Solve $\frac{dy}{dx} + \frac{y}{x} = \frac{1}{(1+\log x+\log y)^2}$

Question

Solve $\frac{dy}{dx} + \frac{y}{x} = \frac{1}{(1+\log x+\log y)^2}$

I tried solving this differential equation, I got stuck after a few steps I got $(a+\log(xy))^2d(xy) = xdx$ I don't know what to do after this

Answer 1

If you write $$\frac{dy}{dx}+\frac{y}{x} = \frac{1}{(1+\log(xy))^2}$$ and substitute $v=xy$, then wonderful things happen. Then $$\frac{dv}{dx} = x\frac{dy}{dx} +y$$ Plug stuff in and the equation becomes $$\frac{1}{x}\frac{dv}{dx}=\frac{1}{(1+\log v)^2},$$ separable.

Answer 2

With $u = xy$, the differential equation becomes separable.

Answer 3

Taking L.C.M on the left side you'll get $$ (xdy + ydx )(1+logxy)^2 = xdx $$ And as we know that $$xdy + ydx = d(xy)$$ so by substituting $$xy=t$$ you'll get $$(1+logt)^2 dt = xdx $$ Integrating both sides $$ \int (1+logt)^2 \mathrm{d}t = (x^2 )/2 + c$$ To solve the L.H.S use your own integration skills (Hint :- Integration by parts )