Solution of first order differential equations by separation of variables 449
8. The rate of cooling of a body is given by
dθ
dt
= kθ, where k is a constant. If θ = 60 ◦ C
when t = 2 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.
[8 m 40 s]
46.5 The solution of equations of the
form
dy
dx
= f (x) · f ( y)
A differential equation of the form
d y
dx
= f (x) · f ( y),
where f (x) is a function of x only and f ( y) is a function
of y only, may be rearranged as
d y
f ( y)
= f (x) dx, and
then the solution is obtained by direct integration, i.e.
dy
f ( y)
=
f (x)dx
Problem 9. Solve the equation 4x y
d y
dx
= y 2 − 1
Separating the variables gives:
4y
y 2 − 1
d y =
1
x
dx
Integrating both sides gives:
4y
y 2 − 1
d y =
1
x
dx
Using the substitution u = y 2 − 1, the general
solution is:
2 ln ( y
2
− 1) = lnx + c
(1)
or
ln( y
2
− 1)
2
− ln x = c
from which, ln
( y 2 − 1) 2
x
= c
and
( y 2 − 1)
2
x
= e
c
(2)
If in equation (1), c = ln A, where A is a different
constant,
then ln( y 2 − 1) 2 = ln x + ln A
i.e. ln( y 2 − 1) 2 = ln Ax
i.e. ( y
2
− 1)
2
= Ax
(3)
Equations (1) to (3) are thus three valid solutions of the
differential equations
4x y
d y
dx
= y
2
− 1
Problem 10. Determine the particular solution of
dθ
dt
= 2e 3t −2θ , given that t = 0 when θ = 0.
dθ
dt
= 2e
3t −2θ
= 2(e
3t
)(e
−2θ
),
by the laws of indices.
Separating the variables gives:
dθ
e −2θ = 2e 3t dt,
i.e. e 2θ dθ = 2e 3t dt
Integrating both sides gives:
e
2θ dθ =
2e
3t dt
Thus the general solution is:
1
2
e
2θ
=
2
3
e
3t
+ c
When t = 0, θ = 0, thus:
1
2
e 0 =
2
3
e 0 + c
from which, c =
1
2
−
2
3
= −
1
6
Hence the particular solution is:
1
2
e
2θ
=
2
3
e
3t
−
1
6
or 3e
2θ
= 4e
3t
− 1
Problem 11. Find the curve which satisfies the
equation x y = (1 + x
2
)
d y
dx
and passes through the
point (0, 1).
Separating the variables gives:
x
(1 + x 2 )
dx =
d y
y
Integrating both sides gives:
1
2 ln (1 + x
2
) = ln y + c
8. The rate of cooling of a body is given by
dθ
dt
= kθ, where k is a constant. If θ = 60 ◦ C
when t = 2 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.
[8 m 40 s]
46.5 The solution of equations of the
form
dy
dx
= f (x) · f ( y)
A differential equation of the form
d y
dx
= f (x) · f ( y),
where f (x) is a function of x only and f ( y) is a function
of y only, may be rearranged as
d y
f ( y)
= f (x) dx, and
then the solution is obtained by direct integration, i.e.
dy
f ( y)
=
f (x)dx
Problem 9. Solve the equation 4x y
d y
dx
= y 2 − 1
Separating the variables gives:
4y
y 2 − 1
d y =
1
x
dx
Integrating both sides gives:
4y
y 2 − 1
d y =
1
x
dx
Using the substitution u = y 2 − 1, the general
solution is:
2 ln ( y
2
− 1) = lnx + c
(1)
or
ln( y
2
− 1)
2
− ln x = c
from which, ln
( y 2 − 1) 2
x
= c
and
( y 2 − 1)
2
x
= e
c
(2)
If in equation (1), c = ln A, where A is a different
constant,
then ln( y 2 − 1) 2 = ln x + ln A
i.e. ln( y 2 − 1) 2 = ln Ax
i.e. ( y
2
− 1)
2
= Ax
(3)
Equations (1) to (3) are thus three valid solutions of the
differential equations
4x y
d y
dx
= y
2
− 1
Problem 10. Determine the particular solution of
dθ
dt
= 2e 3t −2θ , given that t = 0 when θ = 0.
dθ
dt
= 2e
3t −2θ
= 2(e
3t
)(e
−2θ
),
by the laws of indices.
Separating the variables gives:
dθ
e −2θ = 2e 3t dt,
i.e. e 2θ dθ = 2e 3t dt
Integrating both sides gives:
e
2θ dθ =
2e
3t dt
Thus the general solution is:
1
2
e
2θ
=
2
3
e
3t
+ c
When t = 0, θ = 0, thus:
1
2
e 0 =
2
3
e 0 + c
from which, c =
1
2
−
2
3
= −
1
6
Hence the particular solution is:
1
2
e
2θ
=
2
3
e
3t
−
1
6
or 3e
2θ
= 4e
3t
− 1
Problem 11. Find the curve which satisfies the
equation x y = (1 + x
2
)
d y
dx
and passes through the
point (0, 1).
Separating the variables gives:
x
(1 + x 2 )
dx =
d y
y
Integrating both sides gives:
1
2 ln (1 + x
2
) = ln y + c