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Find the general solution of the PDE $ xu_x-xyu_y-y=0 $ for all $u(x,y)$ and find the parametric form of the solution of the PDE which follows the side condition $ **u(s^2,s)=s^3** $ I got part (a) of the solution. The general solution is $ u(x,y)=-xf(ye^x) $ I have the solution that the parametric form of the PDE is $ x(s,t)=s^2e^t $ but i am not sure on how to solve it. |
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You found wrongly about the general solution. $xu_x-xyu_y-y=0$ $xu_x-xyu_y=y$ $u_x-yu_y=\dfrac{y}{x}$ Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example: $\dfrac{dx}{dt}=1$ , letting $x(0)=0$ , we have $x=t$ $\dfrac{dy}{dt}=-y$ , letting $y(0)=y_0$ , we have $y=y_0e^{-t}=y_0e^{-x}$ $\dfrac{du}{dt}=\dfrac{y}{x}=\dfrac{y_0e^{-t}}{t}$ , we have $u(x,y)=y_0\int^t\dfrac{e^{-t}}{t}dt+f(y_0)=ye^x\int^x\dfrac{e^{-t}}{t}dt+f(ye^x)$ $u(s^2,s)=s^3$ : $se^{s^2}\int^{s^2}\dfrac{e^{-t}}{t}dt+f(se^{s^2})=s^3$ $f(se^{s^2})=s^3-se^{s^2}\int^{s^2}\dfrac{e^{-t}}{t}dt$ $\therefore u(x,y)=ye^x\int^x\dfrac{e^{-t}}{t}dt+f(ye^x)$ , where $f(s)$ satisfy $f(se^{s^2})=s^3-se^{s^2}\int^{s^2}\dfrac{e^{-t}}{t}dt$ |
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