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Transform the differential equation $$yu_x(x, y) - xu_y(x, y) = xyu(x, y)$$ by introducing new variables $s = x^2+y^2$ and $t = e^{-x^2/2}$.

Then determine the general solution to the equation.

I think I managed to transform the differential equation? First we find these partial derivatives of $u$

\begin{align} u_x &= \frac{\partial u}{\partial s}2x - \frac{\partial u}{\partial t}tx \\ u_y &= \frac{\partial u}{\partial s}2y \end{align} then we plug them into our differential equation $$ \frac{\partial u}{\partial s}2xy - \frac{\partial u}{\partial t}txy - \frac{\partial u}{\partial s}2xy = xyu(x, y) \implies -\frac{\partial u}{\partial t}txy = xyu(x, y). $$

So our transformed differential equation is $$-\frac{\partial u}{\partial t}txy = xyu(x, y).$$


I have no idea how I'm supposed to find the general solution. The answer for the general solution is $u(x, y) = F(x^2+y^2)\cdot e^{\frac{x^2}{2}}$. How did they arrive at this general solution, and did I do the transformation right?

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You just need to finish. First, you can divide by $xy$. And you need to replace $u(x,y)$ with $u(s,t)$. What you get is $$-\frac{1}{u}\frac{\partial u}{\partial t}=\frac{1}{t}$$ We now search for a solution of the form $F(s)\cdot G(t)$. With some small manipulation, what you get is $$\frac{dG}{G}=-\frac{dt}{t}$$From here $\ln G=-\ln t +\ln C$, where $C$ is some constant (or function) independent of $t$. Taking the exponential, $G(t)=Ct^{-1}$. Now plugging it back into $u$, you can absorb $c$ into the function $F(s)$, and you get $$u(s,t)=F(s)t^{-1}$$ or if you write it in terms of $x$ and $y$: $$u(x,y)=F(x^2+y^2)e^{\frac{x^2}{2}}$$

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  • $\begingroup$ But taking the exponential of $\ln{G} = -\ln{t} + \ln{C}$ should give us $G(t) = t^{-1}C$? How did you get $Ce^{-t}$? And how do you know the solution is of the form $F(s)\cdot G(t)$? $\endgroup$ – wednesdaymiko Aug 13 '18 at 22:49
  • $\begingroup$ oops. I'll fix the error $\endgroup$ – Andrei Aug 13 '18 at 22:52
  • $\begingroup$ Thanks for fixing it! One last thing I don't really get is how did you know that the general solution is of the form $F(s)\cdot G(t)$? $\endgroup$ – wednesdaymiko Aug 13 '18 at 22:58
  • $\begingroup$ You still here? Would love some help from anyone. $\endgroup$ – wednesdaymiko Aug 14 '18 at 0:14
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    $\begingroup$ I'm not sure the separation of variables methods can always be used. It's just a guess (maybe an educated guess sometimes). In this case, the hint was that the differential equation that you get is only a function of $t$. $\endgroup$ – Andrei Aug 14 '18 at 3:24
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you can treat this equation like a differential equation $$-\frac{\partial u}{\partial t}txy = xyu(x, y).$$ $$\frac{\partial u}{\partial t}txy+xyu(x, y)=0$$ $$xy(\frac{\partial u}{\partial t}t+u)=0$$ $$u't+u=0$$ $$(ut)'=0$$ Integrate $$ut=K$$ Note that since it's an equation with partial derivative we have to consider the constant K as a function of s only $$ut =g(s)$$ $$ \implies u(s,t)=t^{-1}g(s)$$ now substitute back the old variables $$u(x,y)=g(x^2+y^2)e^{x^2/2}$$ And g is any function ...

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