Linear Transformation $T\colon V\rightarrow V$ defined by $T(y)=y+e^{-2t}\frac{dy}{dt}$

Question

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.

Answer 1

Your martrix $M$ is o.k.

The equation in the question has a solution $ \iff$ the linear system

$(*)$ $M(x,y,z)^T=(1,2,3)^T$

has a solution. Since $M$ is invertible, the equation $(*)$ has exactly one solution. Hence the equation

$y+e^{-2t}\frac{dy}{dt}=1+2e^{2t}+3e^{4t}$ has xactly one solution.

Answer 2

Since teh detreminant of $M$ is 1 there can be only one solution. Equate $T(a+be^{2t}+ce^{4t})$ to $1+2e^{2}+3e^{4t}$ and you will get $a=13, b=-10,c=3$.

Answer 3

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.