1.

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

Answer» `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,
`=>1/2 int sec^2(v/2) dv = int dx/x^3`
`=>1/2(2)tan(v/2) = -1/2 x^-2+c`
`=>tan(v/2) = -1/(2x^2)+c`
`=>tan(y/(2x)) = -1/(2x^2)+c`


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