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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 $y$.

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    $\begingroup$ $y=c$ is a particular solution but it cannot satisfy the boundary conditions. The general solution will contain the logarithmic integral. $\endgroup$ – Vasya Oct 30 '18 at 18:37
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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 $y(x)$.

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Let $y'=f(y)=>y''=f'(y)y'=ff'=>ff'=f/y=>f\equiv0\ or f'=1/y=>$ $$ \\\int{df}=\int{dy/y} \\f=\ln|y|+const \\y'=ln|y|+const \\x=\int{\frac{dy}{ln|y|+const}} $$

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$$y''=\frac{y'}{y}$$ $$\int y''dy=\int\frac{y'}{y}dy$$ $$y'=\ln|y|+C_1$$ $$y'=\ln|y|+\ln(C_2)$$ $$y'=\ln|C_2y|$$ $$\int\frac{dy}{\ln(y)}=\int dx$$ $$x=\int\frac{dy}{\ln(y)}+C_3$$ and this has no elementary antiderivative

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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?

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