# What is the inverse of $f'(x)-f(x)$?

Consider the following function $$P$$ defined as $$P(f(x))=f'(x)-f(x)$$

What is the inverse of $$P$$?

I can't figure this one out. A hint would be very helpful.

• Did you look at some examples? Oct 13 at 22:05
• What's the domain of $P$? Oct 13 at 22:06
• So it is actually $P(f)=f'-f$ and we are looking for a functional $Q(P(f))=P(Q(f))=f$? Oct 13 at 22:16
• P is not one-to-one. For instance, both e^x and 2e^x are mapped to zero. Is there any more information you’re leaving out? Oct 13 at 22:18
• @John.W $P$ and $Q$ are not invertible in these cases either, since their kernels are still nontrivial. Notice that $P\left(ce^{ax}\right)\equiv 0$ and $Q\left(ce^{-bx}\right)\equiv 0$ for any constant $c$. Oct 13 at 22:26

While there is no inverse, you can find a right inverse. As Rezha Adrian Tanuharja points out, we have $$\tag{1} P(f(x)) = e^x\frac{d}{dx}\left(e^{-x}f(x)\right).$$ But then we can solve for $$f$$ to get $$\tag{2} f(x) = e^x \int e^{-x}P(f(x))\,dx,$$ for some choice of antiderivative. More importantly, $$(2)\Rightarrow(1)$$ for any choice of antiderivative, which means that $$Q(g):= x \mapsto e^x \int_a^x e^{-t}g(t)\,dt$$ satisfies $$P(Q(g))=g$$. Note that $$Q$$ is not a unique solution, since any antiderivative would have worked.
As many have noted, $$P$$ is not injective, so you can't find a well defined left inverse.
There is no inverse. The function is not injective. $$\text P(0) = \text P(e^x)$$.