Question

Solve the differential equation `dy/dx=(y+sqrt(x^2+y^2))/x`

Answer 1

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

Put y = vx

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

`therefore (1) becomes, v+x (dv)/dx=(vx+sqrt(x^2+v^2x^2))/x`

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

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

Integrating, we get,

`int 1/sqrt(1+v^2) dv=int 1/x dx+c_1`

`therefore log |v+sqrt(1+v^2)|=log|x|+logc, where c_1=logc`

`therefore (y+sqrt(x^2+y^2))/x=cx`

`(y+sqrt(x^2+y^2))=cx^2 ` is the general solution