Solve the system
\begin{align*}
y'_1&= \phantom{-2}y_1 \\
y'_2&= -2y_1-4y_2
\end{align*}
I think they want me to solve it by using diagonalization. So far so good. I got the following:
The coefficient matrix for the system is
$$A=
\begin{bmatrix}
\phantom{-}1 & \phantom{-}0 \\
-2 & -4
\end{bmatrix}
$$
Then I find the eigenvalues
$$
\text{det}(A-\lambda I)=
\begin{bmatrix}
\phantom{-}1-\lambda & \phantom{-}0 \\
-2 & -4-\lambda
\end{bmatrix}=
(1-\lambda)(-4-\lambda)
$$
The eigenvalues are $\lambda=1$ and $\lambda=-4$.
Then I will find the matrix $P$ that $A$ is diagonalized by. $P$ is made up by the eigenvectors of $A$. So I find the eigenvectors.
For $\lambda=1$
$$
\begin{bmatrix}
\phantom{-}0 & \phantom{-}0 \\
-2 & -5
\end{bmatrix}
\begin{bmatrix}
x_1 \\
x_2
\end{bmatrix}=
\begin{bmatrix}
0 \\
0
\end{bmatrix}
$$
and for $\lambda=-4$
$$
\begin{bmatrix}
\phantom{-}5 & 0 \\
-2 & 0
\end{bmatrix}
\begin{bmatrix}
x_1 \\
x_2
\end{bmatrix}=
\begin{bmatrix}
0 \\
0
\end{bmatrix}
$$
Then I got stuck. What are the eigenvectors? I also need hints for the rest.
Best Answer
i think you may not want to solve this equation by diagonalising it. this system is decoupled, therefore we can solve it directly. you have $y_1 = c_1e^t.$ subbing this in the second equation gives you, $$y_2' + 4y_2 = -2c_1e^t \tag 1$$ now $(1)$ has $y_2 = c_2e^{-4t}$ for a homogenous solution and $-\frac 25c_1e^t$ for a particular solution. therefore the solution to the system of equation is $$y_1 = c_1e^t, y_12 = c_2e^{-4t} - \frac 25c_1e^t.$$