staffweb.itsligo.ie · contents 1 mathematics review ce1h 2 1.1 partial fractions . . . . . . . . ....

Mathematics

Leo Creedon

1-st year Civil Engineering (Level 8)

February 23, 2009

Contents

1 Mathematics Review CE1H 21.1 Partial Fractions . . . . . . . . . . . . . . . . . . . . . . . . . . . 151.2 Integration using Trigonometric Substitutions and other tricks 22

2 First Order Differential Equations CE1H 272.1 Separable Differential Equations . . . . . . . . . . . . . . . . . 282.2 First Order Linear Differential Equations (The Integrating Fac-

tor Method) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 352.3 Homogeneous Differential Equations . . . . . . . . . . . . . . 44

3 Laplace Transforms CE1H 503.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 513.2 Proof by Induction . . . . . . . . . . . . . . . . . . . . . . . . . 523.3 Inverse Laplace Transforms . . . . . . . . . . . . . . . . . . . . 583.4 Solving Differential Equations using Laplace Transforms . . . 66

4 Second Order Differential Equations 764.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77

5 Complex Numbers 93

6 Matrices 1026.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1036.2 Addition and Subtraction of matrices . . . . . . . . . . . . . . . 1046.3 Scalar Multiplication . . . . . . . . . . . . . . . . . . . . . . . . 1066.4 Zero Matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107

1

CONTENTS 2

6.5 Multiplication of Matrices . . . . . . . . . . . . . . . . . . . . . 1076.6 Transpose of a Matrix . . . . . . . . . . . . . . . . . . . . . . . . 1156.7 Identity Matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . 1166.8 The Inverse of a 2× 2 matrix . . . . . . . . . . . . . . . . . . . . 1176.9 The inverse of a 3× 3 matrix . . . . . . . . . . . . . . . . . . . . 1216.10 Solving equations involving Matrices . . . . . . . . . . . . . . 1246.11 Cramer’s Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1346.12 Eigenvalues and Eigenvectors . . . . . . . . . . . . . . . . . . . 139

7 Partial Differentiation 146

Bibliography

Chapter 1

Mathematics Review CE1H

3

CHAPTER 1. MATHEMATICS REVIEW CE1H 4

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

]+[

Ex−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

.


1.2 Integration using Trigonometric Substitutions

and other tricks

Example 1.80 Prove the formula in the log tables Page

41: ∫1

a2 − x2dx =

1

2aln

∣∣∣∣a + x

a− x

∣∣∣∣ + C.

Using partial fractions, 1a2−x2 = 1

(a+x)(a−x) = Aa+x + B

a−x

⇒ 1 = A(a− x) + B(a + x) ∗x = a ⇒ 1 = 0 + B(2a) ⇒ B = 1

2a

x = −a ⇒ 1 = A(2a) + B.0 ⇒ A = 12a

∴ I =∫

1a2−x2dx =

∫ 12a

a+x +12a

a−xdx

= 12a

∫1

a+x + 1a−xdx = 1

2a{ln |a + x| − ln |a− x|} + C

= 12a ln |a+x

a−x| + C as required.

Example 1.81 (Completing the square)

Find I =∫

1x2+4x+2

dx.

I =∫

1(x+2)2−2

dx. Let u = x + 2, so du = dx.

∴ I =∫

1u2−(

√2)2

du = (−1)∫

1(√

2)2−u2du

= (−1) 12√

2ln |

√2+u√2−u

|+ C by the formula from the previous

Example (log tables page 41).

Thus I = −12√

2ln |

√2+x+2√2−x−2

| + C.

Note that we could also have done this example using par-


tial fractions. Try this at home.

Example 1.82 Find I =∫

13x2+2x+4

dx.

Note that we cannot do this using partial fractions since

3x2+2x+4 is irreducible, so 13x2+2x+4

is already the partial

fractions decomposition of 13x2+2x+4

(check!).

Instead, try completing the square:

I =∫ 1

3x2+2

3x+43dx = 1

3

∫1

(x+13)2+11

9dx = 1

3

∫1

(x+13)2+

√119

2dx.

Let u = x + 13, so du = dx.

∴ I = 13

∫1

u2+√

119

2du = 13

1√119

tan−1

(u√119

)+ C

= 13

√911 tan−1[

√911(x + 1

3)] + C

= 1√11

tan−1( 3√11

x + 1√11

) + C.



1x2+6x+4

dx.

I =∫

1(x+3)2−5

dx. Let u = x + 3, so du = dx.

∴ I = . . .


13x2+2x−4

dx.

I =∫ 1

3x2+2

3x−43dx = 1

3

∫1

(x+13)2−13

9dx.

Example 1.85 (Trigonometric Substitution)

Find I =∫ √

1− x2dx.

Recall that sin2 θ + cos2 θ = 1 (log tables page 9).

Let x = sin θ, so dx = cos θdθ.

∴ I =∫ √

1− sin2 θ cos θdθ

=∫ √

cos2 θ cos θdθ (since sin2 θ + cos2 θ = 1)

=∫

cos θ cos θdθ =∫

cos2 θdθ


= 12[θ + 1

2 sin(2θ)] + C (log tables page 42)

= 12[sin

−1 x + 12 sin(2 sin−1 x)] + C.

Exercise 1.86 Find I =∫ √

a2 − x2dx, where a is a con-

stant.

Let x = a sin θ. ∴ dx = a cos θdθ.


Exercise 1.87 Prove this formula from page 41 of the log

tables: I =∫

1√a2−x2dx = sin−1(x

a) + C.

Chapter 2

First Order Differential EquationsCE1H

28

CHAPTER 2. FIRST ORDER DIFFERENTIAL EQUATIONS CE1H 29

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 CE1H

51

CHAPTER 3. LAPLACE TRANSFORMS CE1H 52

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 CE1H 58∫[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

77

CHAPTER 4. SECOND ORDER DIFFERENTIAL EQUATIONS 78

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.

Example 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)].

Exercise 4.8 Solve the D.E. d2ydx2 − 2dy

dx + 10y = 0.

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


Exercise 4.10 Solve the D.E. d2ydx2 − 12dy

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.

Example 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 sin 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 not exactly on our table,

but we can work it out by replacing x with 2x.

∴ 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.

Chapter 5

Complex Numbers

94

CHAPTER 5. COMPLEX NUMBERS 95

Let z = a + ib, where a and b are real numbers and

i2 = −1. z is called a complex number. a is called the

real part of z and b is called the complex part of z.

Example 5.1 Let z1 = a1 + ib1 and z2 = a2 + ib2. Then

z1+z2 = (a1+a2)+i(b1+b2), z1−z2 = (a1−a2)+i(b1−b2),

z1z2 = (a1+ib1)(a2+ib2) = a1a2+ia1b2+ia2b1+i2b1b2 =

(a1a2 − b1b2) + i(a1b2 + a2b1) (since i2 = −1).

Exercise 5.2 Let z1 = 1+2i and z2 = 3−4i. Find z1+z2,

z1 − z2 and z1z2.

Exercise 5.3 Let z1 = 3+5i and z2 = 6−2i. Find z1+z2,

z1 − z2 and z1z2.


Definition 5.4 Let z = a + ib. Then z̄ = a− ib is called

the conjugate (or complex conjugate) of z.

We can draw a graph of a complex number on a

diagram called an Argand diagram.

So to get a complex conjugate, we just reflect it through

the real (horizontal) axis on its Argand diagram.

Write Re(z) for the real part of z and write Im(z) for

the imaginary part of z.

So Re(a + ib) = a and Im(a + ib) = b.

The length (or mudulus) of a complex number z =


a + ib is |z| = |a + ib| =√

a2 + b2

(by Pythagoras, |z|2 = a2 + b2, so |z| =√

a2 + b2).

Exercise 5.5 Draw on an Argand diagram the set of com-

plex numbers z where Re(z) = 2.


plex numbers z where Im(z) = −3.



plex numbers z where |z| = 1.


plex numbers z where |z + 2− i| = 2.

Next we learn to divide by complex numbers. To

divide by a complex number, multiply above and be-

low by the complex conjugate of the denominator

and then simplify (i.e. z1z2

= z1z2

z̄2z̄2

= . . .)

Example 5.9 Let z1 = 1 + 2i and z2 = 3− 4i.


Then z1z2

= 1+2i3−4i

z̄2z̄2

= (1+2i)(3+4i)(3−4i)(3+4i) = 3+4i+6i+8i2

9+12i−12i+i2(−4)4

= −5+10i9+(−1)(−16) = −5+10i

25 = −525 + 10

25i = −15 + 2

5i.

Exercise 5.10 Let z1 = 5− 3i and z2 = 2 + 4i. Find z1z2

.

The Polar Form of a Complex Number

Let z = a + ib. Thus r =√

a2 + b2 is the length (or

modulus) of z (r = |z|). θ = tan−1 ba is the argument of

z (= arg z). (From the graph, tan θ = ba, so θ = tan−1 b

a.)

Exercise 5.11 Let z = 2 + 5i. Find |z| and arg z.


Note that from the diagram cos θ = ar , so a = r cos θ.

Also, sin θ = br(

oppositehypothenuse), so b = r sin θ. Thus we can

rewrite z = a + ib as z = a + ib = r cos θ + ir sin θ =

r(cos θ + i sin θ) = z. This is called the polar form of

the complex number z.

Exercise 5.12 Find the polar form of z = 4 + 3i.

Exercise 5.13 Find the standard form of the complex num-

ber z if z has length 10 and argument π4 .

Theorem 5.14 (Euler’s Theorem) eiθ = cos θ + i sin θ

Proof: Omit.


Example 5.15 eiπ = −1.

Indeed, eiπ = cos π + i sin π = −1 + i0 = −1.

Next we can use Euler’s Theorem to prove

Theorem 5.16 (DeMoivre’s Theorem) einθ = cos(nθ)+

i sin(nθ) = (cos θ + i sin θ)n.

Proof: einθ = (eiθ)n = (cos θ + i sin θ)n (by Euler’s Theo-

rem). Also,

einθ = ei(nθ) = cos(nθ) + i sin(nθ) (by Euler’s Theorem).

Example 5.17 Let z = 1 +√

3i. Write z in polar form,

find z15 and write your answer in standard (Cartesian)

form.

r = |z| =

√12 +

√3

2=√

1 + 3 = 2.

arg z = θ = tan−1√

31 = π

3 .

∴ z = r(cos θ + i sin θ) = reiθ = 2eiπ3 .

∴ z15 = (2eiπ3 )15 = 215(eiπ

3 )15 = 32, 768e(iπ315) = 32, 768ei5π.

In standard form this is 32, 768(cos(5π) + i sin(5π))

(by DeMoivre’s Theorem) = 32, 768(−1 + i0) = −32, 768.

Exercise 5.18 Let z =√

3 + i. Find z100.

Chapter 6

Matrices

103

CHAPTER 6. MATRICES 104

6.1 Introduction

Definition 6.1 A matrix is a rectangular array of num-

bers.

Example 6.2

(2 5

1 3

)is a matrix.

Example 6.3

3 1 2

2 1 0

0 −5 1

is a matrix.

Example 6.4

(2 1 2

3 5 2

)is a matrix.

Note : A matrix depends on the number of rows and

columns that it contains.

Example 6.5 Example 6.2 is a 2× 2 matrix, Example 6.3

is a 3× 3 matrix and Example 6.4 is a 2× 3 matrix.

Exercise 6.6 What types of matrices are the following :

(i)

(2

1

)(ii)

(2 3)

(iii)

3 1

5 1

2 3

(iv)

3 3 1 0

0 5 1 0

5 1 −1 7


6.2 Addition and Subtraction of matrices

We can only add or subtract matrices of the same

type. When two matrices are of the same type, we

add or subtract componentwise.

Example 6.7 Let A =

(2 2

5 5

)and B =

(1 5

3 −5

). Then

A + B =

(2 + 1 2 + 5

5 + 3 5 + (−5)

)=

(3 7

8 0

).

Example 6.8 Let A =

2 5 3

−1 0 1

3 2 1

and B =

1 1 5

2 −1 3

7 8 9

.

Then

A−B =

2− 1 5− 1 3− 5

−1− 2 0− (−1) 1− 3

3− 7 2− 8 1− 9

=

1 4 −2

−3 1 −2

−4 −6 −8

.

Exercise 6.9 Let A =

(2 5

3 1

), B =

(5 3

−1 2

), C =(

2 5

3 1

)and D =

(2 1 5

3 3 2

). Calculate where possible:


(i) A + B.

(ii) A−B.

(iii) B − C.

(iv) B + A.

(v) D + A.

(vi) A− C.


2 5 3

1 1 1

0 1 1

, B =

3 5 2

5 1 0

0 0 1

and

C =

6 0 3

1 0 0

0 2 0

. Calculate :

(i) A + B.

(ii) A− C.

(iii) B + A.


6.3 Scalar Multiplication

When we are multiplying a matrix by scalar (e.g. a

real or complex number), we simply multiply every

entry in the matrix by the scalar.

Example 6.11 Let A =

(2 2

5 5

)and B =

(1 5

3 −5

). Then

5A + 6B =

(10 10

25 25

)+

(6 30

18 −30

)

=

(16 40

43 −5

)


(2 −1

0 −3

)and B =

(1 4

3 −2

).

Find (i) 3A + 5B

(ii) A− 3B

(iii) 2A− 1

2B.


6.4 Zero Matrix

Definition 6.13 The Zero Matrix (On) is an n× n ma-

trix where each of the entries is zero.

Example 6.14 O2 =

(0 0

0 0

), O3 =

0 0 0

0 0 0

0 0 0

and O4 =

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

.

The zero matrix has an important property. Let A be

any n× n matrix. Then

A + On = A = On + A .

6.5 Multiplication of Matrices

When multiplying matrices, we have to be very care-

ful since we can only multiply matrices that are con-

formable.


Definition 6.15 Let A and B be matrices. Then A and

B are conformable if the number of columns in A are the

same as the number of rows in B.

Note : If two matrices A and B are conformable, this

doesn’t necessarily mean that B and A are conformable.

Example 6.16 A =

(2 5 3

1 2 3

), B =

(2 2

−1 0

), C =

(5 0

0 3

)and D =

3 1 0

0 1 0

0 3 −1

. A is 2 × 3 matrix, B is

2 × 2 matrix , C is 2 × 2 matrix and D is 3 × 3 matrix.

For the matrices A, B, C and D, Which pairs of matrices

are conformable?

• A and B are not conformable, however B and A

are conformable.

• A and C are not conformable, however C and A

are conformable.

• A and D are conformable, however D and A are

not conformable.

• B and C are conformable, also C and B are con-

formable.


• C and D are not conformable, also D and C are

not conformable.

Exercise 6.17 A =

(2 2

1 3

), B =

(2 5

5 4

), C =

(2 1 3

3 4 2

),

D =

2 5

2 1

0 2

, E =

2 1 0

2 −1 9

0 3 2

and F =

0 1 0

1 0 2

1 1 3

.

For the matrices A, B, C, D, E and F , Which pairs of

matrices are conformable?

Let A be an n × p matrix and B be a p × m matrix.

Clearly A and B are conformable. When we multi-

ply the matrix A by the matrix B (AB), the resulting

matrix is a n × m matrix. To find the (i, j)th-entry

of AB, we single out row i from matrix A and col-

umn j from matrix B. Multiply the corresponding


entries from the row and columns and add the re-

sulting products.


(2 2

1 3

), B =

(2 5

−5 4

).

Calculate AB.

Clearly AB is a 2×2 matrix. Thus we have 4 entries in

AB, one in (1st row, 1st column), one in (1st row, 2nd column),

one in (2nd row, 1st column) and one in (2nd row, 2nd column).

(1st row, 1st column)

Take the 1st row in A and the 1st column in B and

multiply corresponding entries and add these together.

i.e. (2)(2) + (2)(−5) = 4− 10 = −6.

(1st row, 2nd column)

Take the 1st row in A and the 2nd column in B and


i.e. (2)(5) + (2)(4) = 10 + 8 = 18.


(2nd row, 1st column)

Take the 2nd row in A and the 1st column in B and


i.e. (1)(2) + (3)(−5) = 2− 15 = −13.

(2nd row, 2nd column)

Take the 2nd row in A and the 2nd column in B

and multiply corresponding entries and add these to-

gether. i.e. (1)(5) + (3)(4) = 5 + 12 = 17.

∴ AB =

(−6 18

−13 17

).

Note that BA =

((2)(2) + (5)(1) (2)(2) + (5)(3)

(−5)(2) + (4)(1) (−5)(2) + (4)(3)

)=(

9 19

−6 2

)6= AB. Thus if A and B are conformable

and B and A are conformable, it is not necessarily

true that AB = BA.


(2 5

3 1

), B =

(3 5 7

−1 3 1

).


Calculate AB.

AB =

((2)(3) + (5)(−1) (2)(5) + (5)(3) (2)(7) + (5)(1)

(3)(3) + (1)(−1) (3)(5) + (1)(3) (3)(7) + (1)(1)

)

=

(6− 5 10 + 15 14 + 5

9− 1 15 + 3 21 + 1

)

=

(1 25 19

8 18 22

)


(2 5

1 1

), B =

(3

−2

).

Calculate AB.

AB =

((2)(3) + (5)(−2)

(1)(3) + (1)(−2)

)

=

(6− 10

3− 2

)

=

(−4

1

).


1 5 1

1 −1 7

9 3 2

, B =

5 2 3

1 2 3

1 1 −1

.

Calculate AB. AB =

CHAPTER 6. MATRICES 1141(5) + 5(1) + 1(1) 1(2) + 5(2) + 1(1) 1(3) + 5(3) + 1(−1)

1(5) +−1(1) + 7(1) 1(2) +−1(2) + 7(1) 1(3) +−1(3) + 7(−1)

9(5) + 3(1) + 2(1) 9(2) + 3(2) + 2(1) 9(3) + 3(3) + 2(−1)

=

5 + 5 + 1 2 + 10 + 1 3 + 15− 1

5− 1 + 7 2− 2 + 7 3− 3− 7

45 + 3 + 2 18 + 6 + 2 27 + 9− 2

=

11 13 17

11 7 −7

50 26 34

.


1 5 1

1 −1 7

9 3 −2

, B =

5 2

1 2

1 1

.

Calculate AB. AB =


=

1(5) + 5(1) + 1(1) 1(2) + 5(2) + 1(1)

1(5) +−1(1) + 7(1) 1(2) +−1(2) + 7(1)

9(5) + 3(1) +−2(1) 9(2) + 3(2) +−2(1)

=

5 + 5 + 1 2 + 10 + 1

5− 1 + 7 2− 2 + 7

45 + 3− 2 18 + 6− 2

=

11 13

11 7

46 22

.


(3

−3

), B =

(1 7

−2 3

), C =

(−2 −1

5 3

), D =

(0 4 3

−6 2 −7

), E =

3 4

2 −7

8 10

, F =

1 −1 2

2 −7 5

8 10 1

, G =

5 0 1

3 −2 3

1 2 −4

and H =

1 0 2 3

2 −7 1 1

8 1 1 −1

.

(a) For the matrices {A, B, C,D,E, F, G, H}, decide which

pairs are conformable.

(b) For each of the pairs of matrices that are conformable,

calculate their product.


6.6 Transpose of a Matrix

Definition 6.24 The transpose of a matrix A denoted by

AT is obtained by converting the rows of A into columns

one at a time in sequence.


1 0 2

2 −7 1

8 1 1

. Then AT =

1 2 8

0 −7 1

2 1 1

.

In general for two conformable matrices A and B:

• (AB)T = BTAT .

Also if A is any matrix, then (AT )T = A .


(3

−3

), B =

(1 7

−2 3

), C =

(−2 −1

5 3

), D =

(0 4 3

−6 2 −7

), E =

3 4

2 −7

8 10

, F =

1 −1 2

2 −7 5

8 10 1

, G =

5 0 1

3 −2 3

1 2 −4

and H =

1 0 2 3

2 −7 1 1

8 1 1 −1

.


(a) For the matrices {A, B, C,D,E, F, G, H}, find the

transpose of each matrix.

(b) Show (BC)T = CTBT .

(c) Show (FG)T = GTF T .

6.7 Identity Matrix

Definition 6.27 The Identity Matrix (In) is an n × n

matrix where each diagonal entry is 1 and each off diagonal

entry is 0.

Example 6.28 I2 =

(1 0

0 1

), I3 =

1 0 0

0 1 0

0 0 1

and I4 =

1 0 0 0

0 1 0 0

0 0 1 0

0 0 0 1

.

Let A be any n× n matrix, then

A · In = A = In · A .


6.8 The Inverse of a 2× 2 matrix

The determinant of a matrix is denoted by |A| and

the adjoint matrix by A∗. Let A =

(a1,1 a1,2

a2,1 a2,2

), then

|A| = a1,1a2,2 − a1,2a2,1 and A∗ =

(a2,2 −a1,2

−a2,1 a1,1

).


(1 5

−7 1

)and B =

(5 1

3 −5

).

Investigate if (i) |AB| = |A||B| (ii) (A + B)∗ = A∗ + B∗

(iii) (AB)∗ = B∗A∗.

(i) |A| = (1)(1)− (5)(−7) = 1+35 = 36. |B| = (5)(−5)−(3)(1) = −25 − 3 = −28 and |A||B| = (36)(−28) =

−1008.

AB =

(1 5

−7 1

)(5 1

3 −5

)=

(5 + 15 1− 25

−35 + 3 −7− 5

)=

(20 −24

−32 −12

).

Thus |AB| = (20)(−12) − (−32)(−24) = −1008. There-

fore |AB| = |A||B| .

(ii) A + B =

(6 6

−4 −4

), ∴ (A + B)∗ =

(−4 −6

4 6

).


A∗ =

(1 −5

7 1

), B∗ =

(−5 −1

−3 5

). ∴ A∗+B∗ =

(−4 −6

4 6

).

Therefore (A + B)∗ = A∗ + B∗ .

(iii) AB =

(20 −24

−32 −12

). ∴ (AB)∗ =

(−12 24

32 20

).

B∗A∗ =

(−5 −1

−3 5

)(1 −5

7 1

)=

(−5− 7 25− 1

−3 + 35 15 + 5

)=(

−12 24

32 20

). Therefore (AB)∗ = B∗A∗ .

In general if A and B are square matrices of the same

type, then

• |AB| = |A||B|

In general if A and B are matrices of the same type,

then

• (A + B)∗ = A∗ + B∗

In general if A and B are conformable matrices, then

• (AB)∗ = B∗A∗ .


Also if A is any matrix, then (A∗)∗ = A .


(2 3

1 4

), B =

(−1 −3

3 5

)and

C =

(1 2

3 4

). Show that (i) |AB| = |A||B| (ii) |BC| =

|B||C| (iii) (A + C)∗ = A∗ + C∗ (iv) (A + B)∗ =

A∗ + B∗ (v) (AC)∗ = C∗A∗ (vi) (BC)∗ = C∗B∗

(vii) (A∗)∗ = A (viii) (B∗)∗ = B.

We only try to invert square matrices. In general

not every square matrix is invertible. If a square ma-

trix has an inverse, it is said to be invertible or non−singular otherwise it is said to be non− invertible or

singular. If a matrix has an inverse, we denote it by

A−1. The inverse matrix has the property:

A · A−1 = In = A−1 · A .

In general the formula for the inverse of an n×n ma-

trix is :

A−1 =1

|A|A∗ .


Thus

A−1 =1

a1,1a2,2 − a1,2a2,1

(a2,2 −a1,2

−a2,1 a1,1

).

where |A| is the determinant of A and A∗ is the ad-

joint matrix of A. Thus if |A| = 0, A is non−invertible.

Also if |A| 6= 0, A is invertible.


(5 3

4 2

). Find A−1.

|A| = 10 − 12 = −2. Also A∗ =

(2 −3

−4 5

). Therefore

A−1 =1

−2

(2 −3

−4 5

).


(3 2

6 4

).

Find |A|. Does A−1 exist?

|A| = (3)(4)− (2)(6) = 12− 12 = 0.

Therefore A−1 does not exist.


(7 3

4 2

), B =

(7 1

3

63 3

), C =

CHAPTER 6. MATRICES 122(1 −1

5 0

)and D =

(2 6

1 3

). For the matrices {A, B, C,D},

calculate (where possible) their inverses.

6.9 The inverse of a 3× 3 matrix

Let A =

a1,1 a1,2 a1,3

a2,1 a2,2 a2,3

a3,1 a3,2 a3,3

. Then |A| = a1,1

∣∣∣∣∣a2,2 a2,3

a3,2 a3,3

∣∣∣∣∣−a1,2

∣∣∣∣∣a2,1 a2,3

a3,1 a3,3

∣∣∣∣∣ + a1,3

∣∣∣∣∣a2,1 a2,2

a3,1 a3,2

∣∣∣∣∣= a1,1[a2,2a3,3−a2,3a3,2]−a1,2[a2,1a3,3−a2,3a3,1]+a1,3[a2,1a3,2−a2,2a3,1] and

A∗ =

+

∣∣∣∣∣a2,2 a2,3

a3,2 a3,3

∣∣∣∣∣ −∣∣∣∣∣a2,1 a2,3

a3,1 a3,3

∣∣∣∣∣ +

∣∣∣∣∣a2,1 a2,2

a3,1 a3,2

∣∣∣∣∣−

∣∣∣∣∣a1,2 a1,3

a3,2 a3,3

∣∣∣∣∣ +

∣∣∣∣∣a1,1 a1,3

a3,1 a3,3

∣∣∣∣∣ −∣∣∣∣∣a1,1 a1,2

a3,1 a3,2

∣∣∣∣∣+

∣∣∣∣∣a1,2 a1,3

a2,2 a2,3

∣∣∣∣∣ −∣∣∣∣∣a1,1 a1,3

a2,1 a2,3

∣∣∣∣∣ +

∣∣∣∣∣a1,1 a1,2

a2,1 a2,2

∣∣∣∣∣

T

.

Note that A−1 = 1|A|A

∗.



−1 1 2

3 0 5

1 7 2

. Find A−1.

|A| = (−1)

∣∣∣∣∣0 −5

7 2

∣∣∣∣∣− (1)

∣∣∣∣∣3 −5

1 2

∣∣∣∣∣ + (2)

∣∣∣∣∣3 0

1 7

∣∣∣∣∣= (−1)[0(2)−−5(7)]−(1)[3(2)−(−5)(1)]+(2)[3(7)−0(2)]

= −35− 11 + 42 = −4.

A∗ =

+

∣∣∣∣∣0 5

7 2

∣∣∣∣∣ −

∣∣∣∣∣3 5

1 2

∣∣∣∣∣ +

∣∣∣∣∣3 0

1 7

∣∣∣∣∣−

∣∣∣∣∣1 2

7 2

∣∣∣∣∣ +

∣∣∣∣∣−1 2

1 2

∣∣∣∣∣ −∣∣∣∣∣−1 1

1 7

∣∣∣∣∣+

∣∣∣∣∣1 2

0 5

∣∣∣∣∣ −∣∣∣∣∣−1 2

3 5

∣∣∣∣∣ +

∣∣∣∣∣−1 1

3 0

∣∣∣∣∣

T

=

+[(0)(2)− (5)(7)] −[(3)(2)− (5)(1)] +[(3)(7)− (0)(2)]

−[(1)(2)− (2)(7)] +[(−1)(2)− (2)(1)] −[(−1)(7)− (1)(1)]

+[(1)(5)− (0)(2)] −[(−1)(5)− (2)(3)] +[(−1)(0)− (1)(3)]

T

=

−35 −1 21

12 −4 8

5 11 −3

T

=

−35 12 5

−1 −4 11

21 8 −3

.


A−1 =1

|A|A∗. Therefore A−1 =

1

−4

−35 12 5

−1 −4 11

21 8 −3

.


−2 7 6

5 1 −2

3 8 4

. Find |A|.

Does A−1 exist ? Explain ?

|A| = (−2)

∣∣∣∣∣1 −2

8 4

∣∣∣∣∣− (7)

∣∣∣∣∣5 −2

3 4

∣∣∣∣∣ + (6)

∣∣∣∣∣5 1

3 8

∣∣∣∣∣= (−2)[(1)(4)−(−2)(8)]−(7)[(5)(4)−(−2)(3)]+(6)[(5)(8)−(3)(1)]

= (−2)[20]− (7)[26] + (6)[37] = −40− 182 + 222 = 0.

∴ |A| = 0, so A−1 doesn’t exist.


2 3 4

−5 5 6

7 8 9

, B =

1 2 3

2 4 4

1 2 5

and C =

−2 1 4

3 5 −7

1 6 2

.

(i) Find |A|, |B| and |C|. (ii) For the matrices {A, B, C},

calculate their inverses where possible.


6.10 Solving equations involving Matrices

If you want to solve the linear equation ax + b = 0,

you calculate that x = −b

a. However to solve AX =

B where A, X and B are matrices, you cannot say

that X = −B

Asince we cannot divide matrices. There-

fore we must solve equations involving matrices a

different way.


(5 2

1 3

), B =

(7 1

−1 0

), C =(

0 1

−1 1

), D =

(1

−1

)and X be a matrix. Solve the

following matrix equations for X :

(i) AX = B

(ii) XB = C

(iii) BX = D

(iv) ABX = C

(v) BXA = C

(vi) XA = B + 2C.


(i) AX = B .

We first pre-multiply (multiply on left) by A−1 on

both sides of the equation.

A−1AX = A−1B

Thus

I2X = A−1B

∴ X = A−1B

We are now ready to calculate X .

X = A−1B = 113

(3 −2

−1 5

)(7 1

−1 0

)= 1

13

(23 3

−12 −1

)= 23

13313

−1213

−113

.

(ii) XB = C .

XB = C

XBB−1 = CB−1 (post-multiply by B−1)

XI2 = CB−1 (BB−1 = I2)

X = CB−1 (XI2 = X)

X = CB−1 =

(0 1

−1 1

)1

1

(0 −1

1 7

)=

(0 1

−1 1

)(0 −1

1 7

)=

1 7

1 8

.


(iii) BX = D .

BX = D

B−1BX = B−1D pre-multiply by B−1

I2X = B−1D BB−1 = I2

X = B−1D I2X = X

X = B−1D =

(0 −1

1 7

)(1

−1

)=

(1

−6

).

(iv) ABX = C .

ABX = C

A−1ABX = A−1C pre-multiply by A−1

I2BX = B−1C AA−1 = I2

BX = A−1D I2B = B

B−1BX = B−1A−1C pre-multiply by B−1

I2X = B−1A−1C BB−1 = I2

X = B−1A−1C I2X = X


X = B−1A−1C =

(0 −1

1 7

)113

(3 −2

−1 5

)(0 1

−1 1

)=

113

(1 −5

−2 33

)(0 1

−1 1

)= 1

13

(5 −4

−33 29

)=

513

−413

−3313

2913

.

(v) BXA = C .

BXA = C

B−1BXA = B−1C pre-multiply by B−1

I2XA = B−1C

XA = B−1C

XAA−1 = B−1CA−1 post-multiply by A−1

XI2 = B−1CA−1

X = B−1CA−1

X = B−1CA−1 =

(0 −1

1 7

)(0 1

−1 1

)113

(3 −2

−1 5

)=

113

(1 −1

−7 8

)(3 −2

−1 5

)= 1

13

(4 −7

−29 54

)=

413

−713

−2913

5413

.


(vi) XA = B + 2C .

XA = B + 2C

XAA−1 = (B + 2C)A−1 post-multiply by A−1

XI2 = (B + 2C)A−1

X = (B + 2C)A−1

X = (B+2C)A−1 =

[(7 1

−1 0

)+

(0 2

−2 2

)]113

(3 −2

−1 5

)=

113

(7 3

−3 2

)(3 −2

−1 5

)= 1

13

(18 1

−11 16

)=

1813

113

−1113

1613

.


−1 1 2

3 0 5

1 7 2

and B =

1

0

−1

.

Solve AX = B for X where X is a matrix.

AX = B

A−1AX = A−1B

I3X = A−1B

X = A−1B


X = A−1B = 1−4

−35 12 5

−1 −4 11

21 8 −3

1

0

−1

= 1−4

−40

−12

24

=

10

3

−6

.


(7 3

4 2

), B =

(1 −1

5 0

), C =

(1 0

1 3

), D =

(1

−2

), E =

2 3 4

−5 5 6

7 8 9

, F =

1

1

−1

and X be a matrix. Solve the following equations for X :

(i) AX = B

(ii) AX = D

(iii) BX = D

(iv) XAB = C

(v) AXB = C

(vi) AX = B + 2C

(vii) EX = F .


Example 6.40 Solve the following simultaneous equations

x + y = 1

x− y = 7

We can rewrite this system of equations as an equa-

tion with matrices as follows:(1 1

1 −1

)(x

y

)=

(1

7

)

Let A =

(1 1

1 −1

), X =

(x

y

)and B =

(1

7

), so we

have AX = B. So solving the original system of

equations is equivalent to solving this matrix equa-

tion for the matrix X .

AX = B

A−1AX = A−1B

I2X = A−1B

X = A−1B


X =

(x

y

)= A−1B = 1

−2

(−1 −1

−1 1

)(1

7

)= −1

2

(−8

6

)=(

4

−3

). Therefore x = 4 and y = −3.

Example 6.41 Solve the following simultaneous equations

3x− y = 11

3x− 2y = 13

(3 −1

3 −2

)(x

y

)=

(11

13

)=⇒ AX = B

where A =

(3 −1

3 −2

), X =

(x

y

)and B =

(11

13

).

AX = B

A−1AX = A−1B

I2X = A−1B

X = A−1B

X =

(x

y

)= A−1B = 1

−3

(−2 1

−3 3

)(11

13

)= −1

3

(−9

6

)=(

3

−2

). Therefore x = 3 and y = −2.


Example 6.42 Solve the following equations

2x + y + z = 8

5x− 3y + 2z = 3

7x + y + 3z = 20

2 1 1

5 −3 2

7 1 3

x

y

z

=

8

3

20

=⇒ AX = B

where A =

2 1 1

5 −3 2

7 1 3

, X =

x

y

z

and B =

8

3

20

.

AX = B

A−1AX = A−1B

I3X = A−1B

X = A−1B


X =

x

y

z

= A−1B = 13

−11 −2 5

−1 −1 1

26 5 −11

8

3

20

=

13

6

9

3

=

2

3

1

. Therefore x = 2 and y = 3 and z = 1.

Exercise 6.43 Solve the following systems of equations

using matrix methods.

(i) 3x− 2y = 7

4x + y = 13

(ii) 2x + 3y = 5

5x− y = −16

(iii) x + y + 2z = 3

4x + 2y + z = 13

2x + y − 2z = 9

(iv) x + y + z = 2

2x + 3y + z = 7

3x− y + 2z = 4


6.11 Cramer’s Rule

Suppose we want to solve a system of n linear equa-

tions in n unknowns:

a11x1 + a12x2 + · · · + a1nxn = b1

a21x1 + a22x2 + · · · + a2nxn = b2

...

an1x1 + an2x2 + · · · + annxn = bn

We can rewrite this in matrix form as AX = B, where

A =

a11 a12 . . . a1n

a21 a22 . . . a2n

... ... ...

an1 an2 . . . ann

, X =

x1

x2

...

xn

and B =

b1

b2

...

bn

.

We want to solve AX = B, to find X .

Definition 6.44 Given a matrix A =

a11 a12 . . . a1n

a21 a22 . . . a2n

... ... ...

an1 an2 . . . ann

,

its i, j-cofactor matrix Cij is the matrix you get when you

remove the ith row and the jth column. This gives us an

n− 1× n− 1 matrix.


For example C11 =

a22 . . . a2n

...

an2 . . . ann

and

C2n =

a11 a12 . . . a1,n−1

a31 a32 . . . a3,n−1

... ... ...

an1 an2 . . . an,n−1

.

Definition 6.45 Given the matrix A above, its determi-

nant is det(A) = |A| =

(−1)1+1a11det(C11)+(−1)1+2a12det(C12)+· · ·+(−1)1+na1ndet(C1n).

Example 6.46 If A =

(a11 a12

a21 a22

)then

det(A) = a11a22 − a12a21.


1 2 3

4 5 6

7 8 9

. Then det(A) = |A| =

1det(C11)− 2det(C12) + 3det(C13) =

1det

(5 6

8 9

)− 2

(4 6

7 9

)+ 3

(4 5

7 8

)=

1(45−48)−2(36−42)+3(32−35) = −3−2(−6)+3(−3) =

−3 + 12− 9 = 0.

Thus det(A) = 0, so A−1 = A∗

det(A) does not exist.


Exercise 6.48 Find det(A), where A =

1 2 −4

1 0 3

2 −1 0

.

det(A) = 1det(C11)− 2det(C12) + (−4)det(C13) =

1(0+3)− 2(0− 6)− 4(−1− 0) = 1(3)− 2(−6)− 4(−1) =

3 + 12 + 4 = 19.

Using determinants we are now ready to state

Cramer’s Rule

Let AX = B be a system of n linear equations in n

unknowns. Let A =

a11 a12 . . . a1n

a21 a22 . . . a2n

... ... ...

an1 an2 . . . ann

, X =

x1

x2

...

xn

and B =

b1

b2

...

bn

. The jth component of X is

xj =det(Bj)

det(A) , where Bj =

a11 a12 . . . b1 . . . a1n

... ... ... ...

an1 an2 . . . bn . . . ann

.

(In Bj, the vector B replaces the jth column of A).


Proof: Omit.

Example 6.49 Use Cramer’s Rule to solve:

x1 + 3x2 = 0

2x1 + 4x2 = 6.

Here AX = B, with A =

(1 3

2 4

), X =

(x1

x2

)and B =(

0

6

). Now B1 =

(0 3

6 4

)and B2 =

(1 0

2 6

)(recall that Bj is just A with the vector B replacing the jth

column of A).

∴ x1 = det(B1)det(A) =

det

0 3

6 4

det

1 3

2 4

= 0−18

4−6 = −18−2 = 9 = x1.

Also, x2 = det(B2)det(A) =

det

1 0

2 6

−2 = 6−0

−2 = −3 = x2.

Example 6.50 Use Cramer’s Rule to solve

x + 3y − 2z = 3

2x + 5y + 2z = 5

3x− y + 4z = 8.


Here AX = B, where A =

1 3 −2

2 5 2

3 −1 4

, X =

x

y

z

and B =

3

5

8

. Write x = x1, y = x2 and z = x3. Now

B1 =

3 3 −2

5 5 2

8 −1 4

, B2 =

1 3 −2

2 5 2

3 8 4

and

B3 =

1 3 3

2 5 5

3 −1 8

.

∴ x = x1 = det(B1)det(A) =

3det

5 2

−1 4

−3det

5 2

8 4

+(−2)det

5 5

8 −1

1det

5 2

−1 4

−3det

2 2

3 4

+(−2)det

2 5

3 −1

= 3(20+2)−3(20−16)−2(−5−40)

(20−(−2))−3(8−6)−2(−2−15) = 3(22)−3(4)−2(−45)22−3(2)−2(−17) = 66−12+90

22−6+34 =14450 = 72

25 = x1 = x.

Also, y = x2 = det(B2)det(A) =

1det

5 2

8 4

−3det

2 2

3 4

+(−2)det

2 5

3 8

50


= (20−16)−3(8−6)−2(16−15)50 = 4−6−2

50 = −450 = −2

25 = x2 = y.

Lastly, we could show that x3 = z = −950 (check).

Exercise 6.51 Use Cramer’s Rule to solve

x + y − 2z = 1

x− y + 3z = 0

2x− y + z = 2.

(You should get that x = 45, y = −5

5 and z = −35 ).

Exercise 6.52 Use Cramer’s Rule to find x

x + 2z = 0

−x + z = 1

2w + x− y + 2z = −1.

3w + x− 2y + z = 3.

(You should get that x = −23).

6.12 Eigenvalues and Eigenvectors

Let A = [aij] be a square n× n matrix.

Let X =

x1

x2

...

xn

6=

0

0...

0

be an n × 1 column vector.

Let λ be a number. Can we have AX = λX ? ∗


If there is a number λ and a vector X which satis-

fies ∗, then we say that λ is an eigenvalue of A and

X is an eigenvector of A. These have important ap-

plications for coupled oscillations and vibrations and

image processing.

How do we find the eigenvalues of A ?

We must have AX = λX , so

AX − λX = On×1 = AX − λIX = (A− λI)X = On×1.

This happens when A− λI is not invertible

(otherwise X = (A− λI)−1.O = O)

i.e. when det(A− λI) = O.

Thus to find the eigenvalues of A we must solve

det(A− λI) = O for λ.

Example 6.53 Find the eigenvalues of A =

(4 −1

2 1

).

|A−λI| = |

(4 −1

2 1

)−

(λ 0

0 λ

)| = |

(4− λ −1

2 1− λ

)| =

(4−λ)(1−λ)−(−1)2 = 4−4λ−λ+λ2+2 = λ2−5λ+6 =

0 = (λ− 2)(λ− 3).

∴ λ = 2 and λ = 3 are the eigenvalues of A.

Check: |A−2λI| = |

(4 −1

2 1

)−

(2 0

0 2

)| = |

(2 −1

2 −1

)| =


2(−1)− (−1)2 = 0 as required. Thus 2 is an eigenvalue of

A.

Exercise 6.54 Find the eigenvalues of A =

(3 1

−1 2

).

Thus λ = 52 +

√3

2 i and λ = 52 −

√3

2 i are the eigencalues

of A.


8 −2 0

4 2 1

3 2 −1

.


⇒ λ3 − 9λ2 + 12λ + 46 = 0. Using numerical tech-

niques the roots can be shown to be λ = −1.593055586,

λ = 5.296527793− 0.9067081753i and λ = 5.296527793 +

0.9067081753i.

Example 6.56 Find the eigenvectors of A =

(4 −1

2 1

).

Recall that in a previous example we saw that the eigen-

values of A are 2 and 3.

Case 1: λ = 2. Find the eigenvector X associated with the

eigenvector λ = 2.

(A − λI)X = 0, i.e. AX − λIX = 0 ⇒ AX = λX ⇒AX = 2X .


∴

(4 −1

2 1

)X = 2X ⇒

(4 −1

2 1

)(x1

x2

)=

(2x1

2x2

)

⇒4x1 − x2 = 2x1

2x1 + x2 = 2x2

⇒2x1 = x2

2x1 = x2

.

Thus the eigenvector X is

(x1

x2

)=

(x1

2x1

)= x1

(1

2

).

Case 2: λ = 3. Find the eigenvector X associated with the

eigenvector λ = 3.

AX = λX ⇒

(4 −1

2 1

)(x1

x2

)=

(3x1

3x2

)

⇒4x1 − x2 = 3x1

2x1 + x2 = 3x2

⇒x1 = x2

x2 = x1

.

Thus the eigenvector X is

(x1

x2

)=

(x1

x1

)= x1

(1

1

).

∴ A =

(4 −1

2 1

)has eigenvalue 2 with associated eigen-

vector

(1

2

)(and its scalar multiples) and has eigenvalue

3 with associated eigenvector

(1

1

)(and its scalar multi-

ples).



(1 2

1 0

)and

their associated eigenvectors.


∴ A =

(1 2

1 0

)has eigenvalue 2 with associated eigen-

vector

(2

1

)(and its scalar multiples) and has eigenvalue

−1 with associated eigenvector

(1

−1

)(and its scalar mul-

tiples).

Chapter 7

Partial Differentiation

147

CHAPTER 7. PARTIAL DIFFERENTIATION 148

Example 7.1 A cylinder with radius r and height h has

volume V = πr2h. Thus V depends on r and h, so V

is a function of two variables. If we keep r constant and

increase h, then V will increase. The rate of change of V

with respect to h is written as[

dVdh

]r

or more commonly as∂V∂h . This is called the partial derivative of V with respect

to h.

∴ ∂V∂h = ∂

∂h(πr2h) = πr2 since we treat r as a constant.

Also, ∂V∂r = ∂

∂r(πr2h) = πh ∂∂r(r

2)

(since we treat h as a constant) = πh2r.

Example 7.2 Let z = f (x, y) = x3 + xy − y2.

Note that z is a function from two variables to a third vari-

able, so it represents a surface in three dimensions.∂z∂x = ∂

∂x(x3 + xy − y2) = 3x2 + y − 0 and∂z∂y = ∂

∂y(x3 + xy − y2) = 0 + x− 2y.

∴ ∂∂x(∂z

∂x) = ∂2z∂x2 = 6x and

∂∂y(

∂z∂y) = ∂2z

∂y2 = −2 .

Also, ∂∂x(∂z

∂y) = ∂2z∂x∂y = 1 and

∂∂y(

∂z∂x) = ∂2z

∂y∂x = 1.

Note: for almost all functions, ∂2z∂x∂y = ∂2z

∂y∂x.

Exercise 7.3 Let z = f (x, y) = x cos y − 3√

x2y + x.


Find ∂z∂x, ∂z

∂y , ∂2z∂x2 , ∂2z

∂y2 , ∂2z∂x∂y and ∂2z

∂y∂x.

Exercise 7.4 Let z = f (x, y) = sin(3xy + ey). Find ∂z∂x

and ∂z∂y .


Example 7.5 Let z = f (x2 + y2).

Show that x∂z∂y − y ∂z

∂x = 0 ∗∂z∂x = ∂z

∂(x2+y2)∂(x2+y2)

∂x = 2x ∂z∂(x2+y2)

.

Similarly, ∂z∂y = ∂z

∂(x2+y2)∂(x2+y2)

∂y = 2y ∂z∂(x2+y2)

.

Thus the LHS of ∗ equals

x∂z∂y − y ∂z

∂x = 2xy ∂z∂(x2+y2)

− 2xy ∂z∂(x2+y2)

= 0 as required.

Exercise 7.6 Let z = f (yx). Show that x∂z

∂x + y ∂z∂y = 0.

∂z∂x = ∂z

∂(yx)

∂(yx)

∂x = ∂z∂(y

x)y(−1)x−2 = −y

x2∂z

∂(yx)

.∂z∂y = . . .

Exercise 7.7 Let z = f (ax + by), where a and b are con-

stants. Show that b∂z∂x − a∂z

∂y = 0.


Example 7.8 Let z = 1x2+y2−1

.

Show that x∂z∂x + y ∂z

∂y = −2z(1 + z) ∗∂z∂x = ∂

∂x((x2 + y2 − 1)−1) =

(−1)(x2 + y2 − 1)−2 ∂∂x(x2 + y2 − 1) =

(−1)(x2 + y2 − 1)−2(2x) = −2x(x2+y2−1)−2 = ∂z

∂x.

Also,∂z∂y = ∂

∂y((x2 + y2 − 1)−1) =

(−1)(x2 + y2 − 1)−2 ∂∂y(x

2 + y2 − 1) = −2y(x2+y2−1)−2 = ∂z

∂y .

∴ x∂z∂x + y ∂z

∂y = −2x2

(x2+y2−1)2+ −2y2

(x2+y2−1)2= LHS of ∗.

RHS of ∗ = −2z(1 + z) = −2(x2+y2−1)

(x2+y2−1x2+y2−1

+ 1x2+y2−1

) =

−2(x2+y2−1)

( x2+y2

x2+y2−1) = −2x2−2y2

(x2+y2−1)2.

∴ LHS = RHS of ∗.

Exercise 7.9 Let z = xyx−y .

Show that x∂z∂x + y ∂z

∂y = z ∗


Show also that x2 ∂2z∂x2 − y2∂2z

∂y2 = 0 ∗ ∗

Next show that z ∂2z∂x∂y = 2∂z

∂x∂z∂y ∗ ∗∗

Definition 7.10 Let z = f (x, y).

Then fx = fx(x, y) = ∂z∂x = zx, fy = fy(x, y) = ∂z

∂y = zy,


fxx = fxx(x, y) = ∂2z∂x2 = zxx, fyy = fyy(x, y) = ∂2z

∂y2 = zyy,

fxy = fxy(x, y) = ∂2z∂x∂y = zxy and fyx = fyx(x, y) =

∂2z∂y∂x = zyx.

For example fx(1, 2) = ∂z∂x evaluated at x = 1, y = 2.

Example 7.11 Let z = f (x, y) = x3 + 2xy + 2y2.

Find fx(1, 2) and fy(1, 2).

fx = ∂z∂x = 3x2 + 2y, so fx(1, 2) = 3(12) + 2(2) = 7.

fy = ∂z∂y = 2x + 4y, so fy(1, 2) = 2(1) + 4(2) = 10.

Exercise 7.12 Let z = f (x, y) = x sin(xy) + 3.

Find fx(1,π2) and fy(1,

π2).

Exercise 7.13 Let z = f (x, y) = xx2+y2 .


Find fx(1, 0) and fy(2, 1).

Example 7.14 Let w = f (x, y, z) = x3y2 + 3 sin(yz) + 2.

Find the first order partial derivatives at (1, 0, π2).

fx = ∂w∂x = 3x2y2 + 0 + 0 = 3x2y2.

∴ fx(1, 0,π2) = 3(12)(02) = 0.

fy = ∂w∂y = x32y + 3 cos(yz)z + 0 = 2x3y + 3z cos(yz).

∴ fy(1, 0,π2) = 2(13)(0) + 3π

2 cos(0π2) = 0 + 3π

2 cos(0) = 3π2 .

fz = ∂w∂z = 0 + 3 cos(yz)y + 0 = 3 cos(yz)y.

∴ fz(1, 0,π2) = 3 cos(0π

2)0 = 0.

Note: dydx = 1

dxdy

but ∂y∂x 6=

1∂x∂y

.

Example 7.15 Here we show that d2xdy2 =

−d2y

dx2

(dydx)3

(6= 1

d2y

dx2

).

d2xdy2 = d

dy(dxdy ) = d

dy

(1dydx

)= dx

dyddx

(1dydx

)(by the Chain Rule)


= dxdy

ddx[(dy

dx)−1] = dxdy (−1)(dy

dx)−2 d2ydx2 = 1

(dydx)

(−1) 1

(dydx)2

d2ydx2 =

−d2y

dx2

(dydx)3

as required.

Example 7.16 Let x = t + 2 sin t and y = cos t.

Find dydx and d2y

dx2 and find their values at t = 0.dydx = dy

dtdtdx (by the Chain Rule) = dy

dt

(1dxdt

)= d

dt(cos t) 1ddt(t+2 sin t)

=

− sin t 11+2 cos t = − sin t

1+2 cos t = dydx.

Thus at t = 0 we get dydx = − sin 0

1+2 cos 0 = −01+2 = 0.

Now d2ydx2 = d

dx(dydx) = d

dx( − sin t1+2 cos t) = d

dt(− sin t

1+2 cos t)dtdx

(by the Cain Rule) = 1

(dxdt )

(1+2 cos t)(− cos t)−(− sin t)(0−2 sin t)(1+2 cos t)2

by the quotient rule.

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