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 asdy
f(y)=f(x)dx,and
then the solution is obtained by direct integration, i.e.
∫
dy
f(y)=∫
f(x)dxProblem 9. Solve the equation 4xydy
dx=y^2 − 1Separating the variables gives:
(
4 y
y^2 − 1
)
dy=1
xdxIntegrating both sides gives:
∫ (
4 y
y^2 − 1
)
dy=∫ (
1
x)
dxUsing 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}
=cand
(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 equations4 xydy
dx=y^2 − 1Problem 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 tdtThus the general solution is:
1
2e^2 θ=2
3e^3 t+cWhent=0,θ=0, thus:
1
2e^0 =2
3e^0 +cfrom which,c=1
2−2
3=−1
6
Hence the particular solution is:1
2e^2 θ=2
3e^3 t−1
6or 3e^2 θ=4e^3 t−^1Problem 11. Find the curve which satisfies the
equationxy=( 1 +x^2 )dy
dxand 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