# Need help solving what should probably be a very simple PDE

Trying to teach myself PDEs, and I'm stumped on what should probably be a very simple exercise:

Solve the equation $3u_{y}+u_{xy}=0$. And I am given the hint to let $v=u_{y}$ (it's a problem from Strauss' intro book).

Now, when you make the suggested substitution, $3u_{y}+u_{xy}=0$ becomes $3v+v_{x}=0$. The only problem is, I don't know how to solve this kind of PDE; the only types of PDE the book has really talked about at this point are ones of the form $au_{x}+bu_{y}=0$, where $a$ and $b$ are constants, and ones of the form $u_{x}+yu_{y}=0$.

What I've tried to do, therefore, is rewrite $v_{x}$ as $\frac{dv}{dx}$, subtract $3v$ from both sides, and turn it into a type of separable ODE type thing. Then, if I do that, and after substituting $v=u_{y}$ back in, I wind up getting that $u_{y}=\exp{(-3x)}\exp{(f(y))}$ Then, I suppose I'd have to integrate both sides with respect to $y$ to get the solution, but I'm worried it will take some messy integration by parts that never stops, and so I think that this method couldn't possibly be right.

Could somebody tell me the RIGHT way to do this problem? Thanks!! :)

• Since you eliminated $y$ for $3v + v_x = 0$, you can treat it as $y_x + 3y = 0$. Use method of constant coefficients. I will post the answer as long as there are no users who post the best answers. – NasuSama Feb 9 '14 at 20:54
• You mean the method of undetermined coefficients? – ALannister Feb 9 '14 at 20:56
• Method of constant coefficients. That is: Set $y = e^{rx}$ where $r$ is any constant. Evaluate the ODE by substitution and algebra. – NasuSama Feb 9 '14 at 20:56
• Gotcha. Will try. – ALannister Feb 9 '14 at 20:57
• Yeah, it's still not working. Anyway, isn't that method only for inhomogeneous equations? Doesn't help me. If I try to get just the homogeneous part, I end up with $C\exp{(-3x)}$, which is exactly the same thing I got before, except that my constant was a constant function, $\exp{(f(y)}$, call it $g(y)$. So, I have that $u_{y}=g(y)\exp{(-3x)}$ - how do I go from that to give me an expression for $u(x,y)$? – ALannister Feb 9 '14 at 21:27

We are given that

$$3u_y + u_{xy} = 0$$

As the hints suggest, we use the substitution: $v = u_y$. Then, as you said before, the new ODE is

$$3v + v_x = 0$$

By the method of constant coefficients, setting $v = e^{rx}$, we have...

\begin{aligned} re^{rx} + 3e^{rx} &= 0\\ r + 3 &= 0\\ r &= -3 \end{aligned}

So we have

$$v(x) = c_1e^{-3x}$$

where $c_1$ is any constant. But since $v = u_y$,

$$u_y(x,y) = c_1e^{-3x}$$

which implies that a solution is

$$u(x,y) = c_1ye^{-3x}$$

Note. $$3u_y+u_{xy}=0 \quad\Longleftrightarrow\quad \frac{\partial}{\partial y}(3u+u_x)=0 \quad\Longleftrightarrow\quad 3u(x,y)+u_x(x,y)=f(x),$$ for some function $f=f(x)$. Next $$\mathrm{e}^{3x}\big(3u(x,y)+u_x(x,y)\big)=\mathrm{e}^{3x}f(x) \quad\Longleftrightarrow\quad \big(\mathrm{e}^{3x}u(x,y)\big)_x =\mathrm{e}^{3x}f(x),$$ which is equivalent to $$\mathrm{e}^{3x}u(x,y) =u(0,y)+\int_0^x \mathrm{e}^{3t}f(t)\,dt,$$ or $$u(x,y) =\mathrm{e}^{-3x}\,u(0,y)+\int_0^x \mathrm{e}^{-3(x-t)}f(t)\,dt,$$ for a function $f$ which can be determined from the initial data.