Higher Engineering Mathematics

448 DIFFERENTIAL EQUATIONS

8. 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 ◦C when

t=5 minutes, determine the time taken forθ

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

dy

dx

=f(x)·f(y),

wheref(x) is a function ofxonly andf(y) is a func-

tion ofyonly, may be rearranged as

dy

f(y)

=f(x)dx,

and then the solution is obtained by direct integra-

tion, 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=0 whenθ=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