CHAPTER 77 SOLUTION OF FIRST-ORDER DIFFERENTIAL
EQUATIONS BY SEPARABLE VARIABLES
EXERCISE 295 Page 814
1. Sketch a family of curves represented by each of the following differential equations:
(a)
(a) If
dy
=6
dx
(b)
dy
= 6 , then y =
dx
dy
= 3x
dx
(c)
dy
=x+2
dx
∫ 6 d x = 6x + c
There are an infinite number of graphs of y = 6x + c; three curves are shown below.
(b) If
dy
= 3 x , then y =
dx
dx
∫ 3x=
3 2
x +c
2
A family of three typical curves is shown below.
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© 2014, John Bird
(c) If
dy
= x + 2 , then y =
dx
∫
( x + 2) d x =
x2
+ 2x + c
2
A family of three typical curves is shown below.
2. Sketch the family of curves given by the equation
dy
= 2x + 3 and determine the equation of one of
dx
these curves which passes through the point (1, 3).
If
dy
= 2 x + 3 , then y =
dx
∫ ( 2 x + 3) d x = x
2
+ 3x + c
If the curve passes through the point (1, 3) then x = 1 and y = 3
Hence,
and
3 = (1) + 3 (1) + c
2
i.e.
c = –1
y = x 2 + 3x − 1
A family of three curves is shown below, including y = x 2 + 3 x − 1 , which passes through the point
(1, 3)
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© 2014, John Bird
EXERCISE 296 Page 816
dy
= cos 4x – 2x
dx
1. Solve:
Since
dy
= cos 4x – 2x
dx
i.e.
y=
2. Solve: 2x
Since 2x
1
sin 4 x − x 2 + c
4
dy
= 3 – x3
dx
dy
= 3 – x3
dx
i.e.
d y 3 − x3
3 x3
3 x2
=
=
−
=
−
dx
2x
2x 2x 2x 2
3
x3
 3 x2 
+c
y = ∫  −  d x = ln x –
2
6
 2x 2 
Hence,
3. Solve:
If
∫ ( cos 4 x − 2 x ) d x
y=
dy
+ x = 3, given y = 2 when x = 1
dx
dy
dy
3 , then
+x=
= 3− x
dx
dx
and y =
If y = 2 when x = 1, then 2 = 3 –
y = 3x −
Hence,
4. Solve: 3
Since 3
and
1
+c
2
∫ ( 3 − x ) d x = 3x −
from which,
x2
+c
2
c= −
1
2
x2 1
−
2 2
π
dy
2
+ sin θ = 0, given y = when θ =
3
dθ
3
dy
+ sin θ = 0
dθ
i.e.
dy
1
= – sin θ
dθ
3
1
1
 1
y = ∫ − sin θ d θ =  −  (− cos θ ) + c = cos θ + c
3
3
 3
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© 2014, John Bird
y=
2
π
when θ =
3
3
2 1
π
hence, = cos + c
3 3
3
2 11
=  +c
3 3 2
i.e.
=
y
Hence,
5. Solve:
If
2 11
i.e.
−  =
c
3 3 2
i.e.
1
2
1
1
cos θ +
3
2
1
dy
+2=x–3
, given y = 1 when x = 0
ex
dx
1
dy
+ 2 = x −3
ex
dx
then 3
If y = 1 when x = 0, then
Thus,
dy
=x − e − x − 2
dx
y=
1=
1
(0 + 1 + 0) + c
3
1  x2
 2
−x
 + e − 2x  +
3 2
 3
6. The gradient of a curve is given by
the point (1,
and
dy 1
= ( x − e− x − 2 )
dx 3
1
1  x2

− x − 2 dx
−x
−
=
x
e
(
)
 + e − 2x  + c
∫
3
3 2

y=
Hence,
If
c=
or
i.e.
c=
2
3
1 2
2

 x − 4x + x + 4 
6
e

dy
x2
+
= 3x. Find the equation of the curve if it passes through
dx
2
1
).
3
dy
d y x2
x2
3 x , then
+
=
= 3x −
dx 2
dx
2
Hence,
y=

∫  3x −
x2 
3x 2 x3
− +c
d x =
2 
2
6
1
 1
If it passes through 1,  , x = 1 and y =
3
 3
Thus,
Hence,
1 3 1
= − +c
3 2 6
y=
from which,
c=
1 3 1
− + =
−1
3 2 6
3 2 x3
x − −1
2
6
1194
© 2014, John Bird
7. The acceleration a of a body is equal to its rate of change of velocity,
dv
. Find an equation for v in
dt
terms of t, given that when t = 0, velocity v = u.
dv
=a
dt
v = ∫ a d=
t at + c
hence,
When t = 0, velocity v = u, hence,
Hence,
u=0+c
from which, c = u
velocity, v = at + u or v = u + at
8. An object is thrown vertically upwards with an initial velocity u of 20 m/s. The motion of the object
follows the differential equation
ds
= u – gt, where s is the height of the object in metres at time t
dt
seconds and g = 9.8 m/s2. Determine the height of the object after 3 seconds if s = 0 when t = 0
If
ds
= u – gt, then
dt
s=
∫ ( u − gt ) d t =ut −
g t2
+c
2
Since s = 0 when t = 0, then c = 0
Hence,
=
s ut −
g t2
and if u = 20 and s = 9.8, then
2
The height when t = 3,
i.e.
s = 3(20) – 4.9 ( 3)
s = 20t –
9.8t 2
2
i.e.
u = 20t – 4.9 t 2
2
height = 60 – 44.1 = 15.9 m
1195
© 2014, John Bird
EXERCISE 297 Page 817
1. Solve:
dy
= 2 + 3y
dx
Rearranging
dy
dy
= 2 + 3y gives: dx =
2 + 3y
dx
dy
∫ d x = ∫ 2 + 3y
Integrating both sides gives:
Thus, by using the substitution u = (2 + 3y),
x=
2. Solve:
If
dy
= 2 cos2 y
dx
dy
= 2 cos 2 y then
dx
dy
= 2d x
cos 2 y
∫ sec
and
i.e.
2
y d y = ∫ 2d x
tan y = 2x + c
3. Solve: (y2 + 2)
If
1
ln(2 + 3y) + c
3
( y2 + 2)
dy
1
= 5y, given y = 1 when x =
dx
2
dy
then
5y
=
dx

y2
+ 2 ln y =
5x + c
2
i.e.
and
2
∫  y + y  d y =
∫ 5d x
and
y = 1 when x =
 y2 + 2 

 d y = 5d x
 y 
1
, hence,
2
1
5
+ 2 ln1 =+ c
2
2
from which,
c=
1 5
− =
−2
2 2
y2
+ 2 ln y =
5x − 2
2
1196
© 2014, John Bird
4. The current in an electric circuit is given by the equation Ri + L
di
= 0, where L and R are constants.
dt
Show that i = Ie–Rt/L, given that i = I when t = 0
If Ri + L
di
di
= 0, then L
= –Ri
dt
dt
and
di
Ri
= −
dt
L
from which,
di
R
= − dt
i
L
Thus,
ln i = −
i = I when t = 0, thus
ln I = c
Hence,
ln i = −
i.e.
ln i – ln I = −
i.e.
ln
∫
and
di
=
i
R
∫ − L dt
Rt
+c
L
Rt
+ ln I
L
Rt
L
i
Rt
= −
I
L
Taking anti-logarithms gives:
Rt
i
−
=e L
I
and
5. The velocity of a chemical reaction is given by
i=Ie
−
Rt
L
dx
= k(a – x), where x is the amount transferred in
dt
time t, k is a constant and a is the concentration at time t = 0 when x = 0. Solve the equation and
determine x in terms of t.
If
dx
= k(a – x), then
dt
dx
= k dt
a−x
dx
and
∫ a − x = ∫ k dt
i.e.
– ln(a – x) = kt + c
t = 0 when x = 0, hence
Thus,
– ln a = c
– ln(a – x) = kt – ln a
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© 2014, John Bird
i.e.
ln a – ln(a – x) = kt
i.e.
 a 
ln 
 = kt
a−x
and
a
= e kt
a−x
i.e.
a
= a−x
ek t
and
x = a – a e− k t
i.e.
i.e.
a e − k t= a − x
=
x a (1 − e − k t )
6. (a) Charge Q coulombs at time t seconds is given by the differential equation R
dQ
Q
+
= 0, where
dt
C
C is the capacitance in farads and R the resistance in ohms. Solve the equation for Q given that
Q = Q 0 when t = 0
(b) A circuit possesses a resistance of 250 × 103 ohms and a capacitance of 8.5 × 10–6 farads, and after
0.32 seconds the charge falls to 8.0 C. Determine the initial charge and the charge after 1 second,
each correct to 3 significant figures.
(a) If R
dQ Q
=0
+
dt C
i.e.
i.e.
dQ
Q
= −
dt
RC
then
∫
dQ
=
Q
1
∫ − RC d t
ln Q = −
t
+k
RC
Q = Q 0 when t = 0, hence, ln Q 0 = k
Hence,
i.e.
i.e.
and
ln Q = −
t
+ ln Q0
RC
ln Q – ln Q 0 = −
ln
t
RC
Q
t
= −
Q0
RC
t
−
Q
= e RC
Q0
and
Q = Q0 e
−
t
CR
(b) R = 250 × 103 Ω , C = 8.5 × 10–6 F, t = 0.32 s and Q = 8.0 C
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© 2014, John Bird
0.32
−6
3
Hence,
=
8.0 Q=
Q0 ( 0.8602 )
0 e 8.5×10 ×250×10
−
8.0
= 9.30 C
0.8602
from which, initial charge, Q 0 =
When t = 1 s, charge, Q = Q0 e
−
t
CR
= 9.30 e
−
1
8.5×10−6 ×250×103
= 5.81 C
7. A differential equation relating the difference in tension T, pulley contact angle θ and coefficient of
friction µ is
dT
= µT. When θ = 0, T = 150 N, and µ = 0.30 as slipping starts. Determine the tension
dθ
at the point of slipping when θ = 2 radians. Determine also the value of θ when T is 300 N.
Since
dT
dT
= µT then
= µ dθ
dθ
T
dT
= ∫ µ dθ
T
and
∫
i.e.
ln T = μθ + c
When θ = 0, T = 150 N, and µ = 0.30, hence ln 150 = (0.30)(0) + c
from which,
c = ln 150
Hence,
ln T = μθ + ln 150
i.e.
ln T – ln 150 = μθ
i.e.
 T 
ln 
 = µθ
 150 
from which,
T
= e µθ
150
and T = 150 e µθ
When θ = 2 radians, tension, T = 150 e( 0.30)(2) = 150 e0.60 = 273.3 N
When T = 300 N,
Hence,
300 = 150 e(0.30)θ
i.e.
300 (0.30)θ
=e
150
i.e.
2 = e(0.30)θ
ln 2 = ln [e(0.30)θ ] = 0.30 θ
from which, contact angle, θ =
1
ln 2 = 2.31 rad
0.30
1199
© 2014, John Bird
8. The rate of cooling of a body is given by
dθ
= k θ , where k is a constant. If θ = 60°C when t = 2
dt
minutes and θ = 50°C when t = 5 minutes, determine the time taken for θ to fall to 40°C, correct to the
nearest second.
If
dθ
= kθ
dt
then
dθ
θ
= k dt
and
∫
dθ
θ
= ∫ k dt
ln θ = kt + c
i.e.
When θ = 60°C, t = 2, i.e.
ln 60 = 2k + c
(1)
When θ = 50°C, t = 5, i.e.
ln 50 = 5k + c
(2)
(1) – (2) gives:
ln 60 – ln 50 = –3k
from which,
1 60
k = − ln
=
−0.06077
3 50
Substituting in (1):
ln 60 = 2(–0.06077) + c
from which,
Hence,
When θ = 40°C,
and
c = ln 60 + 2(0.06077) = 4.2159
ln θ = kt + c = –0.06077t + 4.2159
ln 40 = –0.06077t + 4.2159
time, t =
4.2159 − ln 40
= 8.672 min = 8 min 40 s
0.06077
1200
© 2014, John Bird
EXERCISE 298 Page 820
1. Solve:
Since
dy
= 2y cos x
dx
dy
= 2 cos x d x
y
dy
= 2y cos x then
dx
dy
= 2 cos x d x
y ∫
and
∫
i.e.
ln y = 2 sin x + c
2. Solve: (2y – 1)
dy
If (2y – 1) =
dx
dy
= (3x2 + 1), given x = 1 when y = 2
dx
( 3x 2 + 1) ,
then
∫ ( 2 y − 1) d y = ∫ ( 3x
2
+ 1) d x
i.e.
y 2 − y = x3 + x + c
x =1 when y = 2, hence,
4–2=1+1+c
Thus,
y 2 − y = x3 + x
3. Solve:
from which,
c=0
dy
= e2x–y, given x = 0 when y = 0
dx
dy
= e2x–y = (e2x)(e– y), by the laws of indices
dx
Separating the variables gives:
dy
= e2x dx i.e. e y dy = e2x dx
y
−
e
Integrating both sides gives:
∫e
Thus the general solution is:
When x = 0, y = 0, thus:
y
d y = ∫ e2 x d x
ey =
1 2x
e +c
2
e0 =
1 0
e +c
2
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© 2014, John Bird
from which,
1
1
=
2
2
c=1–
Hence the particular solution is:
1 2x 1
e +
2
2
dy
= 0, given x = 1 when y = 1
dx
4. Solve: 2y(1 – x) + x(1 + y)
If 2y(1 – x) + x(1 + y)
ey =
dy
=0
dx
then
x(1 + y)
dy
= –2y(1 – x) = 2y(x – 1)
dx
Thus,
∫
 1+ y 
 2( x − 1) 

d y = ∫ 
d x
 x 
 y 
i.e.
∫
1 
2

 + 1 d y = ∫  2 −  d x
x

y 
ln y + y = 2x – 2 ln x + c
x = 1 when y = 1, hence,
ln 1 + 1 = 2 – 2 ln 1 + c
Thus,
ln y + y = 2x – 2 ln x – 1
or
ln y + 2 ln x = 2x – y – 1
i.e.
ln y + ln x 2 = 2x – y – 1
and
ln ( x 2 y ) = 2 x − y − 1
5. Show that the solution of the equation
Since
i.e.
i.e.
y2 +1 y d y
then
=
x2 + 1 x d x
∫
y2 +1
ydy
=
is of the form
x2 + 1
x dx
from which, c = –1
 y2 +1 
 2
 = constant
 x +1 
y
x
dy=∫
dx
y2 +1
x2 + 1
1
1
ln ( y 2 =
+ 1)
ln ( x 2 + 1) + c
2
2
1
1


2 + 1 2 − ln x 2 + 1 2 =
ln
y
(
)
(
)

 c


1
i.e.
 y2 +1 2
ln 
 =c
 x2 + 1 
1202
© 2014, John Bird
 y2 +1 
ln 
 =c
 x2 + 1 
or
 y2 +1 
c
 2
 = e = a constant
1
x
+


and
6. Solve xy = (1 – x2)
Since xy = (1 – x2)
dy
for y, given x = 0 when y = 1
dx
dy
dx
then
dy
x
=
dx
y (1 − x 2 )
and
i.e.
xy
dy
=
(1 − x 2 )
dx
∫
dy
 x 
= ∫
d x
y
 1 − x2 
du
 x 
For ∫ 
d x let u = (1 – x2) from which,
= −2 x

dx
 1 − x2 
and dx =
du
−2 x
1 1
1
1
 x 
 x  du
=
− ∫ du =
− ln u =
− ln (1 − x 2 )
dx=
Hence, ∫ 



∫
2 u
2
2
 1 − x2 
 u  −2 x
dy
 x 
= ∫
d x
y
 1 − x2 
i.e. if
∫
then
ln y = –
x = 0 when y = 1, hence,
ln 1 = –
1
ln(1 – x2) + c
2
1
ln 1 + c
2
from which, c = 0
1
ln(1 – x2)
2
Hence the particular solution is:
ln y = –
and
ln y = ln (1 − x 2 )
i.e.
y = (1 − x 2 )
i.e.
y=
−
−
1
2
1
2
1
= 1
(1 − x 2 ) 2
1
(1 − x 2 )
1203
© 2014, John Bird
7. Determine the equation of the curve which satisfies the equation xy
dy
= x2 – 1, and which passes
dx
through the point (1, 2).
dy
Since xy = x 2 − 1 then
dx
∫
y=
dy
∫
x2 −1
=
dx
x
∫
1

 x − d x
x

y 2 x2
= − ln x + c
2
2
i.e.
If the curve passes through (1, 2) then x = 1 and y = 2
hence,
22 12
= − ln1 + c
2
2
Thus,
y 2 x2
3
= − ln x +
2
2
2
or
y2 =
x 2 − 2 ln x + 3
from which, c =
3
2
8. The p.d., V, between the plates of a capacitor C charged by a steady voltage E through a resistor R is
given by the equation CR
dV
+V=E
dt
(a) Solve the equation for V given that at t = 0, V = 0
(b) Calculate V, correct to 3 significant figures, when E = 25 volts, C = 20 × 10–6 farads,
R = 200 × 103 ohms and t = 3.0 seconds.
(a) Since CR
dV
+V =
E then
dt
dV E −V
=
dt
CR
∫
dV
dt
=∫
E −V
CR
i.e.
− ln ( E − V ) =
from which,
At t = 0, V = 0, hence,
Thus,
t
+k
CR
– ln E = k
− ln ( E − V ) =
t
− ln E
CR
t
ln E − ln ( E − V ) =
CR
1204
© 2014, John Bird
 E
ln 
 E −V
and
t

=
 CR
t
E
C
=e R
E −V
i.e.
E
i.e.
e
= E −V
t
CR
E
V =−
E
and
=−
E Ee
t
−
t
CR
e CR
t
−


=
V E 1 − e C R  volts


i.e.
t
3.0
−


−

(b) Voltage, V =
E 1 − e C R  =
25 1 − e 20×10−6 ×200×103



9. Determine the value of p, given that x3
Since x 3
dy
=p–x
dx
then
=
∫ dy
dy
= p − x , and that y = 0 when x = 2 and when x = 6
dx
p−x
dx
=
x3
∫

25 (1 − e −0.75 ) = 13.2 V
=

∫ ( px
∫
 p 1 
 3 − 2 d x
x 
x
− x −2 ) d x
i.e.
=
∫ dy
i.e.
y=
i.e.
−
y=
p 1
+ +c
2x2 x
y = 0 when x = 2, hence,
0=
−
p 1
+ +c
8 2
(1)
y = 0 when x = 6, hence,
0=
−
p 1
+ +c
72 6
(2)
(1) – (2) gives:
1 1  1 1
0=
−p −  + − 
 8 72   2 6 
i.e.
0=
−
i.e.
p 1
from which, p = 3
=
9 3
−3
px −2 x −1
−
+c
−2
−1
p 1
+
9 3
1205
© 2014, John Bird