Higher Engineering Mathematics, Sixth Edition

Solution of first order differential equations by separation of variables 451

whereCpandCvare constants. Givenn=

Cp

Cv

,

show thatpVn=constant.

Separating the variables gives:

Cv

dp

p

=−Cp

dV

V

Integrating both sides gives:

Cv

∫

dp

p

=−Cp

∫

dV

V

i.e. Cvlnp=−CplnV+k

Dividing throughout by constantCvgives:

lnp=−

Cp

Cv

lnV+

k

Cv

Since

Cp

Cv

=n,thenlnp+nlnV=K,

whereK=

k

Cv

.

i.e. lnp+lnVn=Kor ln pVn=K, by the laws of
logarithms.

HencepVn=eK,i.e.pVn=constant.

Now try the following exercise

Exercise 179 Further problems on

equations of the form

dy

dx

=f(x)·f(y)

In Problems 1 to 4, solve the differential equations.

1.

dy

dx

= 2 ycosx [lny=2sinx+c]

2. ( 2 y− 1 )

dy

dx

=( 3 x^2 + 1 ),givenx=1when

y=2. [y^2 −y=x^3 +x]

3.

dy

dx

=e^2 x−y,givenx=0wheny=0.

[

ey=

1

2

e^2 x+

1

2

]

4. 2y( 1 −x)+x( 1 +y)

dy

dx

=0, given x= 1

wheny=1. [ln(x^2 y)= 2 x−y−1]

5. Show that the solution of the equation
   y^2 + 1
   x^2 + 1

=

y

x

dy

dx

is of the form

√(

y^2 + 1

x^2 + 1

)

=constant.

6. Solve xy=( 1 −x^2 )

dy

dx

for y,givenx= 0

wheny=1. [

y=

1

√

( 1 −x^2 )

]

7. Determine the equation of the curve which
   satisfies the equation xy

dy

dx

=x^2 −1, and

which passes through the point (1, 2).

[y^2 =x^2 −2lnx+3]

8. The p.d.,V, between the plates of a capac-
   itor C charged by a steady voltage E
   through a resistorRis given by the equation
   CR

dV

dt

+V=E.

(a) Solve the equation forV given that at

t=0,V=0.

(b) Calculate V, correct to 3 significant

figures, whenE=25V,C= 20 × 10 −^6 F,

R= 200 × (^103) ⎡andt= 3 .0s.
⎣(a)V=E
(
1 −e
−t
CR
)
(b) 13 .2V
⎤
⎦

9. Determine the value of p, given that
   x^3

dy

dx

=p−x,andthaty=0whenx=2and

whenx=6. [3]