Download Differential equations

FS
O
PR
O
PA
G
E
9
R
R
EC
TE
D
Differential
equations
U
N
C
O
9.1 Kick off with CAS
9.2 Verifying solutions to a differential equation
9.3 Solving Type 1 differential equations,
9.4 Solving Type 2 differential equations,
9.5 Solving Type 3 differential equations,
9.6 Solving Type 4 differential equations,
dy
dx
dy
dx
dy
dx
= f (x)
= f (y)
= f (x)g (y)
d 2y
dx2
= f (x)
9.7 Review
c09DifferentialEquations.indd 456
04/07/15 12:30 PM
9.1 Kick off with CAS
U
N
C
O
R
R
EC
TE
D
PA
G
E
PR
O
O
FS
<To Come>
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.
c09DifferentialEquations.indd 457
04/07/15 12:30 PM
9.2
Verifying solutions to a differential equation
Classification of differential equations
PR
O
O
FS
A differential equation (d.e.) is an equation involving derivatives. It is of the form
dy d2y
gax, y, , 2, .....b = 0.
dx dx
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:
dy
d 2y
dy
(a) = ky
(b) a 2 + b
+ cy = 0
dx
dx
dx
d 2y
(d) 2 + n2y = 0
dx
dx
3
b +3
dy
dx
+ 5y = 0
PA
G
(e) xa
dy
E
‥ − tx# + 2x = t
(c) x
(f) D3t x = "x2 + 1
U
N
C
O
R
R
EC
TE
D
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
d 2x
d 2x
as ‥
x = 2 and D2t x = 2 .
dt
dt
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 equations
To 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).
458 Maths Quest 12 SPECIALIST MATHEMATICS VCE Units 3 and 4
c09DifferentialEquations.indd 458
04/07/15 12:30 PM
Verify that y = x3 is a solution of the differential equation
x3
d 2y
dx2
tHinK
−a
dy
dx
2
b + 3xy = 0.
WritE
y = x3
dy
= 3x2
dx
1 Use basic differentiation to find the first derivative.
d 2y
2 Find the second derivative.
dx2
x3
into the LHS of the differential equation.
4 Simplify and expand, so that LHS = RHS = 0, thus proving
the given solution does satisfy the differential equation.
d 2y
dy 2
b + 3xy
dx
dx2
= x3 × (6x) − (3x2) 2 + 3x × (x3)
−a
PR
O
3 Substitute for y, the first derivative and second derivative
= 6x
FS
1
O
WorKeD
eXaMPLe
= 6x4 − 9x4 + 3x4
=0
Given that y = ekx is a solution of the differential equation
2
d 2y
dy
dx
− 8y = 0, find the values of the real constant k.
EC
dx2
−2
TE
D
WorKeD
eXaMPLe
PA
G
E
Differential equations involving unknowns
When 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 satisfies the differential equation.
tHinK
WritE
1 Use the rule for differentiation of exponential
R
d kx
(e ) = kekx, to find the first derivative.
dx
R
functions,
C
O
2 Differentiate again to find the second derivative.
U
N
3 Substitute for y, the first derivative
d 2y
dy
dx
and the second
into the given differential equation.
dx2
4 Take out the common factor.
derivative
5 Factorise the quadratic equation involving
the unknown.
6 Solve the resulting equation for the unknown and
state the answer.
y = ekx
dy
= kekx
dx
d 2y
dx2
= k2ekx
d 2y
dy
− 8y = 0
dx
dx2
k2ekx − 2kekx − 8ekx = 0
−2
ekx 1 k2 − 2k − 8 2 = 0
ekx ≠ 0 ⇒ k2 − 2k − 8 = 0
(k − 4)(k + 2) = 0
When k = 4 or k = −2, then y = ekx is a
solution of
d 2y
dx2
−2
dy
dx
− 8y = 0.
topic 9 DIfferentIaL eQuatIons
c09DifferentialEquations.indd 459
459
04/07/15 12:30 PM
solutions to a differential equation involving products
When 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.
Verify that y = xe −2x is a solution of the differential equation
d 2y
dx
2
+4
dy
dx
+ 4y = 0.
tHinK
WritE
y = xe−2x
dy
d
d
= x (e−2x) + e−2x (x)
dx
dx
dx
dy
= −2xe−2x + e−2x
dx
1 Use the product rule for differentiation to
dy
2 Simplify the first derivative by taking out the
dx
3 Find the second derivative, using the product
rule again.
d 2y
dx2
TE
D
d 2y
4 Simplify the second derivative by taking out
EC
the common factor.
5 Substitute for y, the first derivative
dx2
d
d
(1 − 2x) + (1 − 2x) (e−2x)
dx
dx
= −2e−2x − 2(1 − 2x)e−2x
d 2y
dy
+
4
+ 4y
dx
dx2
= e−2x (4x − 4) + 4e−2x (1 − 2x) + 4xe−2x
and
dx
into the LHS of
O
R
dx2
the differential equation.
= e−2x
d 2y
= e−2x (−2 − 2(1 − 2x))
2
dx
= e−2x (4x − 4)
R
the second derivative
d2 y
dy
= e−2x (1 − 2x)
PA
G
common factor.
PR
O
O
find the first derivative.
FS
3
E
WorKeD
eXaMPLe
= e−2x [(4x − 4) + 4(1 − 2x) + 4x]
= e−2x [4x − 4 + 4 − 8x + 4x]
=0
C
6 Take out the common factor and simplify,
U
N
so that LHS = RHS = 0, thus proving
the given solution does satisfy the
differential equation.
ExErCisE 9.2 Verifying solutions to a differential equation
PraCtisE
1
WE1
x2
Verify that y = x2 is a solution of the differential equation
d 2y
dx2
−a
dy
dx
2
b + 2y = 0.
2 For the differential equation x4
solution.
460
d 2y
dx2
−a
dy
dx
2
b + 4x2y = 0, show that y = x4 is a
Maths Quest 12 sPeCIaLIst MatheMatICs VCe units 3 and 4
c09DifferentialEquations.indd 460
04/07/15 12:30 PM
Given that y = ekx is a solution of the differential equation
d 2y
dy
+
3
− 10y = 0, find the values of the real constant k.
dx
dx2
d2 y
+ 9y = 0, find the
4 If y = cos(kx) is a solution of the differential equation
2
dx
values of the real constant k.
3
WE2
5
Verify that y = xe3x is a solution of the differential equation
d2 y
dy
−
6
+ 9y = 0.
dx
dx2
WE3
dx2
7 a Verify that y = x satisfies the differential equation x
4
b If y = 2x − 3x + 5, show that a
2
c Given the differential equation
dy
dx
2
dy
dx2
2
d2 y
dx2
PR
O
4
b − 8y + 31 = 0.
− 2x
dy
dx
E
Consolidate
+ 4y = 8 sin(2x), find the value of the real constant A.
O
d2 y
FS
6 Given that y = Ax cos(2x) is a solution of the differential equation
−a
dy
dx
2
b + 4x2y = 0.
+ 6y + 6x2 = 0, show that
3x
+ 1 is a solution.
2
d If y = ax3 + bx2 where a and b are constants, show that
d2 y
dy
x2 2 − 4x + 6y = 0.
dx
dx
TE
D
PA
G
y = x3 − 3x2 −
8 aFind the constants a, b and c if y = a + bx + cx2 is a solution of the differential
d2 y
U
N
C
O
R
R
EC
dx2
+2
dy
+ 4y = 4x2.
dx
b Determine the constants a, b, c and d if y = ax3 + bx2 + cx + d is a solution of
d2 y
dy
+ y = x3.
the differential equation 2 + 2
dx
dx
c Show that y = xn is a solution of the differential equation
d2 y
dy 2
x2y 2 − x2 a b + ny2 = 0.
dx
dx
d2 y
dy
−
2x
− 10y = 0 has a solution y = xn find the
d The differential equation x2
2
dx
dx
possible values of n.
equation
9 a Given that x = e3t + e−4t show that
d2x dx
+
− 12x = 0.
dt
dt2
b If y = Ae3x + Be−3x where A and B are constants, show that
c Find the values of real constant k such that y = ekx satisfies
d2 y
+5
d2 y
dx2
− 9y = 0.
dy
− 6y = 0.
dx
dx2
d2 y
dy
+ 4 + 13y = 0
d Find the values of m where m ∈ C if y = emx satisfies
2
dx
dx
Topic 9 Differential ­equations c09DifferentialEquations.indd 461
461
04/07/15 12:30 PM
d2 y
+ 4y = 0.
dx2
b Show that y = A sin(3x) + B cos(3x), where A and B are constants, is a solution
d2 y
of the differential equation 2 + 9y = 0.
dx
c If y = a sin(nx) + b cos(nx), where n is a positive real number, show that
d2 y
+ n2y = 0.
dx2
d2 x
d Given that x = a sin(pt) satisfies
+ 9x = 0, find the value of p.
dt2
d2 y
dy
2
− 2x − 2y = 0.
11 a Show that y = ex satisfies the differential equation
2
dx
dx
2
b Verify that y = cos(x ) satisfies the differential equation
d2y dy
x 2−
+ 4x3y = 0.
dx
dx
PR
O
O
FS
10 aIf y = 3 sin(2x) + 4 cos(2x), show that
d2 y
2
dx
+x
dy
dx
− y = 0.
PA
G
(x2 + 1)
E
c If y = ax + b"x2 + 1 where a and b are constants, show that
d Given that y = loge (x + "x2 − 9), show that (x2 − 9)
d2 y
dx2
+x
dy
dx
= 0.
d2 y
TE
D
12 aShow that y = tan(ax), where a ∈ R \ {0 }, satisfies the differential
= 2a2y(1 + y2).
dx2
b Verify that y = tan2 (ax), where a ∈ R \ {0 }, is a solution of the differential
d2 y
equation 2 = 2a2 (3y2 + 4y + 1).
dx
c Show that y = loge (ax + b), where a, b ∈ R, is a solution of the differential
d2 y
equation 2 + a2e−2y = 0.
dx
13 aVerify that y = tan−1 (2x) is a solution of the differential equation
d2 y
dy
(1 + 4x2) 2 + 8x = 0.
dx
dx
b Show that y = sin−1 (3x) is a solution of the differential equation
d2 y
dy
(1 − 9x2) 2 − 9x = 0.
dx
dx
U
N
C
O
R
R
EC
equation
x
4
c Verify that y = cos−1 a b is a solution of the differential equation
(16 −
462 x2)
d2 y
dx2
−x
dy
dx
= 0.
Maths Quest 12 SPECIALIST MATHEMATICS VCE Units 3 and 4
c09DifferentialEquations.indd 462
04/07/15 12:30 PM
14 aA 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
mg
v=
(1 − e−kt)
k
dy
+ 13y = 0.
dx
dx2
b Find the real constants a and b if x = t(a cos(3t) + b sin(3t)) is a solution of the
d2 x
differential equation 2 + 9x = 6 cos(3t).
dt
E
equation
−6
PR
O
d2 y
O
15 a
Verify that y = e3x cos(2x) satisfies the differential
FS
at a time t where g and k are constants. Show that
dv
+ kv = mg.
dt
b In a transient circuit, the current i amperes at a time
t seconds is given by i = 3e−2t sin(3t).
d2 i
di
Show that
+ 4 + 13i = 0.
2
dt
dt
d2 y
dy
TE
D
+ by = 0, find the values of the real constants a and b.
dx
dx
b Show that y = ekx (Ax + B), where A, B and k are all real constants, is a solution
2
+a
PA
G
16 aGiven that y = xe−3x is a solution of the differential equation
of the differential equation
2
dx
− 2k
dy
dx
+ k2y = 0.
17 aGiven that y = Ax2e−3x is a solution of the differential equation
d2 y
dy
EC
Master
d2 y
2
+6
+ 9y = 10e−3x, find the value of A.
R
dx
dx
b If y = Ax2e−kx is a solution of the differential equation
U
N
C
O
R
d2 y
2
dx
+ 2k
dy
dx
+ k2y = Be−kx, show that B = 2A.
π
sin(x) is a solution of Bessel’s equation,
Äx
d2 y
dy
4x2 2 + 4x + (4x2 − 1)y = 0.
dx
dx
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
d2 y
dy
+ n(n + 1)y = 0 for ∣ x ∣ < 1, and the solutions of the
by (1 − x2) 2 − 2x
dx
dx
c Show that y =
Topic 9 Differential ­equations c09DifferentialEquations.indd 463
463
04/07/15 12:31 PM
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
FS
when n = 3 and n = 4.
b The Legendre polynomials also satisfy many other mathematical properties.
obtain P2 (x) and P3 (x).
O
1 dn 2
[(x − 1) n]. Use this result to
n
n
2 n! dx
1
PR
O
One such relation is Pn (x) =
c The Legendre polynomials also satisfy 3 Pn (x)Pm (x) dx = 0 when m ≠ n and
1
Solving Type 1 differential equations,
dy
= f(x)
dx
TE
D
9.3
PA
G
−1
−1
2
. Verify these results for P2 (x) and P3 (x).
2n + 1
E
2
3 (Pn (x)) dx =
Classifying solutions to a differential equation
U
N
C
O
R
R
EC
The 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,
dy
= f(x)
dx
Direct integration
dy
In this section we solve first-order differential equations of the form
= f(x),
dx
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,
464 Maths Quest 12 SPECIALIST MATHEMATICS VCE Units 3 and 4
c09DifferentialEquations.indd 464
04/07/15 12:31 PM
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.
WorKeD
eXaMPLe
4
a Find the general solution to
dy
+ 12x = 0.
dx
b Find the particular solution of
dy
+ 6x2 = 0, y(1) = 2.
dx
WritE
dy
dx
a
the subject.
dy
+ 12x = 0
dx
dy
= −12x
dx
O
a 1 Rewrite the equation to make
FS
tHinK
PR
O
y = −312xdx
2 Antidifferentiate to obtain y.
y = −6x2 + c
3 Write the general solution in terms of a constant.
dx
b
the subject.
2 Antidifferentiate to obtain y.
y = −36x2dx
y = −2x3 + c
TE
D
3 Express y in terms of x with an arbitrary constant.
y(1) = 2:
⇒ x = 1 when y = 2
2 = −3(1) 2 + c
c=5
4 Substitute and use the given conditions to determine
EC
the value of the constant.
dy
+ 6x2 = 0
dx
dy
= −6x2
dx
E
dy
PA
G
b 1 Rewrite the equation to make
y = 5 − 3x2
R
R
5 Substitute back for c and state the particular solution.
U
N
C
O
finding particular solutions
In Chapter 9, linear substitutions were used to integrate linear expressions. The
example presented here is a review of this process.
WorKeD
eXaMPLe
5
Solve the differential equation (4 − 3x) 2
tHinK
1 Rewrite the equation to make
dy
dx
+ 1 = 0, y(1) = 2.
WritE
dy
dx
the subject.
dy
+1=0
dx
dy
= −1
(4 − 3x) 2
dx
dy
−1
=
dx (4 − 3x) 2
(4 − 3x) 2
topic 9 DIfferentIaL eQuatIons
c09DifferentialEquations.indd 465
465
04/07/15 12:31 PM
2 Antidifferentiate to obtain y.
y=3
3 Use index laws to express the integrand as a
y = −3 (4 − 3x) −2dx
function to a power.
y = −3u−2 −1
du
3
6 Use the properties of indefinite integrals to
y = 133u−2 du
1 n+1
n
with n = −2 so that
3u du = n + 1u
n + 1 = −1, now add in the constant + c.
TE
D
8 Substitute back for x.
y = −13 u−1 + c
PA
G
7 Perform the integration process, using
E
transfer the constant factor outside the front of
the integral sign.
9 Substitute and use the given conditions to
y=−
1
+c
3u
y=−
1
+c
3 1 4 − 3x 2
y(1) = 2
⇒ x = 1 when y = 2
2 = −13 + c
c = 2 + 13
c = 73
O
R
R
EC
determine the value of the constant.
y=
11 Form the lowest common denominator.
y=
10 Substitute back for c, and state the particular
U
N
C
solution. Although this a possible answer, this
result can be simplified.
12 Expand the brackets in the numerator, do not
expand the brackets in the denominator.
13 Simplify and take out common factors
which cancel.
7
−1
+
3(4 − 3x) 3
−1 + 7 1 4 − 3x 2
3 1 4 − 3x 2
−1 + 28 − 21x
y=
3 1 4 − 3x 2
y=
y=
466 PR
O
5 Substitute for u and dx.
FS
of du by inverting both sides.
Let u = 4 − 3x.
du
= −3
dx
dx
1
=−
3
du
1
dx = − du
3
O
4 Use a linear substitution. Express dx in terms
−1
dx
(4 − 3x) 2
27 − 21x
3 1 4 − 3x 2
3 1 9 − 7x 2
3 1 4 − 3x 2
Maths Quest 12 SPECIALIST MATHEMATICS VCE Units 3 and 4
c09DifferentialEquations.indd 466
04/07/15 12:31 PM
y=
14 State the final answer in simplified form.
Note the maximal domain for which the
solution is valid.
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.
9 − 7x
4
for x ≠
4 − 3x
3
Substitute x = 1:
9−7
y=
=2
4−3
6
Solve the differential equation !3x − 5
dy
+ 6 = 0, y(7) = 2, stating the
dx
largest domain for which the solution is valid.
E
WorKeD
eXaMPLe
PR
O
O
FS
stating the domain for which the solution is valid
As seen in the last example, the maximal domain for which the solution is valid is
important. When solving differential equations, unless the solution is defined 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.
WritE
dx
the subject.
EC
2 Antidifferentiate to obtain y.
dy
TE
D
1 Rewrite the equation to make
PA
G
tHinK
3 Use the properties of indefinite integrals to transfer the
R
R
constant factor outside the front of the integral sign.
O
4 Use index laws to express the integrand, as a function
C
to a power.
5 Use a linear substitution. Express dx in terms of du by
U
N
inverting both sides.
dy
+6=0
dx
dy
= −6
!3x − 5
dx
dy
−6
=
dx
!3x − 5
!3x − 5
y=3
−6
dx
!3x − 5
y = −63
1
dx
!3x − 5
y = −63 (3x − 5) 2dx
−1
Let u = 3x − 5.
du
=3
dx
dx 1
=
du 3
dx = 13 du
6 Substitute for u and dx.
y = −63u
7 Use the properties of indefinite integrals to transfer the
y = −23u 2 du
constant factor outside the front of the integral sign.
−12 1
du
3
−1
topic 9 DIfferentIaL eQuatIons
c09DifferentialEquations.indd 467
467
04/07/15 12:31 PM
y = −u 2 + c
y = −4 !u + c
9 Substitute back for x.
y = −4 !3x − 5 + c
1 n+1
n
with n = −12, so that n + 1 = 12,
3u du = n + 1u
and add in the constant + c.
y(7) = 2
⇒ x = 7 when y = 2
2 = −4 !16 + c
c = 18
10 Substitute and use the given conditions to determine
the value of the constant.
dy
−6
for
dx !3x − 5
3x − 5 > 0
13 Solve the inequality for x to state the largest domain
x > 53
The solution y = 18 − 4 !3x − 5 is
valid for x > 53 .
PA
G
for which the solution is valid for the given differential
equation. State the answer.
3x > 5
PR
O
from the differential equation.
=
E
12 Determine the domain for which the solution is valid
O
y = 18 − 4 !3x − 5
11 Substitute back for c and state the particular solution.
FS
1
8 Perform the integration process using
solving first-order differential equations involving inverse
trigonometric functions
EC
1
1 −1 x
dx
=
tan a b + c are important and are used throughout this chapter.
a
a
a2 + x2
R
7
x
x
−1
dx = sin−1 a b + c, 3
dx = cos−1 a b + c and
a
a
2
2
"a − x
dy
+ 2 = 0, y(0) = 0, stating the
dx
largest domain for which the solution is valid.
Solve the differential equation "16 − x2
U
N
tHinK
C
O
WorKeD
eXaMPLe
−
x2
R
3
1
"a2
TE
D
The results 3
1 Rewrite the equation to make
WritE
dy
dx
the subject.
"16 − x2
dy
dx
+ 2 = 0, y(0) = 0
"16 − x2
2 Antidifferentiate to obtain y.
468
dy
= −2
dx
dy
−2
=
dx "16 − x2
y=3
−2
"16 − x2
dx
Maths Quest 12 sPeCIaLIst MatheMatICs VCe units 3 and 4
c09DifferentialEquations.indd 468
04/07/15 12:31 PM
x
y = 2 cos−1 a b + c
4
3 Perform the integration process using
−1
x
dx = cos−1 a b + c.
a
"a2 − x2
4 Substitute and use the given conditions to
determine the value of the constant.
6 Determine the domain for which the solution
is valid from the differential equation.
7 Solve the inequality for x to state the largest
y= 3
−2
"16 − x2
dx
"16 − x2 > 0
x2 < 16
∣x∣ < 4
x
The solution y = 2 cos−1 a b − π is valid
4
for ∣ x ∣ < 4.
PA
G
domain for which the solution and the
differential equation is valid. State the answer.
Find the general solution of
EC
PRactise
WE4
TE
D
Exercise 9.3 Solving Type 1 differential equations,
1 a
O
particular solution.
x
y = 2 cos−1 a b − π
4
PR
O
5 Substitute back for c and state the
FS
y(0) = 0
⇒ x = 0 when y = 0
0 = 2 cos−1 (0) + c
c = −2 cos−1 (0)
π
c = −2 ×
2
c = −π
E
3
dy
dx
dy
= f(x)
dx
+ 12x3 = 0.
dy
+ 6x = 0, y(2) = 1.
dx
dy
2 a Find the general solution of
+ 12 cos(2x) = 0.
dx
dy
b Solve the differential equation
+ 6 sin(3x) = 0, y(0) = 0, and express y in
dx
terms of x.
dy
3 WE5 Solve the differential equation (5 − 4x) 2
+ 1 = 0, y(1) = 2.
dx
dy
4 Solve the differential equation (7 − 4x)
+ 2 = 0, y(2) = 3.
dx
dy
5 WE6 Solve the differential equation !2x − 5
+ 1 = 0, y(3) = 0, stating the
dx
largest domain for which the solution is valid.
dy
6 Solve the differential equation !x
+ 2 = 0, y 1 4 2 = 3, expressing y in terms
dx
of x, and state the largest domain for which the solution is valid.
U
N
C
O
R
R
b Find the particular solution of
Topic 9 Differential ­equations c09DifferentialEquations.indd 469
469
04/07/15 12:31 PM
7
WE7
Solve the differential equation "64 − x2
dy
dx
− 6 = 0, y(4) = 0, stating the
U
N
C
O
R
R
EC
TE
D
PA
G
E
PR
O
O
FS
largest domain for which the solution is valid.
dy
π
8 Solve the differential equation (16 + x2)
+ 4 = 0, y(4) = , stating the largest
dx
4
domain for which the solution is valid.
9 Find the general solution to each of the following
Consolidate
dy
dy
a
− 4x = 3
b
− (3x − 5)(x + 4) = 0
dx
dx
dy
dy
c e2x
d "x2 + 9
+ 6 = 2e4x
−x=0
dx
dx
For questions 10–18, solve each of the differential equations given and state the
maximal domain for which the solution is valid.
dy
dy
10 a 3x
b
− 2x2 = 5, y(1) = 3
= 6(e−3x + e3x), y(0) = 0
dx
dx
dy
dy
π
11 a
− 4 sin(2x) = 0, y(0) = 2
b
+ 6 cos(3x) = 0, ya b = 5
2
dx
dx
dy
dy
12 a
b
− 8 sin2 (2x) = 0, y(0) = 0
− 12 cos2 (3x) = 0, y(0) = 0
dx
dx
dy
dy
1
1
13 a
=
, y(0) = 0
b
+
= 0, y(2) = 1
dx !4x + 9
dx 3 − 2x
dy
dy
8
1
14 a
=
, y(2) = 3
b
=
, y(2) = 5
2
dx (3x − 5)
dx 7 − 4x
dy
dy
15 a (x2 + 9) − 3x = 0, y(0) = 0
b "x2 + 4
+ x = 0, y(0) = 0
dx
dx
dy
dy
16 a (x2 + 6x + 13)
− x = 3, y(0) = 0 b (x2 − 4x + 9) + x = 2, y(0) = 0
dx
dx
dy
dy
17 a sec(2x) + sin3 (2x) = 0, y(0) = 0
b csc(3x) + 9 cos2 (3x) = 0, y(0) = 0
dx
dx
dy
dy
1
18 a
b ex
+ loge (2x) = 4, y a b = 1
+ x = 5, y(0) = 0
dx
2
dx
19 Solve the following differential equations and state the maximal domain for which
Master
the solution is valid.
dy
dy
+ 2x = 3, y(0) = 0
+ 2x = 3, y(0) = 0
a (4x2 + 9)
b "9 − 4x2
dx
dx
20 aIf a > 0 and b ≠ 0, solve the following differential equations, stating the
maximal domains for which the solution is valid.
dy
dy
+ b = 0, y(0) = 0
ii (a2 − x2)
+ b = 0, y(0) = 0
i"a2 − x2
dx
dx
dy
+ 1 = 0, y(0) = 0
iii(a + bx) 2
dx
dy
b Solve the differential equation e2x
+ cos(3x) = 0, y(0) = 0.
dx
470 Maths Quest 12 SPECIALIST MATHEMATICS VCE Units 3 and 4
c09DifferentialEquations.indd 470
04/07/15 12:31 PM
9.4
Solving Type 2 differential equations,
dy
= f(y)
dx
FS
Invert, integrate and transpose
dy
Solving first-order differential equations of the form
= f(y), y(x0) = y0 is studied in
dx
this section. In this situation it is not possible to integrate directly. The first step in the
solution process is to invert both sides.
dx
dx
1
1
From
= , we obtain
=
.
dy dy
dy f(y)
PR
O
x=3
O
dx
Integrate both sides with respect to y to obtain
1
dy + c.
f(y)
PA
G
E
This gives the general solution. The initial condition can be used to find the value of
the constant c. The resulting equation must be rearranged to express y in terms of x,
which gives the particular solution.
8
Find the general solution to the differential equation
EC
WorKeD
eXaMPLe
TE
D
finding general solutions
Finding a general solution means finding the solution in terms of an
arbitrary constant.
tHinK
dy
dx
the subject.
O
R
R
1 Rewrite the equation to make
U
N
C
2 Invert both sides.
3 Integrate both sides.
4 Use the properties of indefinite integrals to
transfer the constant factor outside the front of
the integral sign.
5 Use index laws to express the integrand,
as a power.
WritE
dy
dx
− 4 !y = 0.
dy
− 4 !y = 0
dx
dy
= 4 !y
dx
dx
1
=
dy 4 !y
x=3
1
dy
4 !y
x = 143
1
dy
!y
x = 143y 2dy
−1
topic 9 DIfferentIaL eQuatIons
c09DifferentialEquations.indd 471
471
04/07/15 12:31 PM
1
x = 14 × 21y2 + c
6 Perform the integration process using
1 n+1
n
with n = −12, so that
3u du = n + 1u
n + 1 = 12, and add in the constant of
integration.
1
7 Simplify.
x = 12y2 + c
8 Transpose to make y the subject.
x = 12 !y + c
10 Square both sides and state the answer in
Let A = 2c.
!y = 2x − A
FS
y = (2x − A) 2
E
terms of an arbitrary constant A.
!y = 2x − 2c
O
9 Since c is a constant, 2c is also a constant.
=x−c
PR
O
1
!y
2
9
Solve the differential equation
tHinK
dx
+ (4 − 3y) 2 = 0, y(2) = 1.
WritE
dy
the subject.
EC
1 Rewrite the equation to make
R
R
dx
O
2 Invert both sides.
U
N
C
3 Integrate both sides.
4 Use index laws to express the integrand as a
function to a power.
5 Use a linear substitution. Express dy in terms
of du by inverting both sides.
472
dy
TE
D
WorKeD
eXaMPLe
PA
G
finding particular solutions
Finding particular solutions involves solving the differential equation and expressing
y in terms of x, then finding the value of the constant of integration.
dy
+ (4 − 3y) 2 = 0
dx
dy
= −(4 − 3y) 2
dx
dx
1
=−
dy
(4 − 3y) 2
x=3
−1
dy
(4 − 3y) 2
x = −3 (4 − 3y) −2dy
Let u = 4 − 3y.
du
= −3
dy
dy
1
=−
3
du
1
dy = − du
3
Maths Quest 12 sPeCIaLIst MatheMatICs VCe units 3 and 4
c09DifferentialEquations.indd 472
04/07/15 12:31 PM
6 Substitute for u and dy.
x = −3u−2
7 Use the properties of indefinite integrals to
x = 133u−2 du
transfer the constant factor outside the front
of the integral sign.
−1
du
3
9 Substitute back for y.
x=−
1
+c
3u
x=−
1
+c
3(4 − 3y)
O
1 n+1
n
with n = −2, so that
3u du = n + 1u
n + 1 = −1, and add in the constant + c.
FS
x = −13 u−1 + c
8 Perform the integration process using
y(2) = 1
⇒ x = 2 when y = 1
PR
O
10 Substitute and use the given conditions to
determine the value of the constant.
2 = −13 + c
E
c = 2 + 13
PA
G
c = 73
x=−
11 Substitute back for c.
12 To begin making y the subject, transpose
TE
D
the equation.
13 Form a common denominator on the
EC
right-hand side.
14 Cancel the common factor and invert
R
both sides.
R
15 Rearrange to make y the subject.
7
1
+
3(4 − 3y) 3
7
1
= −x
3(4 − 3y) 3
7 − 3x
1
=
3(4 − 3y)
3
4 − 3y =
1
7 − 3x
3y = 4 −
1
7 − 3x
4(7 − 3x) − 1
7 − 3x
17 Expand the brackets in the numerator.
3y =
28 − 12x − 1
4 − 3x
18 Simplify and take out the common factor.
3y =
O
3y =
16 Express the right-hand side of the equation
U
N
C
with a common denominator.
19 State the final answer in a simplified form
and state the maximal domain.
27 − 12x
7 − 3x
3(9 − 4x)
3y =
7 − 3x
9 − 4x
7
y=
for x ≠
7 − 3x
3
Topic 9 Differential ­equations c09DifferentialEquations.indd 473
473
04/07/15 12:31 PM
find c or rearrange to make y the subject?
When solving these types of differential equations, it is necessary to find 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.
WorKeD
eXaMPLe
10
Solve the differential equation
dy
dx
+ 4y = 0, y(0) = 3.
dy
dx
dy
+ 4y = 0
dx
dy
= −4y
dx
dx
1
=−
dy
4y
the subject.
PR
O
1 Rewrite the equation to make
FS
WritE
O
tHinK
2 Invert both sides.
x = −3
E
3 Integrate both sides.
PA
G
4 Take the constant factor outside the front of the integral sign.
5 Use 3 du = loge ∣ u ∣ + c to express x in terms of y and the
u
1
1
x = −143 dy
y
x = −14 loge a ∣ y ∣ b + c
TE
D
constant of integration c.
1
dy
4y
EC
From this point forward, we have two processes to complete:
find c, and transpose the equation to make y the subject.
Method 1: Find c first, then transpose to make y the subject.
6 Substitute and use the given conditions to determine the value
O
R
R
of the constant.
U
N
C
7 Substitute back for c and take out the common factor.
8 Use the logarithm laws to simplify the expression.
9 Use the definition of the logarithm.
474
y(0) = 3
⇒ x = 0 when y = 3
1
0 = − loge a ∣ 3 ∣ b + c
4
1
c = loge (3)
4
x = −14 loge a ∣ y ∣ b + 14 loge (3)
x = 14 c loge (3) − loge a ∣ y ∣ b d
x = 14 loge q
4x = loge q
e4x =
3
∣y∣
3
r
∣y∣
3
r
∣y∣
Maths Quest 12 sPeCIaLIst MatheMatICs VCe units 3 and 4
c09DifferentialEquations.indd 474
04/07/15 12:31 PM
∣y∣
10 Invert both sides again in attempting to make y the subject.
3
=
1
e4x
= e−4x
11 Because e−4x > 0, the modulus is not needed. State the
y = 3e−4x
particular solution to the differential equation.
Method 2: Make y the subject and then find the constant c.
x = −14 loge a ∣ y ∣ b + c
6 Rearrange to make y the subject.
loge a ∣ y ∣ b = c − x
FS
1
4
O
loge a ∣ y ∣ b = 4c − 4x
Let B = 4c.
loge a ∣ y ∣ b = B − 4x
PR
O
7 Since c is a constant, 4c is also a constant.
∣ y ∣ = eB − 4x
∣ y ∣ = eBe−4x
8 Use the definition of the logarithm.
Let A = eB.
∣ y ∣ = Ae−4x
PA
G
E
9 Since B is a constant, eB is also a constant.
10 Substitute and use the given conditions to determine the value
TE
D
of the constant.
11 Because e−4x > 0, the modulus is not needed. Substitute for A
EC
and state the particular solution to the differential equation.
y(0) = 3
⇒ x = 0 when y = 3
3 = Ae−0
3=A
y = 3e−4x
O
11
U
N
C
WorKeD
eXaMPLe
R
R
stating the domain for which the solution is valid
As 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
+ "16 − y2 = 0, y(0) = 0, stating the
dx
largest domain for which the solution is valid.
tHinK
1 Rewrite the equation to make
the subject.
2 Invert both sides.
WritE
dy
dx
2
dy
+ "16 − y2 = 0
dx
dy
= −"16 − y2
2
dx
dy
−"16 − y2
=
2
dx
dx
−2
=
dy "16 − y2
topic 9 DIfferentIaL eQuatIons
c09DifferentialEquations.indd 475
475
04/07/15 12:31 PM
x=3
3 Integrate with respect to y.
dy
y(0) = 0
⇒ x = 0 when y = 0
0 = 2 cos−1 (0) + c
c = −2 cos−1 (0)
π
c = −2 ×
2
c = −π
y
x = 2 cos−1 a b − π
4
5 Substitute and use the given conditions
PR
O
to determine the value of the constant.
FS
x
dx = cos−1 a b + c.
a
"a − x
2
O
−1
2
"16 − y2
y
x = 2 cos−1 a b + c
4
4 Perform the integration process using
3
−2
6 Substitute back for c.
y
2 cos−1 a b = x + π
4
E
7 Rewrite the equation.
PA
G
y
x+π
cos−1 a b =
4
2
8 Take cosine of both sides to make y
TE
D
the subject.
9 Expand using trigonometric compound
EC
angle formulas.
R
R
O
C
U
N
solution is valid.
12 Solve the inequality for x to state
the largest domain for which the
solution is valid. State the answer.
476 x π
y = 4 cosa + b
2 2
π
π
x
x
y = 4acosa bcosa b − sina bsina b b
2
2
2
2
x
x
y = 4acosa b × 0 − sina b × 1b
2
2
10 State the particular solution.
11 Determine the domain for which the
y
x π
= cosa + b
4
2 2
x
y = −4 sina b
2
y
x+π
cos−1 a b =
4
2
The range of y = cos−1 (x) is [0, π],
but ∣ y ∣ < 4, so
x+π
0<
<π
2
0 < x + π < 2π
−π < x < π
x
The solution y = −4 sina b is valid for −π < x < π.
2
Maths Quest 12 SPECIALIST MATHEMATICS VCE Units 3 and 4
c09DifferentialEquations.indd 476
04/07/15 12:31 PM
Exercise 9.4 Solving Type 2 differential equations,
Find the general solution to the differential equation !y
WE8
2 Find the general solution to the differential equation
Solve the differential equation
4 Solve the differential equation
5
WE10
dy
dx
dy
dx
dy
dx
− tan(2y) = 0.
+ (5 − 4y) 2 = 0, y(1) = 2.
+ 4y − 7 = 0, y(0) = 3.
Solve the differential equation
6 Given the differential equation
+ 4 = 0.
dy
dx
+ 3y = 0, y(0) = 5.
O
WE9
dx
dx
− 5y = 0, y(0) = 3, express y in terms of x.
PR
O
3
dy
dy
FS
PRactise
1
dy
= f(y)
dx
dy
Solve the differential equation "(64 − y2) − 6
= 0, y(0) = 8, stating the
dx
largest domain for which the solution is valid.
dy
8 Solve the differential equation 16 + y2 − 4
= 0, y(0) = 0, stating the largest
dx
domain for which the solution is valid.
9 Find the general solution to each of the following.
dy 4
dy y2
dy
dy y
=
a
=
b
=y+4
c
=
d
dx y2
dx 4
dx
dx 4
WE11
Consolidate
PA
G
E
7
a
dy
dx
dy
TE
D
10 Solve each of the following differential equations.
+ 5y = 0, y(0) = 4
b
dy
dx
dy
− 3y = 0, y(1) = 2
U
N
C
O
R
R
EC
+ 2y = 5, y(0) = 3
b
− 3y + 4 = 0, y(0) = 2
dx
dx
For questions 12–18, solve each of the differential equations given, and where
appropriate state the largest domain for which the solution is valid.
dy
dy
12 a
= !y, y(1) = 4
b
= y2, y(1) = 3
dx
dx
dy
dy
13 a
= 4e2y, y(2) = 0
b
+ 6e3y = 0, y(1) = 0
dx
dx
dy
dy
14 a
= (5 − 2y) 2, y(1) = 3
b
+ (7 − 3y) 2 = 0, y(3) = 2
dx
dx
dy
y
dy
π
1
1
15 a
b
+ 6 coseca b = 0, ya b = 0
= 2 sec(2y), ya b =
8
12
dx
2
3
dx
dy
dy
− !4y + 9 = 0, y(0) = 0
− 4y2 = 9, y(0) = 0
16 a
b
dx
dx
dy
dy
17 a
b
+ 4y = y2, y(0) = 3
− 3y = y2, y(0) = 6
dx
dx
dy
dy
+ 7y = y2 + 12, y(0) = 0
− 6y − y2 = 8, y(0) = 0
18 a
b
dx
dx
11 a
Topic 9 Differential ­equations c09DifferentialEquations.indd 477
477
04/07/15 12:31 PM
dy
+ ky = 0, y(0) = y0.
dx
b Given that a, b and c are constants, solve the differential equations
dy
dy
i
+ ay = b, y(0) = c
ii
+ ay = by2, y(0) = c.
dx
dx
20 a Given that a and b are constants, solve the differential equations
dy
dy
i
= (ay + b) 2, y(0) = 0
ii
= b2y2 + a2, y(0) = 0.
dx
dx
b If a and b are constants with a > b > 0:
dy
i solve the differential equation
= ( y + a)( y + b), y(0) = 0
dx
ii find lim y(x).
FS
MastEr
19 a If k and y0 are constants, solve the differential equation
separation of variables
E
PR
O
Solving Type 3 differential equations,
dy
= f(x)g(y )
dx
dy
PA
G
9.5
O
x→∞
= f(x)g(y), y(x0) = y0 are called variables
dx
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.
dy
For
= f(x)g(y), divide both sides by g(y), since g(y) ≠ 0. This gives
dx
1 dy
= f(x).
g(y) dx
EC
TE
D
Differential equations of the form
1 dy
3g(y) dx dx = 3f(x) dx
1
dy + c1 = 3f(x) dx + c2.
g(y)
1
3g(y) dy = 3f(x) dx + c, since c = c2 − c1.
U
N
C
O
R
Thus, 3
R
Integrate both sides of the equation with respect to x.
WorKeD
eXaMPLe
12
After performing the integration, an implicit relationship between x and y is obtained.
However, in specific cases it may be possible to rearrange to make y the subject.
Find the general solution to the differential equation
tHinK
1 Write the differential equation.
478
dy
dx
=
x+4
.
y2 + 4
WritE
dy
dx
=
x+4
y2 + 4
Maths Quest 12 sPeCIaLIst MatheMatICs VCe units 3 and 4
c09DifferentialEquations.indd 478
04/07/15 12:31 PM
2
3 (y + 4) dy = 3 (x + 4) dx
2 Separate the variables and integrate both sides.
1 3
y
3
3 Perform the integration and add the constant
on one side only.
1 3
y
3
4 The general solution is given as an implicit
+ 4y − 12x2 − 4x = c
FS
equation, as in this case it is impossible to
solve this equation explicitly for y.
+ 4y = 12x2 + 4x + c
13
Solve the differential equation
dx
+ y = 6x2y, y(0) = 1.
WritE
dy
dx
the subject.
TE
D
1 Rewrite the equation to make
2 Factor the RHS.
dy
+ y = 6x2y
dx
dy
= 6x2y − y
dx
dy
= y(6x2 − 1)
dx
PA
G
tHinK
dy
E
WorKeD
eXaMPLe
PR
O
O
finding particular solutions
Finding particular solutions involves solving the differential equation, expressing y in
terms of x where possible, and then finding the value of the constant of integration.
EC
3 Separate the variables and integrate both sides.
1
2
3 y dy = 3 (6x − 1)dx
on one side only.
R
4 Perform the integration and add in the constant loge a ∣ y ∣ b = 2x3 − x + c
R
5 Substitute and use the given conditions to
C
O
determine the value of the constant.
U
N
6 Substitute back for c and use definition of a
logarithm to state the solution explicitly as
y in terms of x. Note that the modulus is not
3
needed, as e2x −x > 0.
y(0) = 1
⇒ x = 0 when y = 1
loge a ∣ 1 ∣ b = 0 + c
c=0
loge a ∣ y ∣ b = 2x3 − x
3
y = e2x −x
stating the domain for which the solution is valid
As previously stated, when solving differential equations it is necessary to state the
largest domain for which the solution is valid.
topic 9 DIfferentIaL eQuatIons
c09DifferentialEquations.indd 479
479
04/07/15 12:31 PM
14
dy
+ 2x"16 − y2 = 0, y(0) = 4, stating the
dx
largest domain for which the solution is valid.
Solve the differential equation
WritE
the subject.
dy
dy
dx
dx
+ 2x"16 − y2 = 0, y(0) = 4
dy
2 Separate the variables and integrate
both sides.
3 Perform the integration and add the
3
dx
−1
"16 − y
2
= −2x"16 − y2
dy = 32xdx
y
cos−1 a b = x2 + c
4
FS
1 Rewrite the equation to make
O
tHinK
PR
O
WorKeD
eXaMPLe
constant on one side only.
y(0) = 4
⇒ x = 0 when y = 4
−1
cos (1) = c
c=0
4 Substitute and use the given conditions to
PA
G
E
determine the value of the constant.
5 Substitute back for c.
6 Take cosine of both sides to make
TE
D
y the subject.
7 Determine the domain for which the
R
EC
solution is valid.
R
8 Solve the inequality for x to state the
C
O
largest domain for which the solution is
valid. State the answer.
y
cos−1 a b = x2
4
y
= cos(x2)
4
y = 4 cos(x2)
y
cos−1 a b = x2
4
The range of y = cos−1 (x) is [0, π], but x ≠ 0 and
1
is defined for ∣ y ∣ < 4, so 0 < x2 < π
2
"16 − y
The solution y = 4 cos(x2) is valid for 0 < x < !π.
U
N
ExErCisE 9.5 Solving Type 3 differential equations,
PraCtisE
480
dy
= f(x)g(y )
dx
x+2
.
dx y3 + 8
dy y2 + 4
= 2 2 .
2 Obtain an implicit relationship of the form f(x, y) = c for
dx
xy
dy
− y = 3x2y, y(0) = 1.
3 WE13 Solve the differential equation
dx
dy
4 Given the differential equation
+ y2 = 2xy2, y(2) = 1, express y in terms of x.
dx
1
WE12
Find the general solution to the differential equation
dy
=
Maths Quest 12 sPeCIaLIst MatheMatICs VCe units 3 and 4
c09DifferentialEquations.indd 480
04/07/15 12:31 PM
5
Solve the differential equation
dy
− 2x"64 − y2 = 0, y(0) = 0, stating the
dx
largest domain for which the solution is valid.
dy
6 Solve the differential equation 2
− x(16 + y2), y(0) = 0, stating the largest
dx
domain for which the solution is valid.
7 Obtain an implicit relationship of the form f(x, y) = c for each of the following
differential equations.
2
xy2ex
dy x2 + 4
dy
xy
dy
x2y2
dy
=
a
=
b
=
c
=
d
dx y3 + 8
dx y2 + 4
dx y2 + 4
dx y2 + 4
FS
Consolidate
WE14
11 a
12 a
dx
dy
dx
dy
dx
dy
+
x
= 0, y(1) = 2
y
b
b
+ 18x3y2 = 0, y(−1) = 2
= y2e2x, y(0) = 1
dy
PR
O
dx
dy
y2
= 0, y(1) = 1
x
E
10 a
dx
dy
−
PA
G
9 a
dy
b
b
+ y = 3x2y, y(0) = 1
b
TE
D
8 a
O
For questions 8–16, solve each of the given differential equations and express y in
terms of x.
b
+ 2xy2 = y2, y(1) = 2
dx
dy
14 a x
+ 2y = y2, y(1) = 1
dx
dy
15 a (4 + x2)
− 2xy = 0, y(0) = 1
dx
dy
16 a
− x(25 + y2) = 0, y(0) = 0
dx
dx
dy
dx
dy
dx
dy
dx
dy
dx
dy
+ 12y2 sin(4x) = 0, y(π) = 1
+ 6y2x2 = 0, y(1) = 3
−
y2
x2
= 0, y(1) = 2
+ 12x5y2 = 0, y(1) = 2
+ 6x2y2 = y2, y(−1) = 2
+ 8x3y4 = y4, y(0) = 1
dx
dy
b x
− 4y = y2, y(1) = 1
dx
y2 + 4 y dy
b
−
= 0, y(0) = 2
x2 + 9 x dx
dy
b
+ 4x"25 − y2 = 0, y(0) = 5
dx
y
dy
dv
17 For each of the following, use the substitution v = to show that
=v+x ,
x
Master
dx
dx
and hence reduce to a separable differential equation and find the solution.
dy
dy
a x
b x
+ 3y = 4x, y(2) = 1
− y = 4x, y(1) = 2
dx
dx
y
dy
dv
18 Use the substitution v = to show that
= v + x . Hence, reduce the
x
dx
dx
dy
+ ay = bx to a separable differential equation and find
differential equation x
dx
dy
the general solution to x + ay = bx for the cases when:
dx
a a = −1
b a ≠ −1.
U
N
C
O
R
R
EC
13 a
Topic 9 Differential ­equations c09DifferentialEquations.indd 481
481
04/07/15 12:31 PM
9.6
Solving Type 4 differential equations,
d 2y
= f(x)
dx2
Integrate twice
d 2y
= f(x)
dx2
are required. This type of differential equation can be solved by direct integration,
d 2y
dy
d dy
since 2 = a b. Integrating both sides with respect to x gives
= f(x) dx + c1.
dx dx
dx 3
dx
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.
WorKeD
eXaMPLe
15
PR
O
O
FS
In this section, solutions of second-order differential equations of the form
Find the general solution to the differential equation
WritE
d2 y
dx2
the subject.
TE
D
2 Integrate both sides with respect to x.
d 2y
+ 36x2 = 0
dx2
d 2y
= −36x2
2
dx
dy
= −36x2dx
dx 3
EC
3 Perform the integration.
+ 36x2 = 0.
PA
G
1 Rewrite the equation to make
dx
2
E
tHinK
d2 y
4 Integrate both sides again with respect to x.
dy
dx
dy
dx
= −12x3 + c1
= 3 (−12x3 + c1) dx
y = −3x4 + c1x + c2
R
5 Perform the integration and state the general
O
R
solution in terms of two arbitrary constants.
U
N
C
finding particular solutions
d2 y
To solve 2 = f(x) and obtain a particular solution, we need two sets of initial
dx
conditions to find the two constants of integration. These are usually of the
form y(x0) = y0 and y′(x1) = y1.
WorKeD
eXaMPLe
16
Solve the differential equation
tHinK
1 Rewrite the equation to make
482
d 2y
dx2
+ 36x = 0, y(1) = 3, y′(1) = 2.
WritE
d2 y
dx2
the subject.
d 2y
+ 36x = 0
dx2
d 2y
= −36x
dx2
Maths Quest 12 sPeCIaLIst MatheMatICs VCe units 3 and 4
c09DifferentialEquations.indd 482
04/07/15 12:31 PM
dy
= −36x dx
dx 3
= −18x2 + c1
3 Substitute and use the given condition to determine the
y′(1)
dy
⇒
dx
2
c1
value of the first constant of integration.
dy
4 Substitute back for c1.
dx
=2
= 2 when x = 1
= −18 + c1
= 20
= −18x2 + 20
FS
2 Integrate both sides with respect to x.
y = 3 (−18x2 + 20) dx
PR
O
O
5 Integrate both sides again with respect to x.
y = −6x3 + 20x + c2
6 Perform the integration.
y(1) = 3
⇒ y = 3 when x = 1
3 = −6 + 20 + c2
c2 = −11
7 Substitute and use the given condition to determine the
PA
G
8 Substitute back for c2 and state the particular solution.
E
value of the second constant of integration.
y = −6x3 + 20x − 11
WorKeD
eXaMPLe
17
TE
D
simplifying the answer
We have seen earlier that answers can often be given in a simplified form.
Solve the differential equation
EC
tHinK
dx2
the subject.
O
R
R
1 Rewrite the equation to make
d 2y
U
N
C
2 Integrate both sides with respect to x.
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.
4 Use a linear substitution. Express dx in terms of
du by inverting both sides.
d 2y
dx
2
+
2
= 0, y(0) = 0, y′(0) = 0.
(2x + 9) 3
WritE
d 2y
dx2
+
2
=0
(2x + 9) 3
d 2y
dx2
dy
dx
dy
dx
=
−2
(2x + 9) 3
=3
−2
dx
(2x + 9) 3
= −23 (2x + 3) −3 dx
Let u = 2x + 9.
du
=2
dx
dx 1
=
du 2
1
dx = du
2
topic 9 DIfferentIaL eQuatIons
c09DifferentialEquations.indd 483
483
04/07/15 12:31 PM
1
= −23u−3 du
dx
2
dy
5 Substitute for u and dx, and simplify.
dy
dx
dy
= −3u−3du
1
= u−2 + c1
dx 2
dy
dx
=
1
+ c1
2(2x + 9) 2
first constant of integration.
1
c1 = −162
dy
8 Substitute back for c1.
dx
EC
12 Perform the integration and add in the second
R
13 Substitute back for u.
R
constant of integration.
O
14 Substitute and use the given condition,
U
N
C
to determine the value of the second constant
of integration.
y = 3a
1
1
−
b dx
2
162
2(2x + 9)
x
1
b dx −
2
162
2(2x + 9)
x
1 −2 1
y = 3 a u b du −
2
162
2
x
1
y = 3u−2du −
4
162
x
1
y = − u−1 −
+ c2
4
162
y=−
x
1
−
+ c2
4(2x + 9) 162
y(0) = 0
⇒ y = 0 when x = 0
1
0 = −36
+ c2
1
c2 = 36
15 Substitute back for c2 and state the particular
y=−
16 Form the lowest common denominator.
y=
solution. Although this is a possible answer, this
result can be simplified.
484 y = 3a
E
TE
D
11 Use the substitution u = 2x + 9 again
1
1
−
2
162
2 1 2x + 9 2
PA
G
9 Integrate both sides again with respect to x.
10 Simplify the integrand.
=
PR
O
y′(0) = 0
dy
⇒ when x = 0,
=0
dx
1
0 = 162 + c1
7 Use the given condition to find the value of the
FS
constant of integration.
O
6 Perform the integration, adding in the first
x
1
1
−
+
4(2x + 9) 162 36
81 − 2x(2x + 9) + 9(2x + 9)
324(2x + 9)
Maths Quest 12 SPECIALIST MATHEMATICS VCE Units 3 and 4
c09DifferentialEquations.indd 484
04/07/15 12:31 PM
81 − 4x2 − 18x + 18x + 81
324(2x + 9)
2
−4x
y=
324(2x + 9)
17 Expand and simplify the numerator
y=
18 State the particular solution in simplest form.
y=
−x2
81(2x + 9)
y
FS
Beam deflections
d 2y
One application of the Type 4 differential equations, 2 = f(x),
dx
is called beam deflection. A cantilever or a beam can be fixed at
one end and have a weight at the other end. The weight at the
unfixed end causes the beam to bend so that the downwards
deflection, y, at a distance x measured along the beam from the
fixed point satisfies a differential equation of this type. In this
situation the maximum deflection occurs at the end of the beam.
Another type of beam deflection is the case of a beam fixed at
both ends. The weight of the beam causes the beam to bend so
that the downwards deflection, y, at a distance x measured along
the beam from the fixed point satisfies a differential equation
of this type. In this situation we can show that the maximum
deflection occurs in the middle of the beam.
y
x
18
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
d 2y
the differential equation
= kx(x − 2L) for 0 ≤ x ≤ 2L, where x is the
dx2
horizontal distance from one end of the beam and k is a constant related to
the stiffness and bending moment of the beam.
R
EC
WorKeD
eXaMPLe
TE
D
PA
G
E
PR
O
O
x
R
a Find the deflection, y, in terms of x and show that the maximum deflection
O
occurs in the middle of the beam.
U
N
tHinK
C
b Find the maximum deflection of the beam.
a 1 Expand.
2 Integrate both sides with respect to x.
3 Perform the integration.
WritE
a
d2 y
= kx(x − 2L)
dx2
= k(x2 − 2Lx)
dy
dx
dy
dx
= k 3 (x2 − 2Lx) dx
=ka
x3
− Lx2 + c1 b
3
topic 9 DIfferentIaL eQuatIons
c09DifferentialEquations.indd 485
485
04/07/15 12:31 PM
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 3a
y=ka
5 Perform the integration.
x4 Lx3
−
+ c1x + c2 b
4
3
Substitute x = 0 when y = 0:
c2 = 0
7 To find the first constant of integration, c1, use Substitute y = 0 when x = 2L:
(2L) 4 L(2L) 3
0 = ka
−
+ 2Lc1 b
4
3
0 = ka
c1 =
PA
G
y = ka
y=
TE
D
9 Find the first derivative.
EC
10 To show that the maximum deflection occurs in the
dy
dx
= 0 when x = L.
O
R
R
middle of the beam, show that
C
b 1 To find the maximum deflection, substitute x = L
U
N
into the result for y.
2 State the maximum deflection of the beam.
16L4 8L4
−
+ 2Lc1 b
4
3
2L3
3
E
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.
PR
O
O
y = 0 when x = 2L and simplify.
FS
6 To find the second constant of integration, c2,
use x = 0 when y = 0.
486 x3
− Lx2 + c1 b dx
3
x4 Lx3 2L3x
−
+
b
4
3
3
k 4
(x − 4Lx3 + 8L2x)
12
dy
k
= (4x3 − 12Lx2 + 8L2)
dx 12
k
= (x3 − 3Lx2 + 2L3)
3
Substitute x = L:
dy k 3
= (L − 3L3 + 2L3)
dx 3
=0
So the maximum deflection occurs in
the middle of the beam.
b ymax = y(L) =
k 3
(L − 4L3 + 8L3)
12
5L3k
12
The maximum deflection occurs in the
5L3k
.
middle of the beam and is
12
=
Maths Quest 12 SPECIALIST MATHEMATICS VCE Units 3 and 4
c09DifferentialEquations.indd 486
04/07/15 12:32 PM
Exercise 9.6 Solving Type 4 differential equations,
PRactise
1
WE15
d 2y
dx2
Find the general solution to the differential equation
2 Find the general solution to the differential equation
3
WE16
Solve the differential equation
d 2y
dx2
d 2y
dx2
= f(x)
d 2y
dx2
+ 30x4 = 0.
+ 36 sin(3x) = 0.
+ 24x2 = 0, y(−1) = 2, y′(−1) = 3.
d 2y
FS
π
π
+ 12 sin(2x) = 0, ya b = 4, y′a b = 6.
4
4
d 2y
12
5 WE17 Solve the differential equation
+
= 0, y(0) = 0, y′(0) = 0.
2
dx
(3x + 16) 3
d 2y
12
6 Solve the differential equation
+
= 0, y(0) = 0, y′(0) = 1
2
dx
"(2x + 9) 3
dx2
PR
O
O
4 Solve the differential equation
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
= k(x2 − Lx) for 0 ≤ x ≤ L where k is a constant.
2
dx
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
= k(L − x) for 0 ≤ x ≤ L where k is a constant. Find the deflection, y, in
dx2
terms of x and hence find the maximum deflection of the beam.
9 Find the general solution to each of the following.
d 2y
d 2y
+
4
=
0
+ (x + 4)(2x − 5) = 0
a x3
b
dx2
dx2
WE18
U
N
C
R
O
R
Consolidate
EC
TE
D
PA
G
E
7
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
b
d2 y
dx2
d2 y
dx2
d2 y
+ 6x = 0, y(1) = 2, y(2) = 3
+ 24x2 = 0, y(1) = 2, y(2) = 3
+ 8(e2x + e−2x) = 0, x = 0,
dy
= 0, y = 0
dx
dx2
d2 y
dy
−2x = 5, x = 0,
b ex
+
4e
= 0, y = 0
dx
dx2
11 a
Topic 9 Differential ­equations c09DifferentialEquations.indd 487
487
04/07/15 12:32 PM
b
14 a
b
d 2y
dx2
d2 y
dx2
d2 y
dx2
d2 y
dx2
d2 y
dx2
+ 64 sin(4x) = 0, y(0) = 4, y′(0) = 8
π
π
+ 27 cos(3x) = 0, ya b = 3, y′a b = 9
6
6
+ 32 sin2 (2x) = 0, y(0) = 0, y′(0) = 0
+ 16 cos2 (4x) = 0, y(0) = 0, y′(0) = 0
=
+
1
, y(0) = 0, y′(0) = 0
(3x + 2) 3
1
"(2x + 9)
3
FS
13 a
dx2
O
b
d 2y
= 0, y(0) = 0, y′(0) = 0
PR
O
12 a
15 aAt all points on a certain curve, the rate of change of gradient is constant.
EC
TE
D
PA
G
E
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 aAt 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.
d2 y
20
17 aSolve
+
= 0, y(0) = 0 and y′(0) = 0.
2
dx
!4x + 9
d2 y
16
b Solve
+
= 0, y(0) = 0 and y′(0) = 0.
dx2 (4x + 9) 2
18 aA diving board of length L is rigidly fixed at one end and has a girl of
R
R
weight W standing at the free end. The downward deflection, y, measured at
a distance x along the beam satisfies the differential equation
U
N
C
O
EI
488 d2 y
dx2
=
W
(L − x) 2 for 0 ≤ x ≤ L .
2
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
d2 y W
L2
E I 2 = ax2 − Lx + b for 0 ≤ x ≤ L
2
6
dx
Maths Quest 12 SPECIALIST MATHEMATICS VCE Units 3 and 4
c09DifferentialEquations.indd 488
04/07/15 12:32 PM
9
x
d
.
B
R=
dx "9 + 4x2
"(9 + 4x2) 3
iiHence, find the general solution to
d2 y
2
dx
+
9
PR
O
"(9 + 4x2) 3
b iIf a and b are positive real constants, show that
d2 y
E
a
x
d
.
B
R=
2
dx "a + bx
"(a + bx2) 3
2
dx
+
1
"(a + bx2) 3
= 0.
U
N
C
O
R
R
EC
TE
D
PA
G
iiHence, find the general solution to
= 0.
O
20 a iShow that
FS
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
Master
following differential equations.
d2 y
1
a
+
= 0, y(0) = 0 and y′(0) = 0
2
dx
(ax + b) 3
d2 y
1
b
+
= 0, y(0) = 0 and y′(0) = 0
2
dx
(ax + b) 2
Topic 9 Differential ­equations c09DifferentialEquations.indd 489
489
04/07/15 12:32 PM
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.
• 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.
the review contains:
• short-answer questions — providing you with the
opportunity to demonstrate the skills you have
developed to efficiently answer questions using the
most appropriate methods
• Multiple-choice questions — providing you with the
opportunity to practise answering questions using
CAS technology
FS
REVIEW QUESTIONS
Units 3 & 4
<Topic title to go here>
Sit topic test
U
N
C
O
R
R
EC
TE
D
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
confidently target areas of greatest need, enabling
you to achieve your best results.
PA
G
E
PR
O
O
Download the Review questions document from
the links found in the Resources section of your
eBookPLUS.
490
Maths Quest 12 sPeCIaLIst MatheMatICs VCe units 3 and 4
c09DifferentialEquations.indd 490
04/07/15 12:32 PM
9 Answers
11a y = 4 − 2 cos(2x)
Exercise 9.2
1 Check with your teacher.
13a y =
2 Check with your teacher.
3 −5, 2
b y =
4 ±3
3
15a y = 2 loge a
8a a = 0, b = −1, c = 1
b a = 1, b = −6, c = 18, d = −24
9a, b Check with your teacher.
c −6, 1
b y = loge q
E
d −2 ± 3i
1
PA
G
b a = 0, b = 1
TE
D
16a a = 6, b = 9
b Check with your teacher.
17a5
EC
b, c Check with your teacher.
R
R
b y = 13 − 3x2
b y = 2(cos(3x) − 1)
O
C
U
N
9a y = 2x2 + 3x + c
c y = e2x + 3e−2x + c
3
9
2x
1
b
b + loge a
3
4
4x2 + 9
"9 − 4x2
3
2x
2
, ∣x∣ <
asin−1 a b − 1b +
2
3
2
3
x
a
20a
i y = −b sin−1 a b, ∣ x ∣ < a
∣ ∣
b
a−x
loge a
b, ∣ x ∣ < a
2a
a+x
−x
a
,x≠−
iii y =
a(a + bx)
b
e−2x
(2 cos(3x) − 3 sin(3x) − 2)
13
Exercise 9.4
5
2
3
1 y = " (B − 6x) 2
1
2 y = 2 sin−1 (Be2x)
15x − 13
11
,x≠
12x − 11
12
1
4 y = 4 (7 + 5e−4x)
3 y =
7
b y = x3 + 2x2 − 20x + c
d y = "x2 + 9 + c
10a y = 3 c 5 loge a ∣ x ∣ b + x2 + 8d , x ≠ 0
1
b y =
b y =
x
a b − π, ∣ x ∣ < 8
8
π
x
− tan−1 a b
2
4
1
2
19a y = tan−1 a
ii y =
9x − 11
5
,x≠
4x − 5
4
1
7
4 y = 3 + 2 loge a ∣ 4x − 7 ∣ b , x ≠ 4
8 y =
r
b y = (x − 4)e−x + 4
15a Check with your teacher.
6 sin−1
− 4x + 9
18a y = 5x − x loge a 2 ∣ x ∣ b − 2, x ≠ 0
11−14 Check with your teacher.
6 y = 11 − 4!x, x > 0
3
"x2
b y = cos3 (3x) − 1
d±3
5 y = 1 − !2x − 5, x >
x2 + 6x + 13
b
13
17a y = −8 sin4 (2x)
10a−c Check with your teacher.
3 y =
x2 + 9
b
9
PR
O
1
16a y = 2 loge a
d –2, 5
2a y = c − 6 sin(2x)
∣ 2x − 3 ∣ b , x ≠ 32
b y = 2 − "x2 + 4
c Check with your teacher.
1a y = c − 3x4
− 3), x >
FS
7 Check with your teacher.
Exercise 9.3
b y = 6x + sin(6x)
−94
O
6 −2
7 y =
1
(!4x + 9
2
1 + 12 loge a
10x − 17
5
,x≠
3x − 5
3
7
b y = 5 − 2 loge a ∣ 7 − 4x ∣ b , x ≠ 4
14a y =
5 Check with your teacher.
18Check with your teacher.
b y = 3 − 2 sin(3x)
12a y = 4x − sin(4x)
5 y = 5e−3x
6 y = 3e5x
x
6
π
π
8 y = 4 tan(x), − < x <
2
2
7 y = 8 cosa b, −6π < x < 0
b y = 2(e3x − e−3x)
Topic 9 Differential ­equations c09DifferentialEquations.indd 491
491
04/07/15 12:32 PM
4 1 3
− x =c
y 3
8 1 2
1
d 2 y2 − − 2ex = c
y
1
8a y =
,x≠0
1 − loge a ∣ x ∣ b
c y −
b y = Aex − 4
x
3
d y = "12x + A
10a y = 4e−5x
b y = 2e3x−3
1
2
11a y = 2 (5 + e−2x)
b y = 3 (2 + e3x)
1
12a y = 4 (x + 3) 2
b y =
1
17
8
> 17
18
13a y = −2 loge (17 − 8x), x <
5x − 8
3
14a y =
,x≠
2x − 3
2
1
15a y = 2 cos−1 (2x), ∣ x ∣ ≤ 3
1
1
b y = 2sin−1 (4x), ∣ x ∣ ≤ 4
16a y = x2 + 3x
3
b y = 2 tan(6x), −
17a y =
18a y =
b y =
7x − 23
10
b y =
,x≠
3x − 10
3
π
π
<x<
12
12
12
3 + e4x
12(ex + 1)
4ex + 3
4(1 − e2x)
e2x − 2
b y =
6
3e−3x − 2
, x ≠ loge (!2)
TE
D
EC
R
O
R
2 y −
2
C
4
+ 8y −
2 tan−1
3 y = ex
3
+x
− 2x = c
y
1
a b+ =c
x
2
1 ± !13
1
,x≠
2
2
3+x−x
!2π
5 y = 8 sin(x2), 0 < x <
2
!2π
6 y = 4 tan(x2), 0 < x <
2
1
1
7a y3 + 4y − x3 − 4x = c
3
3
4 y =
b
492 1 2
y
2
2x2
b y =
2
− 2x + 1
+ 4 loge a ∣ y ∣ b − 12x2 = c
8
,x≠0
x3
2
− 2x + 3
1
3
"
6x4 − 3x + 1
4x4
4
b y =
, x ≠ ±"
5
5 − x4
2
!5π
5x2
b, ∣ x ∣ ≤
2
5
b y = 5 cos(2x2), ∣ x ∣ ≤
4x3
b y =
2
, x ≠ ±1
1 − x2
17a y = x −
i y =
x2
13a y =
3 −x
16a y = 5 tana
U
N
1
y4
12a y = ex
1
b −ax b
be +
a
a
ac
ii y =
(a − bc)eax + bc
Exercise 9.5
3
6x3 − 5
2x
2
10a y =
b y =
4
2−x
9x − 8
2
11a y =
, x ≠ loge (!3)
3 − e2x
2
b y =
6
4x − 3
b y =
15a y = 4 (x4 + 4)
b i y = ac −
b2x
1
,x≠
1 − abx
ab
a
ii y = tan(abx)
b
ab(1 − e−(a − b)x)
b i y =
ae−(a − b)x − b
ii −a
9a y = "5 − x2, ∣ x ∣ < !5
14a y =
19a y = y0ekx
20a
1
4 − 3 cos(4x)
O
loge (18x − 17), x
b y =
PA
G
b y =
−13
3
4
,x≠
4 − 3x
3
PR
O
c y = Ae
4
FS
4
c−x
E
9a y =
b y = 3"2x2 + 9
!2π
2
b y = 2x a 1 + 2 loge a ∣ x ∣ b b , x ≠ 0
18a y = x a c + b loge a ∣ x ∣ b b , x ≠ 0
b y =
bx
c
+ a
a+1 x
Exercise 9.6
1 y = c2 + c1x − x6
2 y = c2 + c1x + 4 sin(3x)
3 y = −2x4 − 5x − 1
3π
2
16
−3x2
5 y =
,x≠−
3
128(3x + 16)
4 y = 3 sin(2x) + 6x + 1 −
9
6 y = 12!2x + 9 − 3x − 36, x > −2
k 4
5kL4
(x − 2Lx2 + L3x),
12
192
3
k
kL
8 y = (3Lx2 − x3),
6
3
7 y =
Maths Quest 12 SPECIALIST MATHEMATICS VCE Units 3 and 4
c09DifferentialEquations.indd 492
04/07/15 12:32 PM
16a Check with your teacher.
b y = 3x3 − 36x − 48
x3 x4
− −
2
6
c y = c2 + c1x + 2 loge a ∣ x ∣ b +
5
5
,x≠0
2x
5
10a y = −x3 + 8x − 5
b y = −2x4 + 31x − 27
11a y = 4 − 2e2x − 2e −2x
11x 41
4
−
9
3
9
y = 4 sin(2x) − 8x + 4
y = 3 cos(3x) + 18x − 3π + 3
y = 1 − cos(4x) − 8x2
y = 18 cos(8x) − 4x2 − 18
b y = 5e −x − e −3x +
13a
b
b y =
9
b−
∣ ax + b ∣
x
b
1
loge a
b− ,x≠−
2
a
b
ab
a
20ai
Check with your teacher.
1
ii y = c2 + c1x − 4"9 + 4x2
PR
O
b
∣ 4x + 9 ∣
4x
, x ≠ −94
9
WL4
W
18a y =
(6L2x2 − 4Lx3 + x4),
24E I
8E I
4
W x2
WL
b y =
(x − L) 2,
24E I
384E I
−x2
b
19a y =
,x≠−
2
a
2b (ax + b)
b y = loge a
d y = c2 + c1x + 2e−x − 9 e− 3x
12a
9
17a y = 30x + 45 − 3"(4x + 9) 3, x > −4
FS
b y = c2 + c1x +
10x2
O
2
x
9a y = c2 + c1x − , x ≠ 0
b i Check with your teacher.
x2
2
,x≠−
8(3x + 2)
3
9
x
b y = !2x + 9 − − 3, x > −
3
2
15a Check with your teacher.
b y = −6x2 − 24x − 20
14a y =
1
"a + bx2
ab
U
N
C
O
R
R
EC
TE
D
PA
G
E
ii y = c2 + c1x −
Topic 9 Differential ­equations c09DifferentialEquations.indd 493
493
04/07/15 12:32 PM