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I'm asked to find the general solution of $$au_{xx}+bu_{xy}=0$$ With $u=u(x,y)$ and $a$, $b$ real constants. I'm just starting with PDE's, haven't seen any resolution technique except for basic changes of variables and factorizations, and sometimes assuming the solutions might be of a certain form and work with that. First thing I tried was writting: $$au_{xx}+bu_{xy}=(au_x+bu_y)_x=0$$ Which implies $$au_x+bu_y=f(y)$$ It should be simpler to solve this PDE, but due to my amateur state I haven't been able to. I've tried a linear change of independent variables to see if I could get some cancellation by choosing proper coeficients $$w=Ax+By$$ $$v=Cx+Dy$$ Substituting back in the equation $$Aau_w+Cau_v+Bbu_w+Dbu_v=f(y)$$ It doesn't look like I could simplify that, plus I'm not sure how to deal with that arbitrary $f(y)$. I tried this change directly in the original second order equation to no avail. Any help?

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You can successfully apply the method of characteristic curves. In your case, you should consider straight lines in the direction of $(a,b)$. So let $\phi$ the straight line through say $(x_0,0)$ with direction $(a,b)$. Then \begin{equation} \phi(t) = (x_0,0) + t(a,b). \end{equation} Let $U(t)=u(\phi(t))$. Then by the chain rule, $U$ is differentiable and \begin{equation} U'(t) = u_x(\phi(t)) a + u_y(\phi(t)) b = f(bt) \end{equation} You can now recover $U(t)$ through integration. There is still a small issue to consider when $b=0$.

EDIT: More information added in response to OP's comments:

Now, let $(x,y) \in \mathbb{R}^2$ be given. We which to recover $u(x,y)$. Assuming $b \not = 0$, we pick $t = y/b$ and set $x_0 = x - at$, such that \begin{equation} (x,y) = (x_0 + at, 0 + tb) = (x_0,0) + t(a,b) = \phi(t). \end{equation} Then by the above reasoning we have \begin{equation} u(x,y) - u(x_0,0) = U(t) - U(0) = \int_0^t f(bs) ds \end{equation} It follows, that \begin{multline} u(x,y) = u(x_0,0) + \int_0^t f(bs)dx = u(x - ay/b,0) + \int_0^{y/b} f(bs)ds \\ = u(x - ay/b,0) + \frac{1}{b} \int_0^y f(\tau) d\tau. \end{multline} We notice that the constant $a$ and $b$ determine the point were we intersect with the $x$ axis and that they figure prominently in the final expression. It is necessary to specify the solution along a curve which crosses the characteristic curves. In this case it was convenient to use the $x$ axis. In general, the characteristic curves are not straight lines, but the principle is the same. You find suitable curves so that you exploit the chain rule.

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    $\begingroup$ I hadn't seen that method yet, however I followed your instructions. By integrating we get that $U$ will be of the form $F(bt)+C$, where $F$ is the primitive of $f$, I suppose we can consider it to be an arbitrary function. So since this is how $u$ is along the line $\phi$, we can assume this is how $u$ is in the entire plane, hence getting $u(x,y)=F(y)+C$.But this solution doesn't look correct to me, it clearly verifies the equation but it's not too general, looks like there should be more general functions also verifying the equation. And also $a$ and $b$ don't even appear, which looks wrong $\endgroup$ – F.Webber Feb 10 '16 at 17:51
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    $\begingroup$ I have added more information. Be wary of how much information you absorb into your constants of integration.. $\endgroup$ – Carl Christian Feb 10 '16 at 18:17
  • $\begingroup$ Alright, I think I've got a clearer idea now, thanks for your time. I'll go ahead and formalize it now to consolidate this ideas. $\endgroup$ – F.Webber Feb 10 '16 at 18:45

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