<s>[INST] Solving the differential equation $y''=\frac{1}{y}$ I tried to solve the differential equation $\frac{d^2y}{dx^2}=\frac{1}{y}$ by assuming $y(0)=1$ and $y'(0)=0$ and finding a taylor expansion for $y$ at $x=0$. By differentiating $\frac{d^2y}{dx^2}$, I got:
$\frac{d^3y}{dx^3}=-\frac{1}{y^2}\frac{dy}{dx}$
$\frac{d^4y}{dx^4}=
\frac{2}{y^3}\frac{dy}{dx}
-\frac{1}{y^2}\frac{d^2y}{dx^2}$
$\frac{d^5y}{dx^5}=
-\frac{6}{y^4}\frac{dy}{dx}
+\frac{4}{y^3}\frac{d^2y}{dx^2}
-\frac{1}{y^2}\frac{d^3y}{dx^3}$
$\frac{d^6y}{dx^6}=
\frac{24}{y^5}\frac{dy}{dx}
-\frac{18}{y^4}\frac{d^2y}{dx^2}
+\frac{6}{y^3}\frac{d^3y}{dx^3}
-\frac{1}{y^2}\frac{d^4y}{dx^4}$
Evaluating those, I got the following values:
$y^{(3)}(0)=0$
$y^{(4)}(0)=1$
$y^{(5)}(0)=4$
$y^{(6)}(0)=-17$
I then created a Python program to automatically derive more values:
$y^{(7)}(0)=84$
$y^{(8)}(0)=-483$
$y^{(9)}(0)=3192$
$y^{(10)}(0)=-23919$
I used the derivatives to get a Taylor series, and I found that the series does not converge when $x>1$. Did I make a mistake? If not, is there an approach I could use to approximate $y(2)$ or other values where $x>1$?
 [/INST] Multiply both sides of $y''= 1/y $ by $ y'$ to get $$y''y' = \frac {y'}{y}$$
Integrate to get $$(y')^2 = 2\ln y $$
which gives us $$y'=\sqrt {2\ln y+K}$$
Separate variables to get $$\int \frac {dy}{\sqrt {2\ln y+K}} =x+C$$
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