Using Linear Transformations to Solve Differential Equation Define $T:P_3 -> P_3$ by $T(f)(t) = 2f(t)+(1-t)f'(t)$.
a) Show that T is a linear transformation.
b) Give the matrix representing T with respect to the "standard basis" {$1,t,t^2,t^3$}.
c) Determine ker(T) and Image(T).
d) Let $g(t) = 1+2t$. Use your answer from (b) to find a solution to the differential eqn $T(f)=g$.
e) What are all the solutions of $T(f)=g$?

Here is my attempt. Please critique my work and help me with the questions I am stuck on. 
a) Was pretty easy. Will spare y'all the details.
b) Let $f(t) = a+bt+ct^2+dt^3.$
$T(f(t)) = 2a+2b+2c+2d+(1-t)(b+2ct+3dt^2)$
$ =(2a+b)+(b+2c)t+3dt^2+(-dt^3)$
So $T=\begin{bmatrix} 2&1&0&0\\0&1&2&0\\0&0&0&3\\0&0&0&-1\end{bmatrix}$.
c) Using (b), we write: 
$2x_1+x_2=0$
$x_2+2x_3=0$
$x_4=0$
so $ker(T) = \begin{bmatrix} 1\\-2\\1\\0 \end{bmatrix}$
Image(T) = span{$\begin{bmatrix} 2\\0\\0\\0 \end{bmatrix}$, $\begin{bmatrix} 0\\2\\0\\0 \end{bmatrix}$, $\begin{bmatrix} 0\\0\\3\\-1 \end{bmatrix}$}.
d) I set:
$\begin{bmatrix} 2&1&0&0\\0&1&2&0\\0&0&0&3\\0&0&0&-1\end{bmatrix} x = \begin{bmatrix} 1\\2\\0\\0\end{bmatrix}$.
and got $x = \begin{bmatrix} 0.25\\0.5\\0.75\\0\end{bmatrix}$.
How do I interpret this?
e) Not sure how to do this part without understanding (d).

Your answer to part (d) is equivalent to finding $f = 0.25 + 0.5 t + 0.75 t^2 + 0t^3$ solves $Tf = g$. 
You can check: 
$$
2 f + (1 - t)f' = 0.5 + t + 1.5 t^2 + 0.5 + 1.5 t - 0.5 t - 1.5 t^2 \\
= 1 + 2t = g.
$$
For the final part, note that, if $f = 0.25 + 0.5 t + 0.75 t^2$ and $f_0 \in \textrm{Ker}(T)$, then $T(f + f_0) = T(f) + T(f_0) = T(f) = g$. That is, you can add any multiple of $1 - 2t + t^2$ to the answer from the last part to yield another solution.
Note that, in the language of differential equations, this is "The general solution to the inhomogeneous linear differential equation $T(f) = g$ is a linear combination of a particular solution $T(f_p) = g$ and a solution of the homogeneous problem $T(f_0) = 0$." (if/when an initial condition is specified, the solution becomes unique)
