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I'm trying to understand the method of characteristics to solve first-order PDEs. As an example in his course, my professor solve this PDE for $u(x,y)$: $x\frac{\partial u}{\partial x}-y\frac{\partial u}{\partial y}=R $

with $u(s,s)=f(s)$ given along the parametrized curve $\Gamma$ defined by $x(s)=s$ and $y(s)=s$.

To determine the characteristics he first writes: $\frac{dx}{x}=-\frac{dy}{y}$ And then immediately jumps to: $\int^x_s \frac{dx'}{x'}=-\int^y_x\frac{dy'}{y'}$

I guess he simplifies a lot the resolution of the simple ODE but I don't really understand how he does that and how he finds the characteristics that way.

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    See http://math.stackexchange.com/questions/86570/explaining-the-method-of-characteristics2012-08-12

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The basic idea of the method of characteristics is to differentiate the unknown $u$ along a path $(x(t),y(t))$: You will find $\frac{d}{dt}u(x(t),y(t))=\dot x u_x+\dot y u_y$ where I use dots for differentiation wrt $t$, and subscripts for partial derivatives.

Now, to get some help from the given PDE, we need the vector $(\dot x,\dot y)$ to be parallel to the vector $(x,-y)$ from the coefficients of the PDE. So long as $x$ and $y$ are nonzero this requirement boils down to $\frac{\dot x}{x}=-\frac{\dot y}{y}$ which, if we multiply by $dt$, becomes $\frac{dx}{x}=-\frac{dy}{y}$ just as your professor noted. Whether you wish to think of $y$ as a function of $x$, or $x$ as a function of $y$, this is the standard form of a separable ODE, and it is solved by integrating as shown.

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    Thanks for the explanation about the method but I still don't understand how to solve the separable ODE with the parameter $s$ (or $t$ as you used) as my professor did (though I have no problem to solve the equation without it). Why does he put $[s,x]$ and $[x,y]$ as domains of integration ? I guess I am mindblocking but could you just show me the steps ?2012-08-12