# Finding the general solution of the PDE $xu_x-xyu_y -u=0$ using a side condition

Find the general solution of the PDE: $$xu_x-xyu_y -u=0$$

I have found it to be: $u(x,y)=-xf(ye^x)$

This PDE has the property that $u(0,y)=0$. Therefore, $u(0,y)$ cannot be arbitrarily prescribed, even though the $y$-axis crosses each characteristic curve only once.

How do I explain this apparent discrepancy by giving the characteristic curves their preferred parametrizations $(x(t),y(t))$ with

$$x'(t)=x(t)\quad \mbox{ and } \ y'(t)=-x(t)y(t)$$

Any help will be appreciated, thank you.

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I don't quite understand what you're trying to explain. Can you clarify? – Tunococ Jan 25 '13 at 3:40

The problem is already apparent in the equation: $xu_x-xyu_y -u=0$: when $x=0$, both coefficients of derivatives vanish, the PDE drops in order from $1$st to $0$th (which is not even a PDE at all). Whenever you see a differential equation changing order like that, expect weirdness in solutions.
As a model, consider the ODE $xu'-u=0$ where $u$ is an unknown function of $x$. The solutions are $u(x)=cx$, for any constant $c$. They all vanish at $x=0$. The theorems about existence and uniqueness for initial value problems do not apply at $x=0$, because they are for equations of the form $u'=F(x,u)$. Here $F(x,u)=u/x$ which is not continuous at $0$.
This is what happens when you work along the characteristics of your PDE. They all get to $x=0$, but since the PDE is degenerate there, one can't solve the initial value problem. What to do: prescribe the values of $u$ somewhere else.