Solve the following differential equation `x^2(dy)/(dx)-xy=1+cos(y/x), x!=0`

Explanation

`x^2dy/dx - xy = 1+cos(y/x)`

`=>dy/dx -y/x = 1/x^2(1+cos(y/x))`

Let `y = vx`, then `dy/dx = v+x(dv)/dx`

Now, our equation becomes,

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

`=>(dv)/(1+cosv) = 1/x^3dx`

`=>1/2sec^2(v/2) dv = dx/x^3`

Integrating both sides,

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