Linear Transformation $T\colon V\rightarrow V$ defined by $T(y)=y+e^{-2t}\frac{dy}{dt}$ 
Consider the set of functions $B=\left\{ {1,e^{2t},e^{4t}}\right\}$. It is given that B is a linearly independent set. Write $V=\text{span}(B)$ and let $T\colon V\rightarrow V$ be the linear transfomration defined by $$T(y)=y+e^{-2t}\frac{dy}{dt}$$
  Using the matrix M of T with respect to the ordered basis $B$, decide whether the equation $$y+e^{-2t}\frac{dy}{dt}=1+2e^{2t}+3e^{4t}$$
  has no solution, one solution or more than one solution $y$ in $V$. Find the solutions (if any).

I managed to determine the matrix $M=\left( {\begin{array}{cc}
   1 & 2 & 0 \\
   0 & 1 & 4 \\
   0 & 0 & 1 \\ 
  \end{array} } \right)$ as
\begin{align*}
T(1)&=1\\
T(e^{2t})&=2+e^{2t} \\
T(e^{4t})&=4e^{2t}+e^{4t} \\
\end{align*}
But how do you determine where the equation in the question has a solution or not? Intuition tells me it has 3 solutions, but I have no valid reasoning to approach this hypothesis.

Solution. The matrix representation $M$ implies that the dimension of the column space and consequently that of $\operatorname{range}T = 3$ thus by the rank-nullity theorem $$\dim\operatorname{null}T = \dim\operatorname{span}\{1,e^{2t},e^{4t}\}- \dim\operatorname{span}\{T(1),T(e^{2t}),T(e^{4t})\}
 = 0$$ implying that $T$ is injective and surjective consequently the equation in question must not only have a solution but must have a unique solution.
I hope you found the above useful.
