# Solve $f'(x)=f(x)^{2}+4$

I'm rusty in ODEs, so this might be simple..

Solve $$f'(x)=f(x)^{2}+4$$

I was able to make a few observations but they don't seem too helpful. First, since $$f(x)^{2}+4\geq 4$$, we have $$f'(x)\geq 4$$ everywhere, so in particular $$f(x)$$ is increasing (this rules out constant functions, for example).

I next considered linear functions. Supposing that $$f(x)=ax+b$$, then $$f'(x)=a$$ so by the given condition we must have $$a=(ax+b)^{2}+4$$ for all $$x$$. In particular, setting $$x=1$$ gives $$a=(a+b)^{2}+4$$ and setting $$x=-1$$ gives $$a=(a-b)^{2}+4$$ In particular this means $$a+b=a-b$$ so that $$b=0$$. In this case setting $$x=0$$ gives $$a=4$$ so our function would have to be $$f(x)=4x$$, but this clearly does not satisfy the desired condition. Therefore we can rule out linear functions.

I feel like I'm approaching this the wrong way because I'm only ruling out specific classes of functions, rather than proving any properties that $$f$$ must have.

Motivation: I am preparing for technical interviews and found this problem here.

Your DE is $$y'=y^2+4$$, which is a separable equation, meaning we can "move" all the $$y$$ terms to one side and the $$x$$ terms on the other, as follows: $$\frac{dy}{dx}=y^2+4 \implies \frac{1}{y^2+4}dy = dx$$ Integrating yields $$\int \frac{1}{y^2+4}dy = \int 1dx \implies \frac 12 \arctan(\frac y2)=x+C$$ and so you can solve for $$y$$: $$y=2\tan(2x+C)$$