Why Can't I Solve y''=3/y²+5? Explained

[Teacame]

If I have a DE like this:
$$y''=\frac{3}{y^{2}}+5$$
Why can't I just integrate both sides to get:
$$y'=-\frac{3}{y^{1}}+5y$$?
And then integrate again to solve for y?

 

[PeroK]

If I have a DE like this:
$$y''=\frac{3}{y^{2}}+5$$
Why can't I just integrate both sides to get:
$$y'=-\frac{3}{y^{1}}+5y$$?
And then integrate again to solve for y?

Because $y''$ means $\frac{d^2y}{dt^2}$

So, you would need to integrate both sides wrt $t$.

If you let $\frac{d^2f}{dy^2} = \frac{3}{y^{2}}+5$

Where $f$ is a function of $y$, then you can simply integrate both sides wrt $y$.

 

[Teacame]

Because $y''$ means $\frac{d^2y}{dt^2}$

So, you would need to integrate both sides wrt $t$.

If you let $\frac{d^2f}{dy^2} = \frac{3}{y^{2}}+5$

Where $f$ is a function of $y$, then you can simply integrate both sides wrt $y$.

Oh that was really dumb of me, didn't think about the notation enough. I'm not supposed to actually solve this, I was just wondering. It looks like the solution is extremely complicated so I probably can't anyway: http://www.wolframalpha.com/input/?i=y''=3/y^2+5

In fact, the solution for just y''=1/y looks complicated: http://www.wolframalpha.com/input/?i=y''=1/y

I haven't actually studied non-linear DEs yet.

 

[pasmith]

ODEs of the form $y''(x) = f(y(x))$ can in principle be solved by multiplying by $y'$ and integrating with respect to $x$ to obtain $$\frac12(y')^2 = F(y)$$ where $F(y) = \frac12 y'(0)^2 + \int_{y(0)}^{y} f(s)\,ds$ is an antiderivative of $f$. The resulting first-order ODE is separable: $$y' = \pm\sqrt{2F(y)}$$ where some care is needed in determining the sign.