Higher Engineering Mathematics

450 DIFFERENTIAL EQUATIONS

Problem 13. For an adiabatic expansion of

agas

Cv

dp

p

+Cp

dV

V

=0,

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, then lnp+nlnV=K,

whereK=

k

Cv

.

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

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

Now try the following exercise.

Exercise 180 Further problems on equa-

tions of the form

dy

dx

=f(x)·f(y)

In Problems 1 to 4, solve the differential

equations.

1.

dy

dx

= 2 ycosx [lny=2 sinx+c]

2. (2y−1)

dy

dx

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

y=2. [y^2 −y=x^3 +x]

3.

dy

dx

=e^2 x−y,givenx=0 wheny=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 forVgiven that at

t=0,V=0.

(b) CalculateV, correct to 3 significant fig-

ures, whenE=25 V, 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, and thaty=0 whenx=2 and

whenx=6. [3]