Finding such $u(x,y)$ that satisfies this differential equation

Question

I am having a hard time finding such $u(x,y)$ that satisfies partial differential equation $$\frac\partial{\partial y}(u_x+u)+2x^2y(u_x+u)=0.$$

Here is everything I know so far: We need to set up $v(x,y)=u_x+u$. I am stuck at this point onwards.

Answer 1

$$\frac\partial{\partial y}(u_x+u)+2x^2y(u_x+u)=0.$$ To answer to the comment request : $$v(x,y)=(u_x+u)$$ $$v_y+2x^2y\:v=0$$ This is a first order ODE since there is no $v_x$ into it. $$\frac{1}{v}\frac{dv}{dy}=-2x^2y$$ Integrate wrt $y$ : $$\ln|v|=-x^2y^2+\text{ an arbitrary constant wrt }y \text{ but function of }x$$ $$v(x,y)=f(x)\: e^{-2x^2y^2}$$ $f(x)$ is an arbitrary function which is a constant of integration wrt $y$. $$v=u_x+u=f(x)\: e^{-2x^2y^2}$$ $$\frac{du}{dx}+u=f(x)\: e^{-2x^2y^2}$$ This is a first order linear ODE for $u$ wrt $x$. Solving leads to $$u(x,y)=e^{-x}\int e^xf(x)\: e^{-2x^2y^2}dx+e^{-x}g(y)$$ $g(y)$ is an arbitrary function.

Without boundary condition one cannot determine $f(x)$ and $g(y)$.