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I have been asked to find the solution of the following PDE: $x^2z_{xx}+2xyz_{xy}+y^2z_{yy}=0$.

I know the pde is of parabolic type. Considering the transformations $\xi = \frac yx$ and $\eta=x$, I have reduced the pde into canonical form. The canonical form is : $\eta ^2 \omega_{\eta \eta}=0$ i.e. $\omega_{\eta \eta}=0$.

Solving this we get $z=\eta f(\xi)+g(\xi)$ i.e. $z=xf(\frac yx)+g(\frac yx)$. But the options that are given as possible answers to the question are:

  1. $z=f(y+cx)+g(y-cx)$
  2. $z=(y+cx)f(y-cx)+g(y-cx)$
  3. $z=(y-cx)f(y+cx)+g(y+cx)$
  4. $z=(y-cx)f(y-cx)+g(y+cx)$

Please somebody help me to solve this problem correctly. Am I doing some mistake?

Megha
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  • Check if any of the alternatives satisfies the PDE. If not, there seems to be an error in the question. – md2perpe Jul 23 '21 at 16:33
  • There is a symmetry between $x$ and $y$ in the PDE, but not in your result, so something seems to be missing in your solution, or at least the solution can be rewritten to be symmetric. – md2perpe Jul 23 '21 at 16:34
  • Your answer seems correct to me. Here the actual fact (and in fact according to me the most suitable way) to solve such kind of problem is to check if any of the alternatives satisfies the given partial differential equation. – nmasanta Jul 23 '21 at 17:59

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