differential nso qu ei at - john wiley & sons nso qu ei at 9.1 kick off with cas 9.2 verifying...

Differential equations

9.1 Kick off with CAS

9.2 Verifying solutions to a differential equation

9.3 Solving Type 1 differential equations, dydx

= f (x)


= f (y)


= f (x)g (y)

9.6 Solving Type 4 differential equations, d

2y

dx2= f (x)

9.7 Review

9c09DifferentialEquations.indd 456 04/07/15 12:30 PM

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Please refer to the Resources tab in the Prelims section of your eBookPlUs for a comprehensive step-by-step guide on how to use your CAS technology.

9.1 Kick off with CAS<To Come>

c09DifferentialEquations.indd 457 04/07/15 12:30 PM

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Verifying solutions to a differential equationClassification of differential equationsA differential equation (d.e.) is an equation involving derivatives. It is of the form

gax, y, dy

dx, d2y

dx2, .....b = 0.

It contains the function y = f(x) as the dependent variable, x as the independent variable, and various derivatives. In this topic, only differential equations that contain functions of one variable, y = f(x) are considered.

Differential equations can be classified according to their order and degree. The order of a differential equations is the order of the highest derivative present. The degree of a differential equation is the degree of the highest power of the highest derivative.

A linear differential equation is one in which all variables including the derivatives are raised to the power of 1.

Some examples of differential equations are:

(a) dy

dx= ky (b) a

d

2y

dx2+ b

dy

dx+ cy = 0

(c) x‥ − tx# + 2x = t (d) d

2y

dx2+ n2y = 0

(e) xady

dxb

3

+ 3 dy

dx+ 5y = 0 (f) Dt

3x = "x2 + 1

Note that (a) and (e) are first order; (b), (c) and (d) are second order; and (f) is third order. Equation (e) has a degree of 3, whereas all the others have a degree of 1. Equations (a), (b), (c) and (d) are linear; (e) and (f) are non-linear. Note also that there are many different notations for derivatives; for example, second-order derivatives can be expressed

as x‥ = d

2xdt2

and D2t x = d

2xdt2

.

Differential equations are extremely important in the study of mathematics and appear in almost every branch of science. Sir Isaac Newton (1642–1727) and Gottfried Wilheim Leibniz (1646–1716) are both credited with the discovery of calculus and differential equations in the 1670s and 1680s.

Verifying solutions to differential equationsTo check that a given solution satisfies the differential equation, use the process of differentiation and substitution. Generally only first- or second-order differential equations and a given solution will be considered in this chapter.

When setting out a proof, it is necessary to show that the left-hand side (LHS) of the equation is equal to the right-hand side (RHS).

9.2

458 Maths Quest 12 sPeCIaLIst MatheMatICs VCe units 3 and 4


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Differential equations involving unknownsWhen we verify a given solution to a differential equation involving algebraic, trigonometric, or exponential functions, there may also be an unknown value that must be determined for which the given solution satisfi es the differential equation.

Verify that y = x3 is a solution of the differential equation

x3

d

2y

dx2− ady

dxb

2

+ 3xy = 0.

tHinK WritE

1 Use basic differentiation to fi nd the fi rst derivative. y = x3

dy

dx= 3x2

2 Find the second derivative.d

2y

dx2= 6x

3 Substitute for y, the fi rst derivative and second derivative into the LHS of the differential equation.

x3

d

2y

dx2− ady

dxb

2

+ 3xy

= x3 × (6x) − (3x2)2 + 3x × (x3)

4 Simplify and expand, so that LHS = RHS = 0, thus proving the given solution does satisfy the differential equation.

= 6x4 − 9x4 + 3x4

= 0

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Given that y = ekx is a solution of the differential equation

d

2y

dx2− 2

dy

dx− 8y = 0, find the values of the real constant k.

tHinK WritE

1 Use the rule for differentiation of exponential

functions, ddx

(ekx) = kekx, to fi nd the fi rst derivative.

y = ekx

dydx

= kekx

2 Differentiate again to fi nd the second derivative.d

2y

dx2= k2ekx

3 Substitute for y, the fi rst derivative dy

dx and the second

derivative d

2y

dx2 into the given differential equation.

d

2y

dx2− 2

dy

dx− 8y = 0

k2ekx − 2kekx − 8ekx = 0

4 Take out the common factor. ekx 1k2 − 2k − 8 2 = 0

5 Factorise the quadratic equation involving the unknown.

ekx ≠ 0 ⇒ k2 − 2k − 8 = 0 (k − 4)(k + 2) = 0

6 Solve the resulting equation for the unknown and state the answer.

When k = 4 or k = −2, then y = ekx is a

solution of d

2y

dx2− 2

dy

dx− 8y = 0.

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topic 9 DIfferentIaL eQuatIons 459


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Verify that y = xe−2x is a solution of the differential equation

d

2y

dx2+ 4

dy

dx+ 4y = 0.

tHinK WritE

1 Use the product rule for differentiation to fi nd the fi rst derivative.

y = xe−2x

dy

dx= x

ddx

(e−2x) + e−2x ddx

(x)

dy

dx= −2xe−2x + e−2x

2 Simplify the fi rst derivative by taking out the common factor.

dy

dx= e−2x(1 − 2x)

3 Find the second derivative, using the product rule again.

d

2y

dx2= e−2x d

dx(1 − 2x) + (1 − 2x) d

dx(e−2x)

d

2y

dx2= −2e−2x − 2(1 − 2x)e−2x

4 Simplify the second derivative by taking out the common factor.

d

2y

dx2= e−2x(−2 − 2(1 − 2x))

= e−2x(4x − 4)

5 Substitute for y, the fi rst derivative dy

dx and

the second derivative d2y

dx2 into the LHS of

the differential equation.

d

2y

dx2+ 4

dydx

+ 4y

= e−2x(4x − 4) + 4e−2x(1 − 2x) + 4xe−2x

6 Take out the common factor and simplify, so that LHS = RHS = 0, thus proving the given solution does satisfy the differential equation.

= e−2x[(4x − 4) + 4(1 − 2x) + 4x]= e−2x[4x − 4 + 4 − 8x + 4x]= 0

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solutions to a differential equation involving productsWhen verifying solutions to a differential equation involving a mixture of algebraic, trigonometric, or exponential functions, it may be necessary to use the product or quotient rules for differentiation.

Verifying solutions to a differential equation1 WE1 Verify that y = x2 is a solution of the differential equation

x2

d

2y

dx2− ady

dxb

2

+ 2y = 0.

2 For the differential equation x4

d

2y

dx2− ady

dxb

2

+ 4x2y = 0, show that y = x4 is a solution.

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3 WE2 Given that y = ekx is a solution of the differential equation

d

2y

dx2+ 3

dy

dx− 10y = 0, find the values of the real constant k.

4 If y = cos(kx) is a solution of the differential equation d2y

dx2+ 9y = 0, find the

values of the real constant k.

5 WE3 Verify that y = xe3x is a solution of the differential equation

d2y

dx2− 6

dy

dx+ 9y = 0.

6 Given that y = Ax cos(2x) is a solution of the differential equation

d2y

dx2+ 4y = 8 sin(2x), find the value of the real constant A.

7 a Verify that y = x4 satisfies the differential equation x4

d2y

dx2− ady

dxb

2

+ 4x2y = 0.

b If y = 2x2 − 3x + 5, show that ady

dxb

2

− 8y + 31 = 0.

c Given the differential equation d2y

dx2− 2x

dy

dx+ 6y + 6x2 = 0, show that

y = x3 − 3x2 − 3x2

+ 1 is a solution.

d If y = ax3 + bx2 where a and b are constants, show that

x2 d2y

dx2− 4x

dy

dx+ 6y = 0.

8 a Find the constants a, b and c if y = a + bx + cx2 is a solution of the differential

equation d2y

dx2 + 2

dy

dx + 4y = 4x2.

b Determine the constants a, b, c and d if y = ax3 + bx2 + cx + d is a solution of

the differential equation d2y

dx2 + 2

dy

dx + y = x3.

c Show that y = xn is a solution of the differential equation

x2y

d2y

dx2 − x2 ady

dxb

2

+ ny2 = 0.

d The differential equation x2

d2y

dx2− 2x

dy

dx− 10y = 0 has a solution y = xn find the

possible values of n.

9 a Given that x = e3t + e−4t show that d2x

dt2+ dx

dt− 12x = 0.

b If y = Ae3x + Be−3x where A and B are constants, show that d2y

dx2− 9y = 0.

c Find the values of real constant k such that y = ekx satisfies d2y

dx2+ 5

dy

dx− 6y = 0.

d Find the values of m where m ∈ C if y = emx satisfies d2y

dx2+ 4

dy

dx+ 13y = 0

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10 a If y = 3 sin(2x) + 4 cos(2x), show that d2y

dx2+ 4y = 0.

b Show that y = A sin(3x) + B cos(3x), where A and B are constants, is a solution

of the differential equation d2y

dx2+ 9y = 0.

c If y = a sin(nx) + b cos(nx), where n is a positive real number, show that d2y

dx2+ n2y = 0.

d Given that x = a sin( pt) satisfies d2x

dt2+ 9x = 0, find the value of p.

11 a Show that y = ex2 satisfies the differential equation

d2y

dx2− 2x

dy

dx− 2y = 0.

b Verify that y = cos(x2) satisfies the differential equation

x

d2y

dx2−

dy

dx+ 4x3y = 0.

c If y = ax + b"x2 + 1 where a and b are constants, show that

(x2 + 1)

d2y

dx2+ x

dy

dx− y = 0.

d Given that y = loge (x + "x2 − 9), show that (x2 − 9)

d2y

dx2+ x

dy

dx= 0.

12 a Show that y = tan(ax), where a ∈ R \{0}, satisfies the differential

equation d2y

dx2= 2a2y(1 + y2).

b Verify that y = tan2(ax), where a ∈ R \{0}, is a solution of the differential

equation d2y

dx2= 2a2(3y2 + 4y + 1).

c Show that y = loge (ax + b), where a, b ∈ R, is a solution of the differential

equation d2y

dx2+ a2e−2y = 0.

13 a Verify that y = tan−1(2x) is a solution of the differential equation

(1 + 4x2)

d2y

dx2+ 8x

dy

dx= 0.

b Show that y = sin−1(3x) is a solution of the differential equation

(1 − 9x2)

d2y

dx2− 9x

dy

dx= 0.

c Verify that y = cos−1ax4b is a solution of the differential equation

(16 − x2)

d2y

dx2− x

dy

dx= 0.



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14 a A parachutist of mass m falls from rest in the Earth’s gravitational field and is subjected to air resistance. The velocity v is given by

v =mg

k (1 − e−kt)

at a time t where g and k are constants. Show that dvdt

+ kv = mg.

b In a transient circuit, the current i amperes at a time t seconds is given by i = 3e−2t sin(3t).

Show that d2i

dt2+ 4

didt

+ 13i = 0.

15 a Verify that y = e3x cos(2x) satisfies the differential

equation d2y

dx2− 6

dy

dx+ 13y = 0.

b Find the real constants a and b if x = t(a cos(3t) + b sin(3t)) is a solution of the

differential equation d2x

dt2+ 9x = 6 cos(3t).

16 a Given that y = xe−3x is a solution of the differential equation

d2y

dx2+ a

dy

dx+ by = 0, find the values of the real constants a and b.

b Show that y = ekx(Ax + B), where A, B and k are all real constants, is a solution

of the differential equation d2y

dx2− 2k

dy

dx+ k2y = 0.

17 a Given that y = Ax2e−3x is a solution of the differential equation

d2y

dx2+ 6

dy

dx+ 9y = 10e−3x, find the value of A.

b If y = Ax2e−kx is a solution of the differential equation

d2y

dx2+ 2k

dy

dx+ k2y = Be−kx, show that B = 2A.

c Show that y = Äπx

sin(x) is a solution of Bessel’s equation,

4x2

d2y

dx2+ 4x

dy

dx+ (4x2 − 1)y = 0.

18 Adrien Marie Legendre (1752–1833) was a famous French mathematician. He made many mathematical contributions in the areas of elliptical integrals, number theory and the calculus of variations. He is also known for the differential equation named after him. Legendre’s differential equation of order n is given

by (1 − x2)d2y

dx2− 2x

dy

dx+ n(n + 1)y = 0 for ∣x ∣ < 1, and the solutions of the

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differential equation are given by the polynomials Pn(x). The first few polynomials are given by

P0(x) = 1

P1(x) = x

P2(x) = 12(3x2 − 1)

P3(x) = 12(5x3 − 3x)

P4(x) = 18(35x4 − 30x2 + 3)

a Verify the solution of the Legendre’s differential equation for the cases when n = 3 and n = 4.

b The Legendre polynomials also satisfy many other mathematical properties.

One such relation is Pn(x) = 12n n!

dn

dxn[(x2 − 1)n]. Use this result to

obtain P2(x) and P3(x).

c The Legendre polynomials also satisfy 3

1

−1

Pn(x)Pm(x) dx = 0 when m ≠ n and

3

1

−1

(Pn(x))2dx = 22n + 1

. Verify these results for P2(x) and P3(x).

Solving Type 1 differential equations, dydx

= f(x)

Classifying solutions to a differential equationThe solution of a differential equation is usually obtained by the process of integration. Because the integration process produces an arbitrary constant of integration, the solutions of a differential equation are classified as follows.

A general solution is one which contains arbitrary constants of integration and satisfies the differential equation.

A particular solution is one which satisfies the differential equation and some other initial value condition, also known as a boundary value, that enable the constant(s) of integration to be found.

In general, the number of arbitrary constants of integration to be found is equal to the order of the differential equation. Throughout this course we study and solve special types of first- and second-order differential equations.

type 1 differential equations, dydx

= f(x)Direct integrationIn this section we solve first-order differential equations of the form

dy

dx= f(x),

y(x0) = y0. Differential equations of this form can be solved by direct integration. Hence, it is necessary to be familiar with all the integration techniques studied so far.

Antidifferentiating both sides gives y = 3f(x) dx + c. This is the general solution,

9.3



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which can be thought of as a family of curves. If we use the given condition x = x0 when y = y0, we can determine the value of the constant of integration c in this particular case, which thus gives us the particular solution.

a Find the general solution to dy

dx+ 12x = 0.

b Find the particular solution of dy

dx+ 6x2 = 0, y(1) = 2.

tHinK WritE

a 1 Rewrite the equation to make dy

dx the subject. a

dydx

+ 12x = 0

dydx

= −12x

2 Antidifferentiate to obtain y. y = −312xdx

3 Write the general solution in terms of a constant. y = −6x2 + c

b 1 Rewrite the equation to make dy

dx the subject. b

dydx

+ 6x2 = 0

dydx

= −6x2

2 Antidifferentiate to obtain y. y = −36x2dx

3 Express y in terms of x with an arbitrary constant. y = −2x3 + c

4 Substitute and use the given conditions to determine the value of the constant.

y(1) = 2:⇒ x = 1 when y = 2

2 = −3(1)2 + cc = 5

5 Substitute back for c and state the particular solution. y = 5 − 3x2

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finding particular solutionsIn Chapter 9, linear substitutions were used to integrate linear expressions. The example presented here is a review of this process.

Solve the differential equation (4 − 3x)2

dy

dx+ 1 = 0, y(1) = 2.

tHinK WritE

1 Rewrite the equation to make dy

dx the subject. (4 − 3x)2

dydx

+ 1 = 0

(4 − 3x)2 dydx

= −1

dydx

= −1(4 − 3x)2

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2 Antidifferentiate to obtain y. y = 3−1

(4 − 3x)2 dx

3 Use index laws to express the integrand as a function to a power.

y = −3(4 − 3x)−2dx

4 Use a linear substitution. Express dx in terms of du by inverting both sides.

Let u = 4 − 3x.dudx

= −3

dxdu

= −13

dx = −13

du

5 Substitute for u and dx. y = −3u−2 −13

du

6 Use the properties of indefinite integrals to transfer the constant factor outside the front of the integral sign.

y = 133u−2

du

7 Perform the integration process, using

3un du = 1

n + 1un + 1 with n = −2 so that

n + 1 = −1, now add in the constant + c.

y = −13

u−1 + c

y = − 13u

+ c

8 Substitute back for x. y = − 13 14 − 3x 2 + c


y(1) = 2⇒ x = 1 when y = 2

2 = −13

+ c

c = 2 + 13

c = 73

10 Substitute back for c, and state the particular solution. Although this a possible answer, this result can be simplified.

y = −13(4 − 3x)

+ 73

11 Form the lowest common denominator. y =−1 + 7 14 − 3x 2

3 14 − 3x 212 Expand the brackets in the numerator, do not

expand the brackets in the denominator.y = −1 + 28 − 21x

3 14 − 3x 2

13 Simplify and take out common factors which cancel.

y = 27 − 21x3 14 − 3x 2

y =3 19 − 7x 23 14 − 3x 2



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14 State the fi nal answer in simplifi ed form. Note the maximal domain for which the solution is valid.

y = 9 − 7x4 − 3x

for x ≠ 43

15 Note that as a check, we can use the given condition to check the value of y. This proves that we have the correct solution.

Substitute x = 1:

y = 9 − 74 − 3

= 2

stating the domain for which the solution is validAs seen in the last example, the maximal domain for which the solution is valid is important. When solving differential equations, unless the solution is defi ned for all values of x, that is for x ∈ R, we are required to state the largest subset of R for which the given differential equation and solution are valid.

Solve the differential equation !3x − 5 dy

dx+ 6 = 0, y(7) = 2, stating the

largest domain for which the solution is valid.

tHinK WritE


dx the subject. !3x − 5

dydx

+ 6 = 0

!3x − 5 dydx

= −6

dydx

= −6

!3x − 5


!3x − 5 dx

3 Use the properties of indefi nite integrals to transfer the constant factor outside the front of the integral sign.

y = −631

!3x − 5 dx

4 Use index laws to express the integrand, as a function to a power.

y = −63(3x − 5)−1

2dx


Let u = 3x − 5.dudx

= 3

dxdu

= 13

dx = 13 du

6 Substitute for u and dx. y = −63u−1

2 13du


y = −23u−12

du

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8 Perform the integration process using

3un du = 1

n + 1un + 1 with n = −1

2, so that n + 1 = 1

2,

and add in the constant + c.

y = −u 12 + c

y = −4!u + c

9 Substitute back for x. y = −4!3x − 5 + c


y(7) = 2⇒ x = 7 when y = 2

2 = −4!16 + cc = 18

11 Substitute back for c and state the particular solution. y = 18 − 4!3x − 5

12 Determine the domain for which the solution is valid from the differential equation.

dy

dx= −6

!3x − 5 for

3x − 5 > 0

13 Solve the inequality for x to state the largest domain for which the solution is valid for the given differential equation. State the answer.

3x > 5

x > 53

The solution y = 18 − 4!3x − 5 is

valid for x > 53.

solving fi rst-order differential equations involving inverse trigonometric functions

The results 31

"a2 − x2 dx = sin−1ax

ab + c, 3

−1

"a2 − x2 dx = cos−1ax

ab + c and

31

a2 + x2 dx = 1

a tan−1ax

ab + c are important and are used throughout this chapter.

Solve the differential equation "16 − x2 dy



tHinK WritE


dx the subject. "16 − x2

dy

dx+ 2 = 0, y(0) = 0

"16 − x2 dydx

= −2

dydx

= −2

"16 − x2


"16 − x2 dx

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3−1


ab + c.

y = 2 cos−1ax4b + c


y(0) = 0⇒ x = 0 when y = 00 = 2 cos−1(0) + cc = −2 cos−1(0)

c = −2 × π2

c = −π

5 Substitute back for c and state the particular solution.

y = 2 cos−1ax4b − π

6 Determine the domain for which the solution is valid from the differential equation.

y = 3−2

"16 − x2 dx

"16 − x2 > 0 x2 < 16

7 Solve the inequality for x to state the largest domain for which the solution and the differential equation is valid. State the answer.

∣x ∣ < 4The solution y = 2 cos−1ax

4b − π is valid

for ∣x ∣ < 4.


= f(x)

1 a WE4 Find the general solution of dy

dx+ 12x3 = 0.

b Find the particular solution of dy

dx+ 6x = 0, y(2) = 1.

2 a Find the general solution of dy

dx+ 12 cos(2x) = 0.

b Solve the differential equation dy

dx+ 6 sin(3x) = 0, y(0) = 0, and express y in

terms of x.

3 WE5 Solve the differential equation (5 − 4x)2 dy

dx+ 1 = 0, y(1) = 2.

4 Solve the differential equation (7 − 4x)

dy

dx+ 2 = 0, y(2) = 3.

5 WE6 Solve the differential equation !2x − 5 dy



6 Solve the differential equation !x

dy

dx+ 2 = 0, y 14 2 = 3, expressing y in terms

of x, and state the largest domain for which the solution is valid.

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7 WE7 Solve the differential equation "64 − x2 dy

dx− 6 = 0, y(4) = 0, stating the


8 Solve the differential equation (16 + x2)

dy

dx+ 4 = 0, y(4) = π

4, stating the largest

domain for which the solution is valid.

9 Find the general solution to each of the following

a dy

dx− 4x = 3 b

dy

dx− (3x − 5)(x + 4) = 0

c e2x

dy

dx+ 6 = 2e4x d "x2 + 9

dy

dx− x = 0

For questions 10–18, solve each of the differential equations given and state the maximal domain for which the solution is valid.

10 a 3x dy

dx− 2x2 = 5, y(1) = 3 b

dy

dx= 6(e−3x + e3x), y(0) = 0

11 a dy

dx− 4 sin(2x) = 0, y(0) = 2 b

dy

dx+ 6 cos(3x) = 0, yaπ

2b = 5

12 a dy

dx− 8 sin2(2x) = 0, y(0) = 0 b

dy

dx− 12 cos2(3x) = 0, y(0) = 0

13 a dy

dx= 1

!4x + 9, y(0) = 0 b

dy

dx+ 1

3 − 2x= 0, y(2) = 1

14 a dy

dx= 1

(3x − 5)2, y(2) = 3 b

dy

dx= 8

7 − 4x, y(2) = 5

15 a (x2 + 9)dy

dx− 3x = 0, y(0) = 0 b "x2 + 4

dy

dx+ x = 0, y(0) = 0

16 a (x2 + 6x + 13)

dy

dx− x = 3, y(0) = 0 b (x2 − 4x + 9)

dy

dx+ x = 2, y(0) = 0

17 a sec(2x)dy

dx+ sin3(2x) = 0, y(0) = 0 b csc(3x)

dy

dx+ 9 cos2(3x) = 0, y(0) = 0

18 a dy

dx+ loge (2x) = 4, ya1

2b = 1 b ex

dy

dx+ x = 5, y(0) = 0

19 Solve the following differential equations and state the maximal domain for which the solution is valid.

a (4x2 + 9)

dy

dx+ 2x = 3, y(0) = 0 b "9 − 4x2

dy

dx+ 2x = 3, y(0) = 0

20 a If a > 0 and b ≠ 0, solve the following differential equations, stating the maximal domains for which the solution is valid.

i "a2 − x2 dy

dx+ b = 0, y(0) = 0 ii (a2 − x2)

dy

dx+ b = 0, y(0) = 0

iii (a + bx)2 dy

dx+ 1 = 0, y(0) = 0

b Solve the differential equation e2x

dy

dx+ cos(3x) = 0, y(0) = 0.

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= f(y)

Invert, integrate and transposeSolving fi rst-order differential equations of the form

dy

dx= f(y), y(x0) = y0 is studied in

this section. In this situation it is not possible to integrate directly. The fi rst step in the solution process is to invert both sides.

From dxdy

= 1dy

dx

, we obtain dxdy

= 1f(y)

.

Integrate both sides with respect to y to obtain

x = 31

f(y) dy + c.

This gives the general solution. The initial condition can be used to fi nd the value of the constant c. The resulting equation must be rearranged to express y in terms of x, which gives the particular solution.

finding general solutionsFinding a general solution means fi nding the solution in terms of an arbitrary constant.

9.4

Find the general solution to the differential equation dy

dx− 4!y = 0.

tHinK WritE


dx the subject.

dydx

− 4!y = 0

dydx

= 4!y

2 Invert both sides. dxdy

= 14!y

3 Integrate both sides. x = 31

4!y dy


x = 143

1!y

dy

5 Use index laws to express the integrand, as a power.

x = 143y

−12dy

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UNCORRECTED PAGE P

ROOFS


3undu = 1n + 1

un + 1 with n = −12, so that

n + 1 = 12, and add in the constant of

integration.

x = 14

× 21y

12 + c

7 Simplify. x = 12y

12 + c

x = 12!y + c

8 Transpose to make y the subject. 12!y = x − c

!y = 2x − 2c

9 Since c is a constant, 2c is also a constant. Let A = 2c.!y = 2x − A

10 Square both sides and state the answer in terms of an arbitrary constant A.

y = (2x − A)2

finding particular solutionsFinding particular solutions involves solving the differential equation and expressing y in terms of x, then fi nding the value of the constant of integration.

Solve the differential equation dy

dx+ (4 − 3y)2 = 0, y(2) = 1.

tHinK WritE


dx the subject.

dydx

+ (4 − 3y)2 = 0

dydx

= −(4 − 3y)2


= − 1(4 − 3y)2

3 Integrate both sides. x = 3−1

(4 − 3y)2 dy

4 Use index laws to express the integrand as a function to a power.

x = −3(4 − 3y)−2dy

5 Use a linear substitution. Express dy in terms of du by inverting both sides.

Let u = 4 − 3y.

dudy

= −3

dy

du= −1

3

dy = −13

du

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UNCORRECTED PAGE P

ROOFS

6 Substitute for u and dy. x = −3u−2 −13

du

7 Use the properties of indefinite integrals to transfer the constant factor outside the front of the integral sign.

x = 133u−2

du


3undu = 1n + 1

un + 1 with n = −2, so that

n + 1 = −1, and add in the constant + c.

x = −13 u−1 + c

x = − 13u

+ c

9 Substitute back for y. x = − 13(4 − 3y)

+ c


y(2) = 1⇒ x = 2 when y = 1

2 = −13

+ c

c = 2 + 13

c = 73

11 Substitute back for c. x = − 13(4 − 3y)

+ 73

12 To begin making y the subject, transpose the equation.

13(4 − 3y)

= 73

− x

13 Form a common denominator on the right-hand side.

13(4 − 3y)

= 7 − 3x3

14 Cancel the common factor and invert both sides.

4 − 3y = 17 − 3x

15 Rearrange to make y the subject. 3y = 4 − 17 − 3x

16 Express the right-hand side of the equation with a common denominator.

3y =4(7 − 3x) − 1

7 − 3x

17 Expand the brackets in the numerator. 3y = 28 − 12x − 14 − 3x

18 Simplify and take out the common factor. 3y = 27 − 12x7 − 3x

3y =3(9 − 4x)

7 − 3x19 State the final answer in a simplified form

and state the maximal domain.y = 9 − 4x

7 − 3x for x ≠ 7

3



UNCORRECTED PAGE P

ROOFS

find c or rearrange to make y the subject?When solving these types of differential equations, it is necessary to fi nd the constant of integration and also rearrange to make y the subject. Sometimes the order in which we do these operations can make the processes simpler.


dx+ 4y = 0, y(0) = 3.

tHinK WritE


dx the subject.

dydx

+ 4y = 0

dydx

= −4y


= − 14y

3 Integrate both sides. x = −314y

dy

4 Take the constant factor outside the front of the integral sign. x = −143

1y

dy

5 Use 31u

du = loge ∣u ∣ + c to express x in terms of y and the

constant of integration c.

x = −14 loge a ∣y ∣ b + c

From this point forward, we have two processes to complete: fi nd c, and transpose the equation to make y the subject.

Method 1: Find c fi rst, then transpose to make y the subject.


y(0) = 3⇒ x = 0 when y = 3

0 = −14

loge a ∣3 ∣ b + c

c = 14

loge (3)

7 Substitute back for c and take out the common factor. x = −14 loge a ∣y ∣ b + 1

4 loge (3)

x = 14c loge (3) − loge a ∣y ∣ b d

8 Use the logarithm laws to simplify the expression. x = 14 loge q 3

∣y ∣r

4x = loge q 3∣y ∣

r

9 Use the defi nition of the logarithm. e4x = 3∣y ∣

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UNCORRECTED PAGE P

ROOFS

10 Invert both sides again in attempting to make y the subject. ∣y ∣3

= 1e4x

= e−4x

11 Because e−4x > 0, the modulus is not needed. State the particular solution to the differential equation.

y = 3e−4x

Method 2: Make y the subject and then fi nd the constant c.

6 Rearrange to make y the subject. x = −14 loge a ∣y ∣ b + c

14 loge a ∣y ∣ b = c − x

loge a ∣y ∣ b = 4c − 4x

7 Since c is a constant, 4c is also a constant. Let B = 4c.

loge a ∣y ∣ b = B − 4x

8 Use the defi nition of the logarithm. ∣y ∣ = eB − 4x

∣y ∣ = eBe−4x

9 Since B is a constant, eB is also a constant. Let A = eB.

∣y ∣ = Ae−4x


y(0) = 3⇒ x = 0 when y = 3

3 = Ae−0

3 = A

11 Because e−4x > 0, the modulus is not needed. Substitute for A and state the particular solution to the differential equation.

y = 3e−4x

stating the domain for which the solution is validAs discussed in the previous section, the solution to a differential equation should include the largest domain for which the solution is valid.

Solve the differential equation 2 dy

dx+ "16 − y2 = 0, y(0) = 0, stating the


tHinK WritE


dx

the subject.2

dydx

+ "16 − y2 = 0

2

dydx

= −"16 − y2

dydx

= −"16 − y2

2

2 Invert both sides.dxdy

= −2

"16 − y2




UNCORRECTED PAGE P

ROOFS

3 Integrate with respect to y. x = 3−2

"16 − y2 dy


3−1


ab + c.

x = 2 cos−1ay

4b + c


y(0) = 0⇒ x = 0 when y = 0

0 = 2 cos−1(0) + cc = −2 cos−1(0)

c = −2 × π2

c = −π

6 Substitute back for c. x = 2 cos−1ay

4b − π

7 Rewrite the equation. 2 cos−1ay

4b = x + π

cos−1ay

4b = x + π

2

8 Take cosine of both sides to make y the subject.

y

4= cosax

2+ π

2b

y = 4 cosax2

+ π2b

9 Expand using trigonometric compound angle formulas.

y = 4acosax2bcosaπ

2b − sinax

2bsinaπ

2bb

y = 4acosax2b × 0 − sinax

2b × 1b

10 State the particular solution. y = −4 sinax2b

11 Determine the domain for which the solution is valid.

cos−1ay

4b = x + π

2The range of y = cos−1(x) is [0, π],but ∣y ∣ < 4, so

0 < x + π2

< π

12 Solve the inequality for x to state the largest domain for which the solution is valid. State the answer.

0 < x + π < 2π−π < x < π

The solution y = −4 sinax2b is valid for −π < x < π.



UNCORRECTED PAGE P

ROOFS


= f(y)

1 WE8 Find the general solution to the differential equation !y dy

dx+ 4 = 0.

2 Find the general solution to the differential equation dy

dx− tan(2y) = 0.

3 WE9 Solve the differential equation dy

dx+ (5 − 4y)2 = 0, y(1) = 2.

4 Solve the differential equation dy

dx+ 4y − 7 = 0, y(0) = 3.


dx+ 3y = 0, y(0) = 5.

6 Given the differential equation dy

dx− 5y = 0, y(0) = 3, express y in terms of x.

7 WE11 Solve the differential equation "(64 − y2) − 6 dy

dx= 0, y(0) = 8, stating the


8 Solve the differential equation 16 + y2 − 4 dy

dx= 0, y(0) = 0, stating the largest


9 Find the general solution to each of the following.

a dy

dx=

y2

4b

dy

dx= y + 4 c

dy

dx=

y

4d

dy

dx= 4

y2

10 Solve each of the following differential equations.

a dy

dx+ 5y = 0, y(0) = 4 b

dy

dx− 3y = 0, y(1) = 2

11 a dy

dx+ 2y = 5, y(0) = 3 b

dy

dx− 3y + 4 = 0, y(0) = 2

For questions 12–18, solve each of the differential equations given, and where appropriate state the largest domain for which the solution is valid.

12 a dy

dx= !y, y(1) = 4 b

dy

dx= y2, y(1) = 3

13 a dy

dx= 4e2y, y(2) = 0 b

dy

dx+ 6e3y = 0, y(1) = 0

14 a dy

dx= (5 − 2y)2, y(1) = 3 b

dy

dx+ (7 − 3y)2 = 0, y(3) = 2

15 a dy

dx+ 6 cosecay

2b = 0, ya1

3b = 0 b

dy

dx= 2 sec(2y), ya1

8b = π

12

16 a dy

dx− !4y + 9 = 0, y(0) = 0 b

dy

dx− 4y2 = 9, y(0) = 0

17 a dy

dx+ 4y = y2, y(0) = 3 b

dy

dx− 3y = y2, y(0) = 6

18 a dy

dx+ 7y = y2 + 12, y(0) = 0 b

dy

dx− 6y − y2 = 8, y(0) = 0

ExErCisE 9.4

PraCtisE

ConsolidatE



UNCORRECTED PAGE P

ROOFS

19 a If k and y0 are constants, solve the differential equation dy

dx+ ky = 0, y(0) = y0.

b Given that a, b and c are constants, solve the differential equations

i dy

dx+ ay = b, y(0) = c ii

dy

dx+ ay = by2, y(0) = c.

20 a Given that a and b are constants, solve the differential equations

i dy

dx= (ay + b)2, y(0) = 0 ii

dy

dx= b2y2 + a2, y(0) = 0.

b If a and b are constants with a > b > 0:

i solve the differential equation dy

dx= ( y + a)( y + b), y(0) = 0

ii fi nd limx→∞

y(x).


= f(x)g(y )

separation of variables

Differential equations of the form dy

dx= f(x)g(y), y(x0) = y0 are called variables

separable equations, as it is possible to separate all the x terms onto one side of the equation and all the y terms onto the other side of the equation.

For dy

dx= f(x)g(y), divide both sides by g(y), since g(y) ≠ 0. This gives

1g(y)

dy

dx= f(x).

Integrate both sides of the equation with respect to x.

31

g(y) dy

dx dx = 3f(x) dx

Thus, 31

g(y) dy + c1 = 3f(x) dx + c2.

31

g(y) dy = 3f(x) dx + c, since c = c2 − c1.

After performing the integration, an implicit relationship between x and y is obtained. However, in specifi c cases it may be possible to rearrange to make y the subject.

MastEr

9.5

Find the general solution to the differential equation dy

dx= x + 4y2 + 4

.

tHinK WritE

1 Write the differential equation.dy

dx= x + 4

y2 + 4




UNCORRECTED PAGE P

ROOFS

2 Separate the variables and integrate both sides. 3(y2 + 4) dy = 3(x + 4) dx

3 Perform the integration and add the constant on one side only.

13y3 + 4y = 1

2x2 + 4x + c

4 The general solution is given as an implicit equation, as in this case it is impossible to solve this equation explicitly for y.

13y3 + 4y − 1

2x2 − 4x = c

finding particular solutionsFinding particular solutions involves solving the differential equation, expressing y in terms of x where possible, and then fi nding the value of the constant of integration.


dx+ y = 6x2y, y(0) = 1.

tHinK WritE


dx the subject.

dy

dx+ y = 6x2y

dy

dx= 6x2y − y

2 Factor the RHS.dy

dx= y(6x2 − 1)

3 Separate the variables and integrate both sides. 31y

dy = 3(6x2 − 1)dx

4 Perform the integration and add in the constant on one side only.

loge a ∣y ∣ b = 2x3 − x + c


y(0) = 1⇒ x = 0 when y = 1

loge a ∣1 ∣ b = 0 + c c = 0

6 Substitute back for c and use defi nition of a logarithm to state the solution explicitly as y in terms of x. Note that the modulus is not needed, as e2x3−x > 0.

loge a ∣y ∣ b = 2x3 − x y = e2x3−x


stating the domain for which the solution is validAs previously stated, when solving differential equations it is necessary to state the largest domain for which the solution is valid.



UNCORRECTED PAGE P

ROOFS


dx+ 2x"16 − y2 = 0, y(0) = 4, stating the


tHinK WritE


dx

the subject.

dy

dx+ 2x"16 − y2 = 0, y(0) = 4

dy

dx= −2x"16 − y2

2 Separate the variables and integrate both sides.

3−1

"16 − y2 dy = 32xdx

3 Perform the integration and add the constant on one side only.

cos−1ay

4b = x2 + c


y(0) = 4⇒ x = 0 when y = 4

cos−1(1) = c c = 0

5 Substitute back for c. cos−1ay

4b = x2

6 Take cosine of both sides to make y the subject.

y4

= cos(x2)

y = 4 cos(x2)

7 Determine the domain for which the solution is valid.

cos−1 ay

4b = x2

The range of y = cos−1(x) is [0, π], but x ≠ 0 and 1

"16 − y2 is defi ned for ∣y ∣ < 4, so 0 < x2 < π

8 Solve the inequality for x to state the largest domain for which the solution is valid. State the answer.

The solution y = 4 cos(x2) is valid for 0 < x < !π.



= f(x)g(y )

1 WE12 Find the general solution to the differential equation dy

dx= x + 2

y3 + 8.

2 Obtain an implicit relationship of the form f(x, y) = c for dy

dx=

y2 + 4

x2y2.


dx− y = 3x2y, y(0) = 1.

4 Given the differential equation dy

dx+ y2 = 2xy2, y(2) = 1, express y in terms of x.

ExErCisE 9.5

PraCtisE



UNCORRECTED PAGE P

ROOFS


dx− 2x"64 − y2 = 0, y(0) = 0, stating the


6 Solve the differential equation 2 dy

dx− x(16 + y2), y(0) = 0, stating the largest


7 Obtain an implicit relationship of the form f(x, y) = c for each of the following differential equations.

a dy

dx= x2 + 4

y2 + 4b

dy

dx=

xy

y2 + 4c

dy

dx=

x2y2

y2 + 4d

dy

dx=

xy2ex2

y3 + 8

For questions 8–16, solve each of the given differential equations and express y in terms of x.

8 a dy

dx−

y2

x= 0, y(1) = 1 b

dy

dx+ 12y2 sin(4x) = 0, y(π) = 1

9 a dy

dx+ x

y= 0, y(1) = 2 b

dy

dx+ 6y2x2 = 0, y(1) = 3

10 a dy

dx+ 18x3y2 = 0, y(−1) = 2 b

dy

dx−

y2

x2= 0, y(1) = 2

11 a dy

dx= y2e2x, y(0) = 1 b

dy

dx+ 12x5y2 = 0, y(1) = 2

12 a dy

dx+ y = 3x2y, y(0) = 1 b

dy

dx+ 6x2y2 = y2, y(−1) = 2

13 a dy

dx+ 2xy2 = y2, y(1) = 2 b

dy

dx+ 8x3y4 = y4, y(0) = 1

14 a x

dy

dx+ 2y = y2, y(1) = 1 b x

dy

dx− 4y = y2, y(1) = 1

15 a (4 + x2)

dy

dx− 2xy = 0, y(0) = 1 b

y2 + 4

x2 + 9−

yx

dy

dx= 0, y(0) = 2

16 a dy

dx− x(25 + y2) = 0, y(0) = 0 b

dy

dx+ 4x"25 − y2 = 0, y(0) = 5

17 For each of the following, use the substitution v =yx to show that

dy

dx= v + x

dvdx

,

and hence reduce to a separable differential equation and find the solution.

a x dy

dx+ 3y = 4x, y(2) = 1 b x

dy

dx− y = 4x, y(1) = 2

18 Use the substitution v =yx to show that

dy

dx= v + x

dvdx

. Hence, reduce the

differential equation x dy

dx+ ay = bx to a separable differential equation and find

the general solution to x dy

dx+ ay = bx for the cases when:

a a = −1 b a ≠ −1.

ConsolidatE

MastEr



UNCORRECTED PAGE P

ROOFS

Solving Type 4 differential equations, d

2ydx2

= f(x)

Integrate twiceIn this section, solutions of second-order differential equations of the form

d

2y

dx2= f(x)

are required. This type of differential equation can be solved by direct integration,

since d

2y

dx2= d

dxady

dxb. Integrating both sides with respect to x gives

dy

dx= 3f(x) dx + c1.

This is now in the Type I form and can be solved by direct integration.

Finding a general solution involves giving the solution in terms of two arbitrary constants, which we usually denote as c1 and c2.

9.6

Find the general solution to the differential equation d2y

dx2+ 36x2 = 0.

tHinK WritE

1 Rewrite the equation to make d2y

dx2 the subject.

d

2y

dx2+ 36x2 = 0

d

2y

dx2= −36x2

2 Integrate both sides with respect to x.dy

dx= 3−36x2dx

3 Perform the integration.dy

dx= −12x3 + c1

4 Integrate both sides again with respect to x.dy

dx= 3(−12x3 + c1) dx

5 Perform the integration and state the general solution in terms of two arbitrary constants.

y = −3x4 + c1x + c2


finding particular solutions

To solve d2y

dx2= f(x) and obtain a particular solution, we need two sets of initial

conditions to fi nd the two constants of integration. These are usually of the form y(x0) = y0 and y′(x1) = y1.

Solve the differential equation d

2y

dx2+ 36x = 0, y(1) = 3, y′(1) = 2.

tHinK WritE

1 Rewrite the equation to make d2y

dx2 the subject.

d

2y

dx2+ 36x = 0

d

2y

dx2= −36x




UNCORRECTED PAGE P

ROOFS

2 Integrate both sides with respect to x.dydx

= 3−36x dx

= −18x2 + c1

3 Substitute and use the given condition to determine the value of the fi rst constant of integration.

y′(1) = 2

⇒ dy

dx= 2 when x = 1

2 = −18 + c1

c1 = 20

4 Substitute back for c1.dy

dx= −18x2 + 20

5 Integrate both sides again with respect to x. y = 3(−18x2 + 20) dx

6 Perform the integration. y = −6x3 + 20x + c2

7 Substitute and use the given condition to determine the value of the second constant of integration.

y(1) = 3⇒ y = 3 when x = 1

3 = −6 + 20 + c2

c2 = −11

8 Substitute back for c2 and state the particular solution. y = −6x3 + 20x − 11

simplifying the answerWe have seen earlier that answers can often be given in a simplifi ed form.

Solve the differential equation d

2y

dx2+ 2

(2x + 9)3= 0, y(0) = 0, y′(0) = 0.

tHinK WritE

1 Rewrite the equation to make d

2y

dx2 the subject.

d

2y

dx2+ 2

(2x + 9)3= 0

d

2y

dx2= −2

(2x + 9)3


dx= 3

−2(2x + 9)3

dx

3 Transfer the constant factor outside the front of the integral and use index laws to express the integrand as a function to a power.

dy

dx= −23(2x + 3)−3 dx


Let u = 2x + 9.dudx

= 2

dxdu

= 12

dx = 12

du




UNCORRECTED PAGE P

ROOFS

5 Substitute for u and dx, and simplify.dy

dx= −23u−31

2 du

dy

dx= −3u−3du

6 Perform the integration, adding in the first constant of integration.

dy

dx= 1

2u−2 + c1

dy

dx= 1

2(2x + 9)2+ c1

7 Use the given condition to find the value of the first constant of integration.

y′(0) = 0⇒ when x = 0,

dy

dx= 0

0 = 1162

+ c1

c1 = − 1162

8 Substitute back for c1.dy

dx= 1

2 12x + 9 22 − 1162

9 Integrate both sides again with respect to x. y = 3a 12(2x + 9)2

− 1162

b dx

10 Simplify the integrand. y = 3a 12(2x + 9)2

b dx − x162

11 Use the substitution u = 2x + 9 again y = 3a12

u−2b 12

du − x162

y = 143

u−2du − x162

12 Perform the integration and add in the second constant of integration.

y = −14

u−1 − x162

+ c2

13 Substitute back for u. y = − 14(2x + 9)

− x162

+ c2

14 Substitute and use the given condition, to determine the value of the second constant of integration.

y(0) = 0⇒ y = 0 when x = 0

0 = − 136

+ c2

c2 = 136

15 Substitute back for c2 and state the particular solution. Although this is a possible answer, this result can be simplified.

y = − 14(2x + 9)

− x162

+ 136

16 Form the lowest common denominator. y =81 − 2x(2x + 9) + 9(2x + 9)

324(2x + 9)



UNCORRECTED PAGE P

ROOFS

17 Expand and simplify the numerator y = 81 − 4x2 − 18x + 18x + 81324(2x + 9)

y = −4x2

324(2x + 9)

18 State the particular solution in simplest form. y = −x2

81(2x + 9)

Beam defl ectionsOne application of the Type 4 differential equations,

d

2y

dx2= f(x),

is called beam defl ection. A cantilever or a beam can be fi xed at one end and have a weight at the other end. The weight at the unfi xed end causes the beam to bend so that the downwards defl ection, y, at a distance x measured along the beam from the fi xed point satisfi es a differential equation of this type. In this situation the maximum defl ection occurs at the end of the beam.

Another type of beam defl ection is the case of a beam fi xed at both ends. The weight of the beam causes the beam to bend so that the downwards defl ection, y, at a distance x measured along the beam from the fi xed point satisfi es a differential equation of this type. In this situation we can show that the maximum defl ection occurs in the middle of the beam.

y

x

y

x

A beam of length 2L rests with its end on two supports at the same horizontal level. The downward deflection, y, from the horizontal satisfies

the differential equation d

2y

dx2= kx(x − 2L) for 0 ≤ x ≤ 2L, where x is the

horizontal distance from one end of the beam and k is a constant related to the stiffness and bending moment of the beam.

a Find the deflection, y, in terms of x and show that the maximum deflection occurs in the middle of the beam.

b Find the maximum deflection of the beam.

tHinK WritE

a 1 Expand. ad2y

dx2= kx(x − 2L)

= k(x2 − 2Lx)


dx= k 3(x2 − 2Lx) dx

3 Perform the integration.dy

dx= k ax3

3− Lx2 + c1b




UNCORRECTED PAGE P

ROOFS

4 Since the beam is fixed at both ends, x = 0 when y = 0, and y = 0 when x = 2L. We cannot determine the first constant of integration at this stage. Integrate both sides with respect to x again.

y = k 3ax3

3− Lx2 + c1b dx

5 Perform the integration. y = k ax4

4− Lx3

3+ c1x + c2b

6 To find the second constant of integration, c2, use x = 0 when y = 0.

Substitute x = 0 when y = 0:c2 = 0

7 To find the first constant of integration, c1, use y = 0 when x = 2L and simplify.

Substitute y = 0 when x = 2L:

0 = k a(2L)4

4−

L(2L)3

3+ 2Lc1b

0 = k a16L4

4− 8L4

3+ 2L c1b

8 Solve for the first constant and substitute back. Simplify the result by taking a common denominator. This gives the deflection, y, in terms of x.

c1 = 2L3

3

y = k ax4

4− Lx3

3+ 2L3x

3b

y = k12

(x4 − 4Lx3 + 8L2x)

9 Find the first derivative.dy

dx= k

12(4x3 − 12Lx2 + 8L2)

= k3

(x3 − 3Lx2 + 2L3)

10 To show that the maximum deflection occurs in the

middle of the beam, show that dy

dx= 0 when x = L.

Substitute x = L:dy

dx= k

3(L3 − 3L3 + 2L3)

= 0So the maximum deflection occurs in the middle of the beam.

b 1 To find the maximum deflection, substitute x = L into the result for y.

b ymax = y(L) = k12

(L3 − 4L3 + 8L3)

= 5L3k12

2 State the maximum deflection of the beam. The maximum deflection occurs in the

middle of the beam and is 5L3k12

.



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Solving Type 4 differential equations, d

2ydx2

= f(x)

1 WE15 Find the general solution to the differential equation d

2y

dx2+ 30x4 = 0.

2 Find the general solution to the differential equation d

2y

dx2+ 36 sin(3x) = 0.

3 WE16 Solve the differential equation d

2y

dx2+ 24x2 = 0, y(−1) = 2, y′(−1) = 3.

4 Solve the differential equation d

2y

dx2+ 12 sin(2x) = 0, yaπ

4b = 4, y′aπ

4b = 6.

5 WE17 Solve the differential equation d

2y

dx2+ 12

(3x + 16)3= 0, y(0) = 0, y′(0) = 0.

6 Solve the differential equation d

2y

dx2+ 12

"(2x + 9)3= 0, y(0) = 0, y′(0) = 1

7 WE18 A beam of length L has both ends simply supported at the same horizontal level and the downward deflection, y, satisfies the differential equation d

2y

dx2= k(x2 − Lx) for 0 ≤ x ≤ L where k is a constant.

a Find the deflection, y, in terms of x and show that the maximum deflection occurs in the middle of the beam.

b Find the maximum deflection of the beam.

8 A cantilever of length L is rigidly fixed at one end and in the unstrained position is horizontal. If a load is added at the free end of the beam, the downward deflection, y, at a distance x along the beam satisfies the differential equation d

2y

dx2= k(L − x) for 0 ≤ x ≤ L where k is a constant. Find the deflection, y, in

terms of x and hence find the maximum deflection of the beam.

9 Find the general solution to each of the following.

a x3

d

2y

dx2+ 4 = 0 b

d

2y

dx2+ (x + 4)(2x − 5) = 0

c x3

d

2y

dx2+ 2x − 5 = 0 d e3x

d

2y

dx2+ 5 = 2e2x

For questions 10–14, solve each of the given differential equations.

10 a d2y

dx2+ 6x = 0, y(1) = 2, y(2) = 3

b d2y

dx2+ 24x2 = 0, y(1) = 2, y(2) = 3

11 a d2y

dx2+ 8(e2x + e−2x) = 0, x = 0,

dy

dx= 0, y = 0

b ex

d2y

dx2+ 4e−2x = 5, x = 0,

dy

dx= 0, y = 0

ExErCisE 9.6

PraCtisE

ConsolidatE



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12 a d

2y

dx2+ 64 sin(4x) = 0, y(0) = 4, y′(0) = 8

b d

2y

dx2+ 27 cos(3x) = 0, yaπ

6b = 3, y′aπ

6b = 9

13 a d2y

dx2+ 32 sin2(2x) = 0, y(0) = 0, y′(0) = 0

b d2y

dx2+ 16 cos2(4x) = 0, y(0) = 0, y′(0) = 0

14 a d2y

dx2= 1

(3x + 2)3, y(0) = 0, y′(0) = 0

b d2y

dx2+ 1

"(2x + 9)3= 0, y(0) = 0, y′(0) = 0

15 a At all points on a certain curve, the rate of change of gradient is constant. Show that the family of curves with this property are parabolas.

b At all points on a certain curve, the rate of change of the gradient is −12. If the curve has a turning point at (−2, 4), find the equation of the particular curve.

16 a At all points on a certain curve, the rate of change of the gradient is proportional to the x-coordinate, Show that the family of curves with this property are cubics.

b At all points on a certain curve, the rate of change of the gradient is 18x. If the curve has a turning point at (–2, 0), find the equation of the particular curve.

17 a Solve d2y

dx2+ 20

!4x + 9= 0, y(0) = 0 and y′(0) = 0.

b Solve d2y

dx2+ 16

(4x + 9)2= 0, y(0) = 0 and y′(0) = 0.

18 a A diving board of length L is rigidly fixed at one end and has a girl of weight W standing at the free end. The downward deflection, y, measured at a distance x along the beam satisfies the differential equation

E I d2y

dx2= W

2(L − x)2

for 0 ≤ x ≤ L .

The deflection and inclination to the horizontal are both zero at the fixed end, and the product E I is a constant related to the stiffness of the beam. Find the formula for y in terms of x and determine the maximum deflection of the beam.

b A uniform beam of length L carries a load of W per unit length and has both ends clamped horizontally at the same horizontal level. The downward deflection, y, measured at any distance x from one end along the beam satisfies the differential equation

E I d2y

dx2= W

2ax2 − Lx + L2

6b for 0 ≤ x ≤ L



UNCORRECTED PAGE P

ROOFS

where W, E and I are constants. Prove that the maximum deflection occurs in the middle of the beam, and determine the maximum deflection of the beam.

19 If a and b are positive real constants, find the particular solution to each of the following differential equations.

a d2y

dx2+ 1

(ax + b)3= 0, y(0) = 0 and y′(0) = 0

b d2y

dx2+ 1

(ax + b)2= 0, y(0) = 0 and y′(0) = 0

20 a i Show that ddxB x

"9 + 4x2R = 9

"(9 + 4x2)3.

ii Hence, find the general solution to d2y

dx2+ 9

"(9 + 4x2)3= 0.

b i If a and b are positive real constants, show that

ddxB x

"a + bx2R = a

"(a + bx2)3.

ii Hence, find the general solution to d2y

dx2+ 1

"(a + bx2)3= 0.

MastEr



UNCORRECTED PAGE P

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studyON is an interactive and highly visual online tool that helps you to clearly identify strengths and weaknesses prior to your exams. You can then con� dently target areas of greatest need, enabling you to achieve your best results.

ONLINE ONLY 9.7 Review www.jacplus.com.au

the Maths Quest review is available in a customisable format for you to demonstrate your knowledge of this topic.

the review contains:• short-answer questions — providing you with the

opportunity to demonstrate the skills you have developed to ef� ciently answer questions using the most appropriate methods

• Multiple-choice questions — providing you with the opportunity to practise answering questions using CAS technology

• Extended-response questions — providing you with the opportunity to practise exam-style questions.

a summary of the key points covered in this topic is also available as a digital document.

REVIEW QUESTIONSDownload the Review questions document from the links found in the Resources section of your eBookPLUS.


Units 3 & 4 <Topic title to go here>

Sit topic test


UNCORRECTED PAGE P

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9 AnswersExErCisE 9.21 Check with your teacher.

2 Check with your teacher.

3 −5, 2

4 ±3


6 −2


8 a a = 0, b = −1, c = 1

b a = 1, b = −6, c = 18, d = −24

c Check with your teacher.

d –2, 5

9 a, b Check with your teacher.

c −6, 1

d −2 ± 3i

10 a−c Check with your teacher.

d ± 311−14 Check with your teacher.

15 a Check with your teacher.

b a = 0, b = 1

16 a a = 6, b = 9

b Check with your teacher.

17 a 5

b, c Check with your teacher.


ExErCisE 9.31 a y = c − 3x4 b y = 13 − 3x2

2 a y = c − 6 sin(2x) b y = 2(cos(3x) − 1)

3 y = 9x − 114x − 5

, x ≠ 54

4 y = 3 + 12 loge a ∣4x − 7 ∣ b , x ≠ 7

4

5 y = 1 − !2x − 5, x > 52

6 y = 11 − 4!x, x > 0

7 y = 6 sin−1ax8b − π, ∣x ∣ < 8

8 y = π2

− tan−1ax4b

9 a y = 2x2 + 3x + c b y = x3 + 72x2 − 20x + c

c y = e2x + 3e−2x + c d y = "x2 + 9 + c

10 a y = 13c5 loge a ∣x ∣ b + x2 + 8 d , x ≠ 0

b y = 2(e3x − e−3x)

11 a y = 4 − 2 cos(2x) b y = 3 − 2 sin(3x)

12 a y = 4x − sin(4x) b y = 6x + sin(6x)

13 a y = 12(!4x + 9 − 3), x > −9

4

b y = 1 + 12 loge a ∣2x − 3 ∣ b , x ≠ 3

2

14 a y = 10x − 173x − 5

, x ≠ 53

b y = 5 − 2 loge a ∣7 − 4x ∣ b , x ≠ 74

15 a y = 32 loge ax2 + 9

9b

b y = 2 − "x2 + 4

16 a y = 12 loge ax2 + 6x + 13

13b

b y = loge q 3

"x2 − 4x + 9r

17 a y = −18sin4(2x)

b y = cos3(3x) − 1

18 a y = 5x − x loge a2 ∣x ∣ b − 32, x ≠ 0

b y = (x − 4)e−x + 4

19 a y = 12

tan−1a2x3b + 1

4 loge a 9

4x2 + 9b

b y = 32asin−1a2x

3b − 1b + "9 − 4x2

2, ∣x ∣ < 2

3

20 a i y = −b sin−1axab , ∣x ∣ < a

ii y = b2a

loge a ∣ a − xa + x ∣ b , ∣x ∣ < a

iii y = −xa(a + bx)

, x ≠ −ab

b y = e−2x

13(2 cos(3x) − 3 sin(3x) − 2)

ExErCisE 9.41 y = "3 (B − 6x)2

2 y = 12sin−1(Be2x)

3 y = 15x − 1312x − 11

, x ≠ 1112

4 y = 14(7 + 5e−4x)

5 y = 5e−3x

6 y = 3e5x

7 y = 8 cosax6b , −6π < x < 0

8 y = 4 tan(x), −π2

< x < π2



UNCORRECTED PAGE P

ROOFS

9 a y = 4c − x

b y = Aex − 4

c y = Ae

x

4 d y = "3 12x + A

10 a y = 4e−5x b y = 2e3x−3

11 a y = 12(5 + e−2x) b y = 2

3(2 + e3x)

12 a y = 14(x + 3)2 b y = 3

4 − 3x, x ≠ 4

313 a y = −1

2 loge (17 − 8x), x < 17

8

b y = −13 loge (18x − 17), x > 17

18

14 a y = 5x − 82x − 3

, x ≠ 32

b y = 7x − 233x − 10

, x ≠ 103

15 a y = 2 cos−1(2x), ∣x ∣ ≤ 13

b y = 12sin−1(4x), ∣x ∣ ≤ 1

4

16 a y = x2 + 3x

b y = 32 tan(6x), − π

12< x < π

12

17 a y = 12

3 + e4xb y = 6

3e−3x − 2

18 a y =12(ex + 1)

4ex + 3

b y =4(1 − e2x)

e2x − 2, x ≠ loge (!2)

19 a y = y0ekx

b i y = ac − babe−ax + b

a

ii y = ac(a − bc)eax + bc

20 a i y = b2x1 − abx

, x ≠ 1ab

ii y = ab

tan(abx)

b i y =ab(1 − e−(a − b)x)

ae−(a − b)x − bii −a

ExErCisE 9.5

1 y4

4+ 8y − x2

2− 2x = c

2 y − 2 tan−1ay

2b + 1

x= c

3 y = ex3 + x

4 y = 1

3 + x − x2, x ≠ 1 ± !13

2

5 y = 8 sin(x2), 0 < x < !2π2

6 y = 4 tan(x2), 0 < x < !2π2

7 a 13y3 + 4y − 1

3x3 − 4x = c

b 12y2 + 4 loge a ∣y ∣ b − 1

2x2 = c

c y − 4y

− 13x3 = c

d 12y2 − 8

y− 1

2ex2 = c

8 a y = 1

1 − loge a ∣x ∣ b, x ≠ 0

b y = 14 − 3 cos(4x)

9 a y = "5 − x2, ∣x ∣ < !5

b y = 3

6x3 − 5

10 a y = 2

9x4 − 8b y = 2x

2 − x

11 a y = 2

3 − e2x, x ≠ loge (!3)

b y = 2

4x6 − 3

12 a y = ex3−x b y = 2

4x3 − 2x + 3

13 a y = 2

2x2 − 2x + 1b y = 1

"3 6x4 − 3x + 1

14 a y = 2

1 − x2, x ≠ ±1 b y = 4x4

5 − x4, x ≠ ±"4 5

15 a y = 14(x4 + 4) b y = 2

3"2x2 + 9

16 a y = 5 tana5x2

2b , ∣x ∣ ≤ !5π

5

b y = 5 cos(2x2), ∣x ∣ ≤ !2π2

17 a y = x − 8

x3, x ≠ 0

b y = 2xa1 + 2 loge a ∣x ∣ b b , x ≠ 0

18 a y = xac + b loge a ∣x ∣ b b , x ≠ 0

b y = bxa + 1

+ cxa

ExErCisE 9.61 y = c2 + c1x − x6

2 y = c2 + c1x + 4 sin(3x)

3 y = −2x4 − 5x − 1

4 y = 3 sin(2x) + 6x + 1 − 3π2

5 y = −3x2

128(3x + 16), x ≠ −16

3

6 y = 12!2x + 9 − 3x − 36, x > −92

7 y = k12

(x4 − 2Lx2 + L3x), 5kL4

192

8 y = k6

(3Lx2 − x3), kL3

3



UNCORRECTED PAGE P

ROOFS

9 a y = c2 + c1x − 2x

, x ≠ 0

b y = c2 + c1x + 10x2 − x3

2− x4

6

c y = c2 + c1x + 2 loge a ∣x ∣ b + 52x

, x ≠ 0

d y = c2 + c1x + 2e−x − 59e− 3x

10 a y = −x3 + 8x − 5

b y = −2x4 + 31x − 27

11 a y = 4 − 2e2x − 2e −2x

b y = 5e −x − 49

e −3x + 11x3

− 419

12 a y = 4 sin(2x) − 8x + 4

b y = 3 cos(3x) + 18x − 3π + 3

13 a y = 1 − cos(4x) − 8x2

b y = 18 cos(8x) − 4x2 − 1

8

14 a y = x2

8(3x + 2), x ≠ −2

3

b y = !2x + 9 − x3

− 3, x > −92


b y = −6x2 − 24x − 20


b y = 3x3 − 36x − 48

17 a y = 30x + 45 − 53"(4x + 9)3, x > −9

4

b y = loge a∣4x + 9 ∣

9b − 4x

9, x ≠ −9

4

18 a y = W24E I

(6L2x2 − 4Lx3 + x4), WL4

8E I

b y = W x2

24E I(x − L)2,

WL4

384E I

19 a y = −x2

2b2(ax + b), x ≠ −b

a

b y = 1

a2 loge a

∣ax + b ∣b

b − xab

, x ≠ −ba

20 a i Check with your teacher.

ii y = c2 + c1x − 14"9 + 4x2

b i Check with your teacher.

ii y = c2 + c1x − 1ab

"a + bx2



UNCORRECTED PAGE P

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differential nso qu ei at - john wiley & sons nso qu ei at 9.1 kick off with cas 9.2 verifying...

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