.V.P. d2ydx2 + 2dy

dx + 3y = 0, given

that y(0) = 2 and y′′(0) = 3.

Leo Creedon CE3 Predegree Mathematics These notes are in ... · Mathematics Leo Creedon CE3...

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