PDE $u_xy + 2u_x = 2x(1+2y)$ with BD $: u(0,y)=0, u(x,0)=x\exp(-x^2)$

Question

Find a solution of the boundary value problem
PDE: $u_xy + 2u_x = 2x(1+2y)$
BD $: u(0,y)=0, u(x,0)=x\exp(-x^2)$

Answer 1

I am suspecting the equation is actually $$u_{xy} + 2 u_{x} = 2 x(1+2y)$$ If that's the case, then this is an exercise in calculus. First rewrite $$u_{xy} + 2 u_{x} = \frac{\partial}{\partial x}\big(u_y + 2 u\big)= 2 x(1+2y)$$ Set $w(x,y) = u_y(x, y) + 2\, u(x,y)$. Then the equation becomes $$w_x = 2x (1+2y).$$ Integrated once with respect to $x$ you get $$w(x,y) = x^2(1+2y) + f(y)$$ for some smooth function $f(y)$. Since $u(0,y) = 0$, its first derivative with respect to $y$ can be calculated and is $u_y(0,y) = 0$ and thus $$w(0, y) = u_y(0,y) + u(0,y) = 0.$$ Consequently $$w(0,y) = f(y) \equiv 0$$ and therefore $$u_y(x,y) + 2\, u(x,y) = w(x,y) = x^2(1+2y)$$ so we end up with the differential equation (linear and ordinary) $$u_y + 2\,u = x^2(1+2y)$$ To solve it, multiply both sides with $e^{2y}$ $$e^{2y} \, u_y + 2 \, e^{2y}\,u = x^2(1+2y)e^{2y}$$ which by the product rule is the same as $$\frac{\partial}{\partial y}\big(e^{2y} \, u\big) = x^2(1+2y)e^{2y}$$ Integrate both sides $$e^{2y} \, u(x, y) = g(x) + x^2 \, \int \, (1+2y) e^{2y} \, dy$$ $$u(x, y) = g(x) \, e^{-2y} + x^2 \,e^{-2y} \, \left(\int \, (1+2y) e^{2y} \, dy\right)$$ where $g(x)$ is an arbitrary smooth function. Perform the integration and get $$u(x, y) = g(x) \, e^{-2y} + x^2 \,e^{-2y} \, e^{2y} \, y = g(x) \, e^{2y} + x^2 y $$ Finally use the second boundary condition to determine $$u(x,0) = g(x) = x e^{-x^2}$$ The solution is $$u(x,y) = x e^{-x^2 - 2y} + x^2 y$$