How do you solve equations of $ f'(t)=f(t)^a$?

Question

I was doing some math for fun and I came across equations of this form: $f'(t)=f(t)^a $. I understand how to solve it for case $a=1$ -it's just $f(t)=Ae^t+C$. And I've hear that for $a=2$ the solution is $f(t)=-\frac{1}{t-c}$. But I haven't been able to find a general method for solving all cases. I'm mostly self taught and I've only take up to Calc 2 at my university. If anyone knows how to do it, I'd greatly appreciate it.

Answer 1

Separate the variables: $$\frac{df}{dt}= f^ a\Rightarrow f^{-a}df = dt \Rightarrow \int f^{-a}df = \int dt$$ If $a\ne1$: $$\frac{f^{1-a}}{1-a} = t + C\Rightarrow f^{1-a} = (1-a)t + C\Rightarrow f=\left\{(1-a)t + C\right\}^{\frac{1}{1-a}}$$ else: $$\frac{df}{dt}=f\Rightarrow f = C\exp x$$ For example, let $a=2$. Using formula we get: $$f=\{-t + C\}^{-1} = \frac{1}{C - t}$$