Help solving $2y'-4y\left(\frac{y''}{y'}\right)=\log\left(\frac{y''}{y'}\right)$

Question

I'm at the begginig of a differential equation course, and I'm having trouble finishing this problem. I'm asked to solve the equation $$2y'-4y\left(\frac{y''}{y'}\right)=\log\left(\frac{y''}{y'}\right). \ \ \ \text{(I)}$$ I'm used the variable change $p=y'$, $\ \dot p=\frac{dp}{dy}$, $ \ y''=\dot pp$, I get the equation: $$p=2y\dot p+\frac{1}{2}\log \dot p. \ \ \ \text{(II)}$$ That last one is a Lagrange differential equation. I solved it finding a parametric solution using the variable change $\dot p = u$, and using that I got the first order linear equation $$\frac{dy}{du}+\frac{2y}{u}=\frac{-1}{2u^2}.$$ Solving it by integrating factor, I finally found the parametric values for $y$ and $p$, and they are $$ \left\{ \begin{array}{rcl} y(u,C)&=&\frac{C}{u^2}-\frac{1}{2u} \\ p(u,C)&=&\frac{2C}{u^2}-1+\frac{1}{2}\log u \end{array} \right. $$ This two equations are the solution to the equation (II), but I'm asked to solve the equation (I). Since $p=y'(x)$ in the original equation (I), I integrate the expression of $p$ with respect to $x$, and I get $$y=\int\left(\frac{2C}{u^2}-1+\frac{1}{2}\log u\right)\ dx + K$$ $$\Downarrow$$ $$\boxed{y(x)=x\left(\frac{2C}{u^2}-1+\frac{1}{2}\log u\right) + K}$$ This may be my solution, but something feels wrong about it, since my original $y$ in equation (I) is some $y(x)$, but what I get is a single equation $y(x,u)$ ($C,K$ are constants), and I'm not sure if I'm forgetting another expression to make it have sense. I feel that the solution to (I) must be a parametric one, but I only get one equation for $y$ in function of $x$ and $u$. Is the boxed equation the right solution? If not, where am I wrong? Any help will be appreciated.

Answer 1

I'll try to formally trace your solution. First you apply the trick $y'(x)=p(y(x))$ with the dot indicating the $y$ derivative. This gives correctly equation (II). Next you set $\dot p(y)=u$ and compute $y$ as a function of $u$, $y=v(u)$, giving the combined equation $y(x)=v(\dot p(y(x)))$, that is, $v$ is the inverse function to $u$. This transforms (II) into $$ p(v(u))=2v(u)u+\frac12\ln(u) $$ and taking the $u$ derivative \begin{align} \dot p(v(u))v'(u)=2v'(u)u+2v(u)+\frac1{2u} &\implies 0=v'(u)+\frac{2v(u)}{u}+\frac1{2u^2}\tag{IIIa} \\ &\implies 0=[u^2v'(u)+2uv(u)]+\frac12 \\ &\implies C=u^2v(u)+\frac{u}2\tag{IIIb} \end{align} and thus $$ p(v(u))=2\left(\frac{C}u-\frac12\right)+\frac12\ln(u) $$ At this point you may note that there is a different power in the denominator as $uv(u)=\frac Cu-\frac12$.

Though the connection may be weak, there is still dependence between $x$ and $u$. You can not set $u$ constant in the $x$-integration of $y'(x)=p(y(x))$. Rather you would attempt to get $x$ also as a function $x=w(u)$, so that from $y(w(u))=v(u)$ one gets the equation $y'(w)w'(u)=p(v(u))w'(u)=v'(u)$ where $v'(u)$ is known from (III), giving $$ x=\int w'(u)du=\int\frac{v'(u)}{p(v(u))}du=\int\frac{-\frac{2C}{u^3}+\frac1{2u^2}}{2\frac{C}u-1+\frac12\ln(u)}du $$ which does not give the impression of an easy solution.

Answer 2

Differentiating $p=2y\dot p+\log\sqrt{\dot p}$ gives $\dot p=2\dot p+2y\ddot p+\ddot p/(2\dot p)$ so $(1+4yq)\dot q=-2q^2$ where $q=\dot p$. As this is exact, the general solution is $c=f(y)$ where $f_q=1+4yq$ and $f_y=2q^2$. Thus $f(y)=d+q+2yq^2$ so $q=(-1\pm\sqrt{1+Cy})/4y$ and substituting back into $p=y'$ yields $x=\int\frac{dy}{\int q(y)\,dy}$.