# Solving an ODE with two dependent variables

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.

• What makes you believe this has an explicit solution? Apr 20, 2021 at 6:37
• @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? Apr 20, 2021 at 6:46

$$\theta''y+2\theta'y'+2y'+2y=0$$ $$\theta''+2(\theta'+1)\frac{y'}{y}+2=0$$ $$\frac{y'}{y}=-\frac{\theta''+2}{2(\theta'+1)}$$ $$\ln|y|=-\int \frac{\theta''+2}{2(\theta'+1)} dx+\text{constant}$$ 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.

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,

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