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After transforming the PDE: $$y^{5}u_{xx}-y.u_{yy}+2u_y=0$$ into canonical form, I get: $$-4y^{5}.v_{\xi\eta}+2y^{2}.(v_{\xi}+v_{\eta})=0$$ where $$ \begin{cases} \xi=\frac{y^3}{3}-x\\ \eta=\frac{y^3}{3}+x \end{cases} $$I have no idea finding the general solution. Hope any help from u guys!

gymbvghjkgkjkhgfkl
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Godgog Arsenal
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  • If you are going to use the canonical form, you need to transform the $y$ in $y^5$ and $y^2$ to the $\xi,\eta$ coordinates – Alex Jones Apr 09 '17 at 04:58
  • following your hints, I have transformed it into: $2.\sqrt[3]{\frac{9}{4}.(\xi+\eta)^2}$.$v_{\xi\eta}-(v_\xi+v_\eta)=0$ (since $y>0$ is given). But it still remains complicated for me – Godgog Arsenal Apr 10 '17 at 01:48
  • Divide thru the canonical form by $-2y^2$, so that you just need to solve for $2y^3$. Add your equations for $\xi$ and $\eta$ and multiply $3$ on each side to get $2y^3=3(\xi+\eta)$. With all this, the canonical form is $3(\xi+\eta)v_{\xi\eta}=(v_\xi+v_\eta)$. Alternatively, divide the original equation by $y^3$, and note that $\frac{\partial}{\partial y}(-y^{-2}u_y) = -y^{-2}u_{yy}+2y^{-3}u_y$, so that your equation becomes $y^2u_{xx} = \frac{\partial}{\partial y}(y^{-2}u_y)$. Not sure if either of these are useful, but it's all I can see right now. – Alex Jones Apr 10 '17 at 07:19

1 Answers1

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After change variables

$$\xi=y^3+3x,\quad \eta=y^3-3x$$

we get $$u_{\xi\eta}=0$$

Then

$$u=f(\xi)+g(\eta),$$ $$u=f(y^3+3x)+g(y^3-3x).$$

This is general solution.

Aleksas Domarkas
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