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I am asked to solve $$au_x+bu_y+cu=0$$ I am tempted to first solve $au_x+au_y=0$ which has characteristic lines $C=ay-bx$ and thus a solution to this is given by $$u(x,y)=f(ay-bx)$$ where $f$ is an arbitrary function. Then substituting back into the original equation yields $$au_x+bu_y+cu=0+cu=cf(ay-bx)=0$$ implying that I have merely found the trivial solution $u(x,y)=f(ay-bx)=0$.

So far the book I am using has only explained the method of characteristic equations and I have solved various difficult ones like $\sqrt{1-x^2}u_x+u_y=0$. So I am guessing that I should be able to solve $au_x+bu_y+cu=0$ using this method combined with maybe some clever thinking. I might be able to use the fact that the directional derivative of $u$ along the lines $C=ay-bx$ is $-cu$ and so maybe along these lines $u=e^{-cf(ay-bx)}$ or something. If anyone has any suggestions I would be thankful.

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I think I just found a solution $$u=Ke^{-c\left(\frac{1}{2}\left(x+y\right)\right)}$$ where $K$ is an arbitrary constant. This was sort of trial and error though, so anyone who has an interesting way to arrive here speak up please :)! – Slugger Feb 14 '13 at 20:50
Your solution doesn't hold for arbitrary $a$ and $b$. You can check by simple substitution. It's valid only for $a = b = \frac 12$ – Kaster Feb 14 '13 at 23:59
up vote 2 down vote accepted

Make a change of variables $$ t = bx + ay \\ p = bx - ay $$ So $$ u_x = b(u_t + u_p) \\ u_y = a(u_t - u_p) $$ After substituting in PDE $$ ab(u_t + u_p) + ab(u_t - u_p) + cu = 2ab\ u_t + cu = 0 $$ It can be easily integrated $$ \frac {u_t}u = -\frac c{2ab} \\ \ln u = -\frac c{2ab}t+f(p) \\ u = F(p)e^{-\frac c{2ab}t} $$ or, in initial variables $$ u = F(bx-ay)e^{-\frac c{2ab}(bx+ay)} $$ where $F(x) = e^{f(x)}$


If you use $$ t = ax + by \\ p = bx - ay \\ $$ so equation is $$ (a^2+b^2)u_t + cu = 0 \\ \ln u = -\frac c{a^2+b^2}t \\ u = F(p)e^{-\frac {ct}{a^2+b^2}} \\ u = F(bx - ay)e^{-\frac c{a^2+b^2} (ax + by)} $$ which is also a solution.

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Sorry, I made a mistake. Fixed it though. – Kaster Feb 14 '13 at 23:26
Very clear answer, thanks! – Slugger Feb 16 '13 at 14:35

let $v(x,y)=e^{cx}u(x,y)$ and compute $v_x,v_y$ then we have$$v_x+v_y=0$$ by geometric method we have$$ v(x,y)=e^{\frac{-c}{a}}f(y-x)$$ easily conclude $$u(x,y)=e^{\frac{-c}{a}}f(\frac{y}{b}-\frac{x}{a})$$

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