Let $f$ be such that $f'(x)=xf(x)$ Use induction to proove that:
$f^{(n)}(x) = xf^{(n−1)}(x) + (n − 1)f^{(n−2)}(x)$
holds for all $n \geq 2$.
How do I even prove that this holds for n=2? I've tried taking the second derivative and plugging in n = 2, but what I get makes no sense. I'm then asked to relate this result with maclaurin series and a differential eq. Have made some progress on this, but i cannot figure out how to do the induction step.
As mentioned in the comments for $n=2$ you just need to differentiate and you get $f^{(2)}(x) = f(x) + x f'(x) = x f^{(2-1)}(x) + (2-1) f^{(2-2)}(x)$ which is what you wanted.
For the induction step differentiate the formula $f^{(n)}(x) = xf^{(n−1)}(x) + (n − 1)f^{(n−2)}(x)$: $$f^{(n+1)}(x) = f^{(n−1)}(x) + xf^{(n)}(x) + (n − 1)f^{(n−1)}(x)= xf^{(n)}(x) + (n)f^{(n−1)}(x)$$ and we are done.
According to your assumption if $f'(x)=xf(x)$ then, firstly $f(x)=0$ is one solution. And, $$f'(x)=xf(x) \implies \frac{f'}{f}=x$$ Note that $\frac{f'}{f}$ is the logarithmic derivative of $f$. So, $$\frac{d}{dx}\ln f=x$$ Integration on both sides, $$\ln f(x)=\frac{x^2}{2}$$ $$f(x)=e^{\frac{x^2}{2}}$$ Now I believe you can proceed by yourself.
