It is asked to solve the PDE

$$u_x + u_y + u = e^{x+2y}$$

My attempt: We have that $$\frac{dx}{1}=\frac{dy}{1}=\frac{du}{e^{x+2y}-u} $$

$\Rightarrow \frac{dy}{dx}=1 \Rightarrow y=x+C_1 \iff C_1=y-x$

$\Rightarrow \frac{du}{dx}= e^{x+2y}-u \iff \dot{u}+u=e^{x+2y} \iff \\ \frac{d}{dx}(e^x u) = e^{2x+2y} \iff u = e^{-x} ( \frac{1}{2}e^{2x+2y} + C_2)$

As $u$ is constant along its characteristics:

$$u(x,y)=\frac{e^{-x}}{2}(e^{2x+2y} + 2C_2(y-x))$$

where $C_2$ is any function of one variable. But wolfram gave the following answer

$$u(x,y)=\frac{e^{-x}}{4}(e^{2x+2y} + 4C_2(y-x))$$

as you can check here

What am I doing wrong?


You write $$ \frac{d}{dx}(e^xu)=e^{2x+2y} $$ and then integrate with respect to $x$. If you first replace $y$ by $x+C_1$ in the right-hand side, and then integrate, you integrate $$ e^{4x+2C_1} $$ which indeed gives a division by $4$. The constant in front of $C_2$ does not matter, since $C_2$ is arbitrary.


The first passage already seems wrong to me: $$\frac{dx}{1}=\frac{du}{e^{x+2y}-u}\iff u_x= e^{x+2y}-u$$ so where is $u_y$?

In fact if you add it to the other one, which is $$\frac{dy}{1}=\frac{du}{e^{x+2y}-u}\iff u_y= e^{x+2y}-u$$ you get the wrong relation $u_x+u_y+2u=2e^{x+2y}$, which is probably where the factor $2$ comes from.

  • $\begingroup$ The $u$ in these equations is $u(s,s+c)$ which can also be written as $u(x,x+c)$ so that $\frac{d}{dx}u(x,x+c)=u_x+u_y=e^{x+2y}-u$. $\endgroup$ – Lutz Lehmann Feb 2 '18 at 8:06

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