Question

Here's a fun little differential equation

Answer 1

\[y''=y^2\]

Answer 2

ud need initial conditions

Answer 3

y(0)=? y'(0)=?

Answer 4

i dunno what to do about the y^2

Answer 5

Well, looks like this ended up being harder than I thought, I figured after I did this first step:

Play around with the chain rule:
\[y''=\frac{d y'}{dx} = \frac{dy'}{dy}\frac{dy}{dx} = \frac{dy'}{dy}y'\]

\[\int y' dy' = \int y^2 dy\]
\[\frac{1}{2} (y')^2 = \frac{1}{3} y^3 + \frac{1}{2}C\]
For some reason I just sorta thought the next step would be easier:

\[\frac{dy}{dx} = \sqrt{\frac{2}{3} y^3 + C}\]


\[\int \frac{dy}{\sqrt{\frac{2}{3} y^3 + C}} = x+K\]

But yeah that's actually not really 'easy' Oh well lol.