Toronto Math Forum

7.5 p 405,# 12
Solve the following equation.
\begin{equation*}
x' = \begin{pmatrix} 3 & 2 & 4 \\ 2 & 0 & 2 \\ 4 & 2 & 3 \end{pmatrix}x
\end{equation*}

First, find the eigenvalues.

\begin{equation*}
\begin{pmatrix} 3-\lambda & 2 & 4 \\2 &  -\lambda & 2 \\ 4 & 2 & 3-\lambda\end{pmatrix} = 0  \notag
\end{equation*}

The characteristic equation is:
\begin{gather}
\lambda^3 - 6\lambda^2 - 15\lambda - 8 = 0\\
(1 + \lambda)( (1 + \lambda)(8 - \lambda) = 0\\
\end{gather}

Therefore, the eigenvalues are $ \lambda = 8,\ \lambda = -1,\ \lambda = -1$.

For $ \lambda = 8$:
\begin{equation*}
\begin{pmatrix}
-5 & 2 & 4 
\\2 & -8 & 2
\\ 4 & 2 & -5\end{pmatrix}
\begin{pmatrix} x \\ y \\ z \end{pmatrix} =
\begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}
\end{equation*}
The corresponding eigenvector is $v_1 = \begin{pmatrix} 2 \\ 1 \\ 2 \end{pmatrix}$

For $ \lambda = -1$:
\begin{equation*}
\begin{pmatrix}
4 & 2 & 4 
\\2 & 1 & 2
\\ 4 & 2 & 4\end{pmatrix}
\begin{pmatrix} x \\ y \\ z \end{pmatrix} =
\begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}
\end{equation*}
The corresponding eigenvectors are $v_2 = \begin{pmatrix} 1 \\ -2 \\ 0 \end{pmatrix}$ and $v_3 = \begin{pmatrix} 0 \\ -2 \\ 1 \end{pmatrix}$


Therefore, the general solution is:

\begin{equation*}x = c_1\begin{pmatrix} 2 \\ 1 \\ 2 \end{pmatrix}e^{8t} + c_2\begin{pmatrix} 1 \\ -2 \\ 0 \end{pmatrix}e^{-t} + c_3\begin{pmatrix} 0 \\ -2 \\ 1 \end{pmatrix}e^{-t} \end{equation*}

2.1 p. 40, #18
\begin{gather}
ty' + 2y = \sin t,\\
y(Ï€/2) = 1
\end{gather}
First, change the equation by dividing every term by $t$.
\begin{equation}
y' + (2/t)y = (\sin (t))/t
\end{equation}
Then, solve for the integrating factor.
\begin{equation}
μ = e^{\int (2/t)\,dt} = e^{2\ln(t)\,} = t^2
\end{equation}
Multiply every term in equation by μ.
\begin{equation}
t^2y' + 2ty = t\sin(t)
\end{equation}
Integrate both sides of the equation.
\begin{gather}
\int[t^2y]' = \int{t\sin(t)}dt\implies
t^2y = −t\cos(t) − \int{-\cos(t)}dt= −t\cos(t) + \sin(t) + C\implies y = (−t \cos(t) +\sin (t) + C)/t^2
\end{gather}
Substitute the initial value to solve for C.
\begin{gather}
C = (Ï€/2)^2-1
\end{gather}
Therefore, the solution is:
\begin{gather}
y = (−t \cos(t) +\sin (t) + (π/2)^2-1)/t^2
\end{gather}