differential equation

448 DIFFERENTIAL EQUATIONS
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 = f (x) · f ( y)
dx
A differential equation of the form dy = f (x) · f (y),
dx
where f (x) is a function of x only and f (y) is a function
of y only, may be rearranged as dy
f (y) = f (x)dx,
and then the solution is obtained by direct integration,
i.e.
∫
Problem 9.
∫ dy
f (y) =
f (x)dx
Solve the equation 4xy dy
dx = y2 −1
Separating the variables gives:
( ) 4y
y 2 dy = 1 − 1 x dx
Integrating both sides gives:
∫ ( ) ∫ ( )
4y
1
y 2 dy = dx
− 1
x
Using the substitution u = y 2 − 1, the general
solution is:
2ln(y 2 − 1) = ln x + c (1)
or ln (y 2 − 1) 2 − ln x = c
{ (y 2 − 1) 2 }
from which, ln
= c
x
( y 2 − 1) 2
and
= e c (2)
x
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
4xy dy
dx = y2 − 1
Problem 10. Determine the particular solution
of dθ
dt = 2e3t−2θ , given that t = 0 when θ = 0.
dθ
dt = 2e3t−2θ = 2(e 3t )(e −2θ ),
by the laws of indices.
Separating the variables gives:
dθ
e −2θ = 2e3t 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 e2θ = 2 3 e3t + c
When t = 0, θ = 0, thus:
1
2 e0 = 2 3 e0 + c
from which, c = 1 2 − 2 3 =−1 6
Hence the particular solution is:
1
2 e2θ = 2 3 e3t − 1 6
or 3e 2θ = 4e 3t − 1
Problem 11. Find the curve which satisfies the
equation xy = (1 + x 2 ) dy and passes through the
dx
point (0, 1).
Separating the variables gives:
x dy
(1 + x 2 dx =
) y