(1)
Clearly, (1) is a homogenous differential equation To solve it we make substitution
y = vx
Differentiating y = vx with respect to x, we get
dy
dx  =
dv
v
x dx
+
or
dv
v
+x dx
 =
2
1
2
vv
+
or
dv
x dx =
2
1
2
vv
−
122
v
−vdv
 = dx
x
or
22
v1
v −dv
 =
xdx
−
Therefore
22
v1
∫v −dv
 =
x1 dx
−∫
or
log |v2 – 1 | = – log |x| + log |C1|
or
log |(v2 – 1) (x)| = log |C1|
or
(v2 – 1) x = ± C1
Replacing v by y
x  , we get
2
2
1
y
x
x
⎛
−⎞
⎜
⎟
⎝
⎠
 = ± C1
or
(y2 – x2) = ± C1 x or x2 – y2 = Cx
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MATHEMATICS
406
EXERCISE 9 5
In each of the Exercises 1 to 10, show that the given differential equation is homogeneous
and solve each of them 1