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I have a PDE

$$U_{xy} = 2xyU_{y}$$ with $U(1,y) = e^y + 1$ and $U(x,0) = 1/x^2$.

My attempt to solve is as follows:

Let $V(x,y) = U_y(x,y)$. Note that \begin{align} U(1,y) &= e^y + 1\\ U_y(1,y) &= e^y\\ V(1,y) &= e^y \end{align}

Substituting, \begin{align} \renewcommand{\d}{\mathrm{d}} V_x(x,y) &= 2xyV(x,y)\\ \int_{V(1,y)}^{V(x,y)}\frac{\d V}{V} &= \int_{1}^{x}2xy \d x\\ \ln V - \ln e^y &= x^2y - 1^2y\\ V &= e^{x^2y}\\ U_y &= e^{x^2y}\\ \int_{0}^{y}U_y \d y &= \int_0^y e^{x^2y} \d y\\ U(x,y) - U(x,0) &= \dfrac{e^{x^2y}}{x^2} - \dfrac{e^{0}}{x^2}\\ U(x,y) &= \dfrac{e^{x^2y}}{x^2} - \dfrac{1}{x^2} + \dfrac{1}{x^2}\\ U(x,y) &= \dfrac{e^{x^2y}}{x^2} \end{align}

However, despite satisfying the PDE. My answer gives wrong initial condition along $x=1$ (should be $e^y + 1$ but I got $e^y$). I couldn't quite figure out where I make the mistake.

Karn Watcharasupat
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    Is it possible there is a typo in your question? The boundary conditions don't agree in the point $(1,0)$. If you substitute $y=0$ in the first condition, you get $U(1,0) = e^0 + 1 = 2$, but if you substitute $x=1$ in the other one, one gets $U(1,0)=\frac{1}{1^2}=1$. What is the domain you want to solve this equation on? – Ernie060 Nov 17 '19 at 19:48
  • @Ernie060 I'm suspecting that too. The domain is not stated in the question so I supposed it's $\mathbb{R}^2$. I tried solving by doing indefinite integration and found inconsistencies at $(1,0)$ as well. Perhaps there might be a mistake in the question itself after all :/ – Karn Watcharasupat Nov 17 '19 at 20:07

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