How to solve the differential equation $y''=\frac{y'}{y}$ I want to solve the initial value problem:
$$y''=\frac{y'}{y},\;\ y'(x=0)=1,\;\ y(x=0)=e$$
I have attempted to integrate on both sides but this results in a logarithm term on the r.h.s., which causes problems when I divide on both sides by it. I am not sure whether it is possible to isolate $y$.      
 A: Let us consider your initial value problem:
$$y''(x)=\frac{y'(x)}{y(x)},\;\ y'(0)=1,\;\ y(0)=e$$
Integrate on both sides with respect to $x$. Recognize that $y''(x)\ dx=dy'(x)$ and $y'(x)\ dx=dy(x)$:
$$\int y''(x)\ dx=\int \frac{y'(x)}{y(x)}dx\rightarrow\int dy'(x)=\int \frac{dy(x)}{y(x)}\rightarrow y'(x)=\ln(y(x))+\text{C}_1$$
Let $y'(0)=1$ and $y(0)=e$ and derive that $\text{C}_1=0$.
It follows that
$$y'(x)=\ln(y(x))\rightarrow \frac{y'(x)}{\ln(y(x))}=1$$
Integrate on both sides with respect to $x$. Recognize that  $y'(x)\ dx=dy(x)$:
$$\int \frac{y'(x)}{\ln(y(x))}dx=\int dx\rightarrow\int\frac{dy(x)}{\ln(y(x))}=\int dx\implies\text{li}(y(x))=x+\text{C}_2$$
Let $y(0)=e$ and derive that $\text{C}_2=\text{li}(e)$.
Therefore,
$$\text{li}(y(x))=x+\text{li}(e)$$
Here, $\text{li}(z)$ denotes the logarithmic integral.
This function does not have an inverse function, so we can't isolate $y(x)$.
A: Let $y'=f(y)=>y''=f'(y)y'=ff'=>ff'=f/y=>f\equiv0\ or f'=1/y=>$
$$
\\\int{df}=\int{dy/y}
\\f=\ln|y|+const
\\y'=ln|y|+const
\\x=\int{\frac{dy}{ln|y|+const}}
$$
A: $$y''=\frac{y'}{y}$$
$$\int y''dy=\int\frac{y'}{y}dy$$
$$y'=\ln|y|+C_1$$
$$y'=\ln|y|+\ln(C_2)$$
$$y'=\ln|C_2y|$$
$$\int\frac{dy}{\ln(y)}=\int dx$$
$$x=\int\frac{dy}{\ln(y)}+C_3$$
and this has no elementary antiderivative
A: Write your equation in the form
$$\frac{dy(x)}{dx}-\frac{d^2y(x)}{dx^2}y(x)=0$$ and Substitute $$v(y)=\frac{dy(x)}{dx}$$
Can you proceed?
