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Find the general solution of the PDE $2u_x-3u_y=x$ , where $u=u(x,y)$ . We have to use the method of characteristics. I know that the slope is $-\dfrac{3}{2}$ and the characteristic lines are $-3x-2y=\text{constant}$ . After doing a change of variables $w=-3x-2y$ and $z=y$ , the PDE becomes $-3V_z=\dfrac{w+2z}{-3}$ . I am not sure on how to go on after this |
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Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example: $\dfrac{dx}{dt}=2$ , letting $x(0)=0$ , we have $x=2t$ $\dfrac{dy}{dt}=-3$ , letting $y(0)=y_0$ , we have $y=-3t+y_0=-\dfrac{3x}{2}+y_0$ $\dfrac{du}{dt}=x=2t$ , letting $u(0)=f(y_0)$ , we have $u(x,y)=t^2+f(y_0)=\dfrac{x^2}{4}+f\left(\dfrac{3x}{2}+y\right)$ |
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Assume that $u(x,y)=\frac14x^2+g(3x+2y,2x-3y)$ for some function $g:(s,t)\mapsto g(s,t)$, then $u_x=\frac12x+3g_s+2g_t$ and $u_y=2g_s-3g_t$ hence $u:(x,y)\mapsto u(x,y)$ solves the PDE iff $2(\frac12x+3g_s+2g_t)-3(2g_s-3g_t)=x$, that is, $g_t=0$. Thus, the function $u:(x,y)\mapsto u(x,y)$ solves the PDE iff $u(x,y)=\frac14x^2+g(3x+2y)$ for some function $g$. |
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