Question

For the differential equation find a particular solution satisfying the given condition:

`2xy + y^2 - 2x^2  dy/dx = 0; y = 2`   when x  = 1

Answer 1

The given equation

`2xy + y^2 - 2x^2 dy/dx = 0`

or `dy/dx = (2xy + y^2)/(2x^2)`

`= y/x + 1/2(y/x)^2`         ....(i)

Clearly, this equation is a differential equation.

∴ putting y = vx

`dy/dx = v + x (dv)/dx`   ....(in equation (i))

`v + x (dv)/dx = v + 1/2 v^2`

`=> x (dv)/dx = 1/2 v^2`

`=> 2 xx 1/v^2 (dv) = 1/x dv`

On integrating,

`2 int 1/v^2 dv = int 1/x dv - 2/v = log |x| + C`

So, on substituting `(y/x)` in place of v,

`- (2x)/y = log |x| + C`     ....(ii)

Given y = 2 if x = 1                 ...[from equation (ii)]

`(- 2)/2 = log1 + C`

- 1 = 0 + C

⇒ C = - 1

On putting C = – 1 in equation (ii)

`(- 2x)/y = log |x| - 1`

y = `(2x)/(1 - log |x|)`