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I am trying to solve $$u_x + u_y + u = e^{x+2y} $$

I started this problem by using the coordinate method. I set $$t = x+y$$ $$p = x-y$$(Skipping a couple of steps) I got to $$u_t + \frac{1}{2}u = \frac{1}{2}e^{\frac{3}{2}t-\frac{1}{2}p}$$

and this is where I get stuck. My questions is: is this just an ODE for which I need to multiply by an integrating factor and then solve?

vincentvega
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2 Answers2

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This is a first-order linear ODE. Yes, you can solve it with an integrating factor; undetermined coefficients is another method that works.

Robert Israel
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Just use the method of characteristics:

$$\mathrm{d}x = \mathrm{d}y = \mathrm{d}u/\left(e^{x+2y}-u\right),$$ where the first identity tells us $x-y = c_1$, where $c_1$ is a constant. Substitute $x = c_1+y$ to come up with the problem for $u$ as a function of $y$, fixed $x$ on the characteristic:

$$ \mathrm{d}u = \mathrm{d}u/(e^{c_1+3y} - u), $$ which rearranges to:

$$ \frac{\mathrm{d}u}{\mathrm{d} y} + u = e^{c_1+3y}, $$ which can be solved up to a constant of integration, say $c_2$. After that, put $c_2 = f(c_1)$ and you are done. Can you take it from here?

Cheers!

Dmoreno
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  • You can take a first look at it at http://en.wikipedia.org/wiki/Method_of_characteristics. For a reference, I'm afraid I'm not following any specific one, just what I remember from my notes. I guess any reference on partial differential equations will contain a section devoted to this method. – Dmoreno Jun 01 '15 at 01:44