Solution of first order differential equations by separation of variables 451
where C p and C v are constants. Given n =
C p
C v
,
show that pV n = constant.
Separating the variables gives:
C v
dp
p
=−C p
dV
V
Integrating both sides gives:
C v
d p
p
= −C p
dV
V
i.e.
C v ln p = −C p ln V + k
Dividing throughout by constant C v gives:
ln p = −
C p
C v
ln V +
k
C v
Since
C p
C v
= n, then ln p +n ln V = K ,
where K =
k
C v
.
i.e. ln p +ln V n = K or ln pV n = K , by the laws of
logarithms.
Hence pV n = e K , i.e. pV n = 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.
d y
dx
= 2y cos x
[ln y = 2 sin x + c]
2. (2y − 1)
d y
dx
= (3x 2 + 1), given x = 1 when
y = 2.
[ y 2 − y = x 3 + x]
3.
d y
dx
= e 2x−y , given x = 0 when y = 0.
e y =
1
2
e 2x +
1
2
4. 2y(1 − x) + x(1 + y)
d y
dx
= 0, given x = 1
when y = 1.
[ln (x 2 y) = 2x − y − 1]
5. Show that the solution of the equation
y 2 + 1
x 2 + 1
=
y
x
d y
dx
is of the form
y 2 + 1
x 2 + 1
= constant.
6. Solve x y = (1 − x 2 )
d y
dx
for y, given x = 0
when y = 1.
y =
1
(1 − x 2 )
7. Determine the equation of the curve which
satisfies the equation x y
d y
dx
= x 2 − 1, and
which passes through the point (1, 2).
[ y 2 = x 2 − 2 ln x + 3]
8. The p.d., V , between the plates of a capacitor C charged by a steady voltage E
through a resistor R is given by the equation
CR
dV
dt
+ V = E.
(a) Solve the equation for V given that at
t = 0, V = 0.
(b) Calculate V , correct to 3 significant
figures, when E =25V, C = 20 ×10 −6 F,
R = 200 ×10 3 and t = 3.0 s.
⎡
⎣ (a) V = E
1 − e
−t
C R
(b) 13.2 V
⎤
⎦
9. Determine the value of p, given that
x
3 d y
dx
= p − x, and that y = 0 when x = 2 and
when x = 6.
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