Toronto Math Forum

I think the first problem was from the textbook
$
\mathbf x' = \left(\begin{array}{ccc}
1 & 1 & 1 \\
2 & 1 & -1 \\
0 & -1 & 1 \\
\end{array}\right) \mathbf x \\
$
The characteristic polynomial is
$
-4 + 3k^2 - k^3 = 0 \\
k = 2, 2, -1
$
The eigenvectors are
$
\lambda = -1,
\left(\begin{array}{c}
-3 \\
4 \\
2
\end{array}\right),
\lambda = 2,
\left(\begin{array}{c}
-0 \\
-1 \\
1
\end{array}\right)

$
Jordan decomposition yields the similarity transform of
$
\mathbf T = \left(\begin{array}{ccc}
-3 & 0 & -1 \\
4 & -1 & -1 \\
2 & 1 & 0
\end{array}\right)
$
Thus the solution is
$
\mathbf x = c_1 \left(\begin{array}{c} -3 \\ 4 \\ 2 \end{array}\right) e^{-t} + c_2 \left(\begin{array}{c} 0 \\ -1 \\ 1 \end{array}\right) e^{2t} + c_3 \left( \left(\begin{array}{c} 0 \\ -1 \\ 1 \end{array}\right) t e^{2t} + \left(\begin{array}{c} -1 \\ -1 \\ 0 \end{array}\right) e^{2t}\right)
$

I think the question was
$$
y'''-y= 2 \sin t
$$
solution:
$$
r^3 - 1 = 0 \\
(r+1)(r^2+r+1) = 0\\
r = -1, \frac{-1 \pm \sqrt{3} i}{2}\\
y_h = c_1 e^{-t} + c_2 e^{ \frac{-1 +\sqrt{3} i}{2}t} + c_3 e^{ \frac{-1 - \sqrt{3} i}{2}t} \\
y_p = A \sin t + B \cos t \\
y_p' = A \cos t - B \sin t \\
y_p'' = -A \sin t - B \cos t \\
y_p''' = - A \cos t + B \sin t \\
A = -1, B = 1\\
y  = c_1 e^{-t} + c_2 e^{ \frac{-1 +\sqrt{3} i}{2}t} + c_3 e^{ \frac{-1 - \sqrt{3} i}{2}t} + \cos t - \sin t
$$

2)
guess $ y = A e^{rt}$, $z = B e^{rt}$
$$
A r^2 + K A + L\left(A-B\right) = 0 \\
B r^2 + K B + L\left(B-A\right) = 0 \\

\left(
\begin{array}{cc}
r^2 + K + L & - L \\
-L & r^2 + K + L \\
\end{array} \right)
\left( \begin{array}{c}
A \\
B
\end{array}
\right) = 0

$$

nontrivial solution exists if there is no inverse to the ugly matrix, therefore its determinant is 0

$$
\left( r^2 + K + L \right)^2 + L^2 = 0
$$

expand and use quadratic equation
$$
r^2 = K - L \pm L
$$

therefore the frequencies are $\pm K$, $\pm\left(K-2L\right)$

$y = \{A \sin(\omega t) + B \cos(\omega t) \mid A, B \in \mathbb{R} \}$
let me plot that for you

a)

$$
y'' - w^2 y = 0 \\
z = y', y'' = \frac{dz}{dx} = \frac{dz}{dy} \frac{dy}{dx} = \frac{dz}{dy} z \\
\frac{dz}{dy} z = w^2 y \\
\frac{1}{2}z^2=\frac{1}{2}w^2 y^2 + C_1 \\
$$
\begin{equation}
\frac{dy}{dx} = \sqrt{w^2 y^2 + C_1} \\
\end{equation}
factor out $w^2$ from $C_1$
$$
\frac{dy}{dx} = w \sqrt{y^2 + C_1} \\
\frac{dy}{ \sqrt{y^2 + C_1}} = w dx \\
$$
use trig substitution
$$
y = \sqrt{C_1} \tan u, \; dy=\sqrt{C_1} \sec^2 u\;du \\
\int \frac{\sqrt{C_1} \sec^2 u \; du}{\sqrt{C_1 \left( \tan^2 u + 1 \right)}} = \int wdx \\
\int \frac{\sqrt{C_1} \sec^2 u \; du}{\sqrt{C_1 \sec^2 u }} = \int wdx \\
\int \sec u \; du = \int wdx \\
\ln\left( \tan u + \sec\; u \right) = wx + C_2

$$
note $\sec \theta = \sqrt{1 + \tan^2 \theta}$ and $u = \arctan \frac{y}{\sqrt{C_1}}$
$$
\ln\left( \sqrt{y^2 + C_1 } + y \right) = wx + C_2 \\
\sqrt{y^2 + C_1 }  = e^{wx + C_2} - y \\
y^2 + C_1  = \left( e^{wx + C_2} - y\right)^2 \\
$$
expand and simplify
$$
y = \left( e^{2\left(wx+C_2\right)} - C_1 \right) e^{-\left(wx+C_2\right)} \frac{1}{2}
$$

b)
insert IC to (2) to get $C_1 = -w^2$
$$
\frac{dy}{dx} = w \sqrt{y^2-1} \\
\frac{1}{\sqrt{y^2-1}}  dy = w dx \\
\ln \left( \sqrt{ y^2 -1} + y \right) = wx + C_2 \\$$

insert IC to get $C_2 = \ln\left( i \right)$.
simplify using euler's identity
$$
e^{i\frac{\pi}{2}} = i \\
i\frac{\pi}{2} = \ln i \\
$$

$$
y = \left( e^{2\left(wx+i\frac{\pi}{2} \right)} + 1\right) e^{-\left(wx+i\frac{\pi}{2} \right)} \frac{1}{2}
$$

Fixed, I think.

$$
y'+\frac{2}{t} y = \frac{\sin t}{t} \\
\mu y' + \frac{2 \mu}{t} y = \mu \frac{\sin t}{t} \\
\mu = t^2 \\
\frac{d}{dt} t^2 y = t \sin t \\
t^2 y = \int t \sin t dt \\
y = \frac{\sin t - t \cos t + C}{t^2}
$$
use initial value
$$
1 = \frac{4\left(1 - 0 + C\right)}{\pi^2}\\
C= \frac{\pi^2-4}{4}\\
y = \frac{\sin t - t \cos t + \frac{\pi^2-4}{4}}{t^2}\\
y = \frac{4\sin t - 4t \cos t + \pi^2-4}{4t^2}
$$

Changes were made per your suggestions.