Question

Solve the differential equation :

`y+x dy/dx=x−y dy/dx`

Answer 1

`y+x dy/dx=x−y dy/dx`

`x dy/dx + y dy/dx=x−y`

`⇒dy/dx=(x−y)/(x+y)  `  ......(1)

`Let F(x, y) =(x−y)/(x+y)`

`F(λx, λy) = λF(x, y)`
Therefore, F(x, y) is a homogeneous function of degree zero.

Let `y=vx`

`dy/dx=v+x (dv)/dx`

Substituting the value of y and dy/dx in (1) we get,

`v + x (dv)/dx=(x−vx)/(x+vx)=(1−v)/(1+v)`

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

`(1+v)/(v^2+2v−1)dv=−dx/x`

Integrating both sides, we have

`1/2 log∣(y^2/x^2)+(2y)/x−1∣+log|x|=logc`

`⇒log∣(y^2/x^2)+(2y)/x−1∣+2log|x|=2logc`

`⇒log((y^2/x^2)+(2y)/x−1)(x^2)=logc^2`

`⇒((y^2+2yx−x^2)/x^2)(x^2) = c^2`

`⇒y^2+2yx−x^2=C           (where C=c^2)`