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Here $\theta\equiv\theta(x),y\equiv y(x)$. Need to solve $\theta''y+2\theta'y'+2y'+2y=0$ for $y$, possibly in terms of $\theta$ and $x$.

I tried to apply "method of grouping" by multiplying through with $y$, \begin{align*} \theta''y^2+2\theta'y'y+2y'y+2y^2=0\\ d(\theta'y^2+y^2)+2y^2=0 \end{align*}

Cannot figure out how to deal with the $2y^2$ above. Trying to avoid some kind of "implicit" solution if possible.

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    $\begingroup$ What makes you believe this has an explicit solution? $\endgroup$ Apr 20, 2021 at 6:37
  • $\begingroup$ @NinadMunshi nothing. It would make my work easier. This is part of a bigger problem. Anyway, if implicit solution is the only option, what would that look like? $\endgroup$ Apr 20, 2021 at 6:46

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$$\theta''y+2\theta'y'+2y'+2y=0$$ $$\theta''+2(\theta'+1)\frac{y'}{y}+2=0$$ $$\frac{y'}{y}=-\frac{\theta''+2}{2(\theta'+1)}$$ $$\ln|y|=-\int \frac{\theta''+2}{2(\theta'+1)} dx+\text{constant}$$ Given a function $\theta(x)$ the function $y(x)$ is : $$y(x)=C\:\exp\left(-\int \frac{\theta''+2}{2(\theta'+1)} dx \right)$$ In the general case this cannot be simplified in term of $\theta(x)$ without integral.

Of course, if $\theta(x)$ is known explicitly and if the integral can be explicitly expressed, then $y(x)$ is obtained explicitly.

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Hint:

Multiplying by $y$,

$$\theta''y^2+2\theta'yy'+2yy'+2y^2=0$$ can be simplified to $$(\theta'y^2+y^2)'+2y^2=0$$ or, with $\phi=(\theta+x)'$ and $z=y^2$, $$(\phi z)'+2z=0.$$

From this,

$$\theta=-\int\frac2{y^2}\left(\int y^2dx+c\right)dx+c'-x.$$

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