Find a general solution of the PDE for $u=u(x,y)$ by using ODE techniques.

Question

Find a general solution of the PDE for $u=u(x,y)$ by using ODE techniques.

This is a simple one, but I got stuck towards the end. I wanted to use separation of variables because it was the easiest choice. Anyway, the problem is

$u_x-2u=0$

So, clearly I'm in $C^1$ for first order partial derivative and anything in $C^k$ must be continuous. Now, I rewrite the equation like this

$\frac{du}{dx}-2u=0$

Adding $2u $ to both sides, I get

$\frac{du}{dx}=2u$

Now I multiply the $dx$ and divide the $2u$

$\frac{du}{2u} = dx$

Now this is where I am stuck... I know that by integrating to both sides I get

$\frac{1}{2}ln \mid u \mid = x+C$

I feel like using an exponential log law to make it

$ln \mid u \mid^2=x+C$

and take the exponential, but the problem is that I would have

$e^{ln \mid u \mid^2}=e^{x+C}$

That's $u^2 = e^{x+C}$!!!

and I don't want that.

The answer is $u(x,y) = f(y)e^{2x}$, but I sort of am not sure where they got the $e^{2x}$ unless I'm applying the log law wrong. Also, I'm curious to know what $f(y)$ is doing here, but then again that could have something to do with $C^1$ because $f(y) $ is required to be a $C^1$ function.

Answer 1

$\textbf{hint}$ The first step is the issue when you integrate a single variable you have freedom of a constant, however in multivariable calculus, a function that is "constant" is also $g(y)$ with respect to x, since the derivative is zero.

Answer 2

The mistake is the line after "I feel like using an exponential log law to make it"

It is instead $ln |u|^{\frac{1}{2}}$. Anyway, what you should instead do is multiply both sides by 2 and raise both sides to e.

Oh also C is a function of y, not a real number. Partial integration, I think it's called.

Try it out. Or:

${\frac{1}{2}}ln |u| = x + C(y)$

$ln |u| = 2x+2C(y)$

$|u| = e^{2x+2C(y)}$

$u = +/- e^{2x+2C(y)}$

$u = f(y)e^{2x}$

where $f(y) =+/- e^{2C(y)} $