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I am trying to figure out where my solution went wrong. I am off by a factor of two.

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

I first found that the characteristic curves are determined by $$\frac{dy}{dx} = 1 \implies y-x = C.$$

I then solved the ODE $$\frac{du}{dx} + u = e^{x+2y}$$

I found $u = \frac12 e^{x+2y} + e^{-x}K(C)$ giving $$u = \frac{e^{x+2y}}{2} + e^{-x}F(y-x)$$ as the general solution where $F$ is an arbitrary function. Where does my work go wrong?

The end solution should be half of what it currently is.

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  • $\begingroup$ Were you given any initial data? $\endgroup$ – Mattos Jan 29 '15 at 2:57
  • $\begingroup$ And from your solution for $u$ you actually get $$ u_x + u_y + u = 2\exp(x + 2y) - \exp(-x)F(x-y)$$ not $$ u_x + u_y + u = 2\exp(x + 2y) $$ $\endgroup$ – Mattos Jan 29 '15 at 3:01
  • $\begingroup$ Your problem lies in the fact that you solved the ODE $$ \frac{du}{dx} + u = \exp(x + 2y) $$ incorrectly. $\endgroup$ – Mattos Jan 29 '15 at 3:27
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$$\begin{align} u_x + u_y + u &= e^{x + 2y} \\ \implies u_x + u_y &= e^{x + 2y} - u \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (*)\\ \end{align} $$

Setting $u = u(x(s),y(s))$ we find

$$\begin{align} \frac{d}{ds} u &= \frac{\partial u}{\partial x} \cdot \frac{dx}{ds} + \frac{\partial u}{\partial y} \cdot \frac{dy}{ds} \\ &= \frac{\partial u}{\partial x} \cdot 1 + \frac{\partial u}{\partial y} \cdot 1 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1) \\ &= e^{x + 2y} - u \end{align}$$

Where $(1)$ comes from our original PDE at $(*)$. Equating, we find

$$\begin{align} \frac{dy}{ds} &= 1 \\ \frac{dx}{ds} &= 1 \implies \frac{dx}{dy} = 1 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2) \\ \frac{du}{ds} &= e^{x + 2y} - u \implies \frac{du}{dy} + u = e^{x + 2y} \ \ \ \ \ \ \ \ \ \ \ \ \ (3) \\ \end{align}$$

Solving $(2)$ and $(3)$

$$x(y) = x_0 + y \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (4)$$

$$\begin{align} \frac{du}{dy} + u &= e^{x + 2y} \\ &= e^{x_0 + 3y} \\ \implies (e^{y}u)' &= e^{x_0 + 4y} \\ \implies e^{y}u &= \frac{e^{x_0 + 4y}}{4} + f(x_0) \\ \implies u &= e^{-y} \bigg(\frac{e^{x_0 + 4y}}{4} + f(x_0) \bigg) \\ &= \frac{e^{x_0 + 3y}}{4} + e^{-y}f(x_0) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (5) \\ \end{align}$$

and using $(4) \implies x_0 = x - y$ we find

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

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