Higher Engineering Mathematics, Sixth Edition

(Nancy Kaufman) #1

Solution of first order differential equations by separation of variables 449



  1. The rate of cooling of a body is given by

    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

dt

=2e^3 t−^2 θ, given thatt=0whenθ=0.


dt

=2e^3 t−^2 θ= 2 (e^3 t)(e−^2 θ),

by the laws of indices.
Separating the variables gives:

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

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