Solve Repeated Roots ODE: $y' = \begin{pmatrix}1 & 2\\ 0 & 1\end{pmatrix}y$

[Dustinsfl]

$y' = \begin{pmatrix}1 & 2\\ 0 & 1\end{pmatrix}y$
The characteristic equation is
$$
\lambda^2 - 2\lambda + 1 = (\lambda - 1)^2 = 0.
$$
So the eigenvalues are $\lambda_{1,2} = 1$.
Solving $(1 - \lambda)y_1 + 2y_2 = 0\iff y_2 = -\dfrac{1}{2}(1 - \lambda)y_1$, we have
$$
y = \begin{pmatrix} 1\\ -\frac{1}{2}(1 - \lambda)\end{pmatrix}.
$$
Then $\mathbf{y_1} = e^t\begin{pmatrix} 1\\ 0\end{pmatrix}$.
To find $\mathbf{y_2}$, let $P = \begin{pmatrix}p_1\\ p_2\end{pmatrix}$.
We must now solve
$$
(A - 1I)P = \begin{pmatrix} 1\\ 0\end{pmatrix}\iff \begin{pmatrix} 0 & 2\\ 0 & 0\end{pmatrix}P = \begin{pmatrix} 1\\ 0\end{pmatrix}.
$$
We are left with the equation $0p_1 + 2p_2 = 1$.
So $p_1 = 0$ and $p_2 = \dfrac{1}{2}$.
Then $\mathbf{y_2} = te^t\begin{pmatrix} 1\\ 0\end{pmatrix} + e^t\begin{pmatrix} 0\\ \frac{1}{2}\end{pmatrix}$.
Thus, the general solution is
$$
y = C_1 e^t\begin{pmatrix} 1\\ 0\end{pmatrix} + C_2\left[te^t\begin{pmatrix} 1\\ 0\end{pmatrix} + e^t\begin{pmatrix} 0\\ \frac{1}{2}\end{pmatrix}\right].
$$

Correct?

 

[Sudharaka]

$y' = \begin{pmatrix}1 & 2\\ 0 & 1\end{pmatrix}y$
The characteristic equation is
$$
\lambda^2 - 2\lambda + 1 = (\lambda - 1)^2 = 0.
$$
So the eigenvalues are $\lambda_{1,2} = 1$.
Solving $(1 - \lambda)y_1 + 2y_2 = 0\iff y_2 = -\dfrac{1}{2}(1 - \lambda)y_1$, we have
$$
y = \begin{pmatrix} 1\\ -\frac{1}{2}(1 - \lambda)\end{pmatrix}.
$$
Then $\mathbf{y_1} = e^t\begin{pmatrix} 1\\ 0\end{pmatrix}$.
To find $\mathbf{y_2}$, let $P = \begin{pmatrix}p_1\\ p_2\end{pmatrix}$.
We must now solve
$$
(A - 1I)P = \begin{pmatrix} 1\\ 0\end{pmatrix}\iff \begin{pmatrix} 0 & 2\\ 0 & 0\end{pmatrix}P = \begin{pmatrix} 1\\ 0\end{pmatrix}.
$$
We are left with the equation $0p_1 + 2p_2 = 1$.
So $p_1 = 0$ and $p_2 = \dfrac{1}{2}$.
Then $\mathbf{y_2} = te^t\begin{pmatrix} 1\\ 0\end{pmatrix} + e^t\begin{pmatrix} 0\\ \frac{1}{2}\end{pmatrix}$.
Thus, the general solution is
$$
y = C_1 e^t\begin{pmatrix} 1\\ 0\end{pmatrix} + C_2\left[te^t\begin{pmatrix} 1\\ 0\end{pmatrix} + e^t\begin{pmatrix} 0\\ \frac{1}{2}\end{pmatrix}\right].
$$

Correct?

I have checked and found nothing wrong. (Yes)