Differentiating the matrix with exponential function?

Question

I need to show that

$$ae^{\lambda t}(1,0)+be^{\lambda t}(t,1)$$ is a general solution of

$$ X'=\begin{bmatrix} \lambda & 1 \\ 0 & \lambda \end{bmatrix}X $$

I guess that the derivative of the first part is just $a \lambda e^{\lambda t } (1.0)$, but I am not sure how to differentiate the second part.

Answer 1

HINT: Write the solutions as

\begin{eqnarray} x(t) &=& ae^{\lambda t} + b t e^{\lambda t} \\ y(t) &=& be^{\lambda t} \end{eqnarray}

And show they solve the problem

\begin{eqnarray} \frac{dx}{dt} &=& \lambda x(t) + y(t) \\ \frac{dy}{dt} &=& \lambda y(t) \end{eqnarray}

Answer 2

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Note that $$be^{\lambda t}\pmatrix {t \\ 1}=be^{\lambda t}(t \pmatrix{ 1 \\0}+ \pmatrix {0 \\ 1})=be^{\lambda t}t \pmatrix{ 1 \\0}+be^{\lambda t} \pmatrix {0 \\ 1}$$

You know how to differentiate that ?

Answer 3

Well, for general solution of $$ X'=MX $$ ($M$ a constant matrix), the technique is always the same.

1. consider an invertible solution (i.e. not zero !) of $$ C'=MC $$ (with $C$ matrices with variable coefficients) like $C=e^{tM}$
2. The general solution is of the form $X=e^{tM}X_0$ where $X_0$ is the "initial vector" (i.e. at $t=0$).

Here your "multiplier" is $$ M=\begin{bmatrix} \lambda & 1 \\ 0 & \lambda \end{bmatrix}= \begin{bmatrix} \lambda & 0 \\ 0 & \lambda\end{bmatrix}+ \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} $$ and, taking into account that scalar matrices ($\lambda I$) commute with everybody, you can apply $e^{A+B}=e^{A}e^{B}$ when $A,B$ commute. Then $$ e^{tM}=e^{\begin{bmatrix} t\lambda & 0 \\ 0 & t\lambda\end{bmatrix}+ \begin{bmatrix} 0 & t \\ 0 & 0 \end{bmatrix}}= e^{\begin{bmatrix} t\lambda & 0 \\ 0 & t\lambda\end{bmatrix}} e^{\begin{bmatrix} 0 & t \\ 0 & 0 \end{bmatrix}}= \begin{bmatrix} e^{t\lambda} & 0 \\ 0 & e^{t\lambda}\end{bmatrix}\begin{bmatrix} 1 & t \\ 0 & 1 \end{bmatrix}= \begin{bmatrix} e^{t\lambda} & te^{t\lambda} \\ 0 & e^{t\lambda}\end{bmatrix} $$ Hope this helps.