Solution of first order differential equations by separation of variables 449
- The rate of cooling of a body is given by
dθ
dt
=kθ,wherekis a constant. Ifθ= 60 ◦C
when t=2 minutes and θ= 50 ◦Cwhen
t=5minutes, determine the time taken forθ
to fall to 40◦C, correct to the nearest second.
[8m 40s]
46.5 The solution of equations of the
form
dy
dx
=f(x)·f(y)
A differential equation of the form
dy
dx
=f(x)·f(y),
wheref(x)is afunction ofxonly andf(y)is afunction
ofyonly, may be rearranged as
dy
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 4xy
dy
dx
=y^2 − 1
Separating the variables gives:
(
4 y
y^2 − 1
)
dy=
1
x
dx
Integrating both sides gives:
∫ (
4 y
y^2 − 1
)
dy=
∫ (
1
x
)
dx
Using the substitution u=y^2 −1, the general
solution is:
2ln(y^2 −1)=lnx+c (1)
or ln(y^2 − 1 )^2 −lnx=c
from which, ln
{
(y^2 − 1 )^2
x
}
=c
and
(y^2 −1)^2
x
=ec (2)
If in equation (1),c=lnA,whereAis a different
constant,
then ln(y^2 − 1 )^2 =lnx+lnA
i.e. ln(y^2 − 1 )^2 =lnAx
i.e. (y^2 −1)^2 =Ax (3)
Equations (1) to (3) are thus three valid solutions of the
differential equations
4 xy
dy
dx
=y^2 − 1
Problem 10. Determine the particular solution of
dθ
dt
=2e^3 t−^2 θ, given thatt=0whenθ=0.
dθ
dt
=2e^3 t−^2 θ= 2 (e^3 t)(e−^2 θ),
by the laws of indices.
Separating the variables gives:
dθ
e−^2 θ
=2e^3 tdt,
i.e. e^2 θdθ=2e^3 tdt
Integrating both sides gives:
∫
e^2 θdθ=
∫
2e^3 tdt
Thus the general solution is:
1
2
e^2 θ=
2
3
e^3 t+c
Whent=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^2 θ=
2
3
e^3 t−
1
6
or 3e^2 θ=4e^3 t−^1
Problem 11. Find the curve which satisfies the
equationxy=( 1 +x^2 )
dy
dx
and passes through the
point (0, 1).
Separating the variables gives:
x
( 1 +x^2 )
dx=
dy
y
Integrating both sides gives:
1
2 ln(^1 +x
(^2) )=lny+c