# Solve differential equation $1 + y'^2 = yy''$

I'm trying to solve this differential equation, here $y$ is a function with variable $x$:

$1 + y'^2 = yy''$

Here is my solution:

$$\left(\frac{y'}{y}\right)' = \left(\frac{y''y - y'^2}{y^2}\right) = \frac{1}{y^2}$$ Let $z = \frac{y'}{y}$. So from above equation, we have: $$z' = \frac {1}{y^2} \Rightarrow z = \frac{-1}{y} + C_1 \Rightarrow \frac{y'}{y} = \frac{-1}{y} + C_1 \Rightarrow y' = C_1y - 1$$ $$\Rightarrow \int\frac{dy}{C_1y - 1} = \int\frac{dx}{x} \Rightarrow \ln(C_1y - 1) = C_1x + C_2$$

This solution is different with the solution in my book, which is $y = C_1\cosh {x+C_1 \over C_2}$. So mysolution is wrong, but I can't find where the mistake is. Can anyone help me to point it out, and can we use my path to solve this equation? Thanks for taking your time

• If you differentiate your expression for $z$ I get $\frac{y'}{y^2}$, which is not the same as $\frac{1}{y^2}$. Oct 28, 2015 at 3:40
• the equation $y'=C_1 y-1$, the solution is not logarithm. But it is also not $cosh$. I think it is $\frac{d}{d x}(\frac{y'}{y})=\frac{1}{y^2}=\frac{d}{dy}(\frac{-1}{y}+C_1)$ , so $\frac{y'}{y} \neq (\frac{-1}{y}+C_1)$ Oct 28, 2015 at 6:23
• It does not seem that $y = c_1\cosh( {x+c_1 \over c_2})$ satisfies the differential equation. Oct 28, 2015 at 8:57

When you write $z'=1/y^2$, you are differentiating with respect to $x$. But then you integrate with respect to $y$.
The independent variable $x$ does not appear explicitly. Let $y'=p$ and consider $p$ as a function of $y$. Then $$y''=\frac{dp}{dx}=\frac{dp}{dy}\,\frac{dy}{dx}=p\,\frac{dp}{dy}.$$ The equation becomes $$1+p^2=y\,p\,\frac{dp}{dy}\implies\frac{p\,dp}{1+p^2}=\frac{dy}{y}.$$ Integrating we get $$\frac12\log(1+p^2)=\log y+C\implies p=y'=\pm\sqrt{C^2\,y^2-1}.$$ The solution is $$\int\frac{dy}{\sqrt{C^2\,y^2-1}}\,dy=\pm x+K.$$
• Oh, great. So I guessed the solution from my book(which is $y = C_1 \cosh{x + C_1 \over C_2}$ is wrong. Thanks a lot Oct 28, 2015 at 17:44