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 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
]
- 2y( 1 −x)+x( 1 +y)
dy
dx
=0, given x= 1
wheny=1. [ln(x^2 y)= 2 x−y−1]
- 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.
- Solve xy=( 1 −x^2 )
dy
dx
for y,givenx= 0
wheny=1. [
y=
1
√
( 1 −x^2 )
]
- 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]
- 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
⎤
⎦
- Determine the value of p, given that
x^3
dy
dx
=p−x,andthaty=0whenx=2and
whenx=6. [3]