Finding inverse of an operator on $C^{1}[0,1]$

Question

Assume an operator $T: C^{1}[0, 1] \rightarrow C[0,1]$, such that

$$ Tx(t) = (t+1)x'(t) - x(t)$$

How to find an inverse operator $T^{-1}$?

So far I've seen that for operator $Zx(t):=\frac{1}{t+1}Tx(t)=x'(t) - \frac{1}{t+1}x(t)$, we can express it as

$$ \left(\frac{1}{1+t} \right) Zx = \left(\frac{1}{1+t}x\right)'$$

However, I don't know how to proceed from here.

Answer 1

Your operator $T$ is not invertible because it is not injective.

Solving for $x$ in $Tx \equiv 0$ gives

$$0 = (Tx)(t) = (t+1)x'(t) - x(t) \implies \frac{dx}{x} = \frac{dt}{t+1} \implies x(t) = C(t+1), C \in \mathbb{R}$$

Hence for $x(t) =C(t+1)$, we get $Tx \equiv 0$, which shows $\ker T \ne \{0\}$.