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Can someone help me to solve this equation for $y$ and $z$?

\begin{equation} y(x_1)^2[\dfrac{3}{a_1^2}(\dfrac{y''(x_1)}{y(x_1)} + \dfrac{y'(x_1)^2}{y(x_1)^2}) + \dfrac{2}{a_2^2}( \dfrac{a_3 a_2^2 z(x_2)}{4 a_4^4} + 2 z(x_2) z''(x_2) + 3 z'(x_2))]=C \end{equation}

where $a_1$, $a_2$, $a_3$,$a_4$ and $C$ are important constant, $z$, $y$, $x_1$ and $x_2$ are variables.

With separation we can also find two equation:

\begin{equation} \begin{aligned} \dfrac{3}{a_1^2} \left(\frac{y''\left(x_1\right)}{y\left(x_1\right)}+\frac{y'\left(x_1\right){}^2}{y\left(x_1\right){}^2}\right)-C y\left(x_1\right){}^{-2}=k \end{aligned} \end{equation}

and \begin{equation} \begin{aligned} \dfrac{2}{a_2^2} \left(\frac{a_3 a_2^2 z\left(x_2\right)}{4 a_4^4}+2 z\left(x_2\right) z''\left(x_2\right)+3 z'\left(x_2\right)\right)=k, \end{aligned} \end{equation}

where $k$ is separation constant.

Thank you for your help.

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  • $\begingroup$ In the particular case where $k=0$ then the $y$ equation simplifes to $y''y + y'^2 = D \implies \frac{1}{2}[y^2]'' = D$ which is easy to solve. Don't see how to do it for a general $k$ though... $\endgroup$ – Winther Feb 6 '17 at 20:48
  • $\begingroup$ thank you for this idea! but i nead solution with k. $\endgroup$ – user413534 Feb 7 '17 at 14:21
  • $\begingroup$ No one has any Idea ? :( $\endgroup$ – user413534 Feb 9 '17 at 11:07
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I've forgotten square, so right equation looks like this: \begin{equation} y(x_1)^2[\dfrac{3}{a_1^2}(\dfrac{y''(x_1)}{y(x_1)} + \dfrac{y'(x_1)^2}{y(x_1)^2}) + \dfrac{2}{a_2^2}( \dfrac{a_3 a_2^2 z(x_2)^2}{4 a_4^4} + 2 z(x_2) z''(x_2) + 3 z'(x_2)^2)]=C \end{equation}

where $a_1$, $a_2$, $a_3$,$a_4$ and $C$ are important constant. z,y,$x_1$ and $x_2$ are variables.

with separation we can also find two equation:

\begin{equation} \begin{aligned} \dfrac{3}{a_1^2} \left(\frac{y''\left(x_1\right)}{y\left(x_1\right)}+\frac{y'\left(x_1\right){}^2}{y\left(x_1\right){}^2}\right)-C y\left(x_1\right){}^{-2}=k \end{aligned} \end{equation}

and

\begin{equation} \begin{aligned} \dfrac{2}{a_2^2} \left(\frac{a_3 a_2^2 z\left(x_2\right)^2}{4 a_4^4}+2 z\left(x_2\right) z''\left(x_2\right)+3 z'\left(x_2\right)^2\right)=k \end{aligned} \end{equation}

The problem in this site, i can not change what i wrote!!!

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  • $\begingroup$ You can edit anything you type. Do you need help with that? $\endgroup$ – The Count Feb 12 '17 at 18:27

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