Toronto Math Forum

4) $x'=\left(\begin{array}{cc}
3 & 3\\
-2 & -1
\end{array}\right)x$

a)

$det(A-\lambda I)=det(\begin{array}{cc}
3-\lambda & 3\\
-2 & -1-\lambda
\end{array})=(3-\lambda)(-1-\lambda)+6=0$

$\lambda^{2}-2\lambda+3=0$

$(\lambda-1)^{2}=-2$

\$\lambda=1\pm\sqrt{2}i$

\$\lambda=1+\sqrt{2}i$

$(\begin{array}{ccc}
2-\sqrt{2}i & 3 & 0\\
-2 & -2-\sqrt{2}i & 0
\end{array}) = (\begin{array}{ccc}
2-\sqrt{2}i & 3 & 0\\
0 & 0 & 0
\end{array}) $

$v=(\begin{array}{c}2+\sqrt{2}i\\-2\end{array})$

$e^{(1+\sqrt{2}i)t}(\begin{array}{c}
2+\sqrt{2}i\\
-2
\end{array})=e^{t}(cos\sqrt{2}t+isin\sqrt{2}t)(\begin{array}{c}2+\sqrt{2}i\\-2\end{array}) = e^{t}(\begin{array}{c}
2cos\sqrt{2}t+2isin\sqrt{2}t+\sqrt{2}icos\sqrt{2}t-\sqrt{2}sin\sqrt{2}t\\
-2cos\sqrt{2}t-2isin\sqrt{2}t
\end{array})=e^{t}(\begin{array}{c}
2cos\sqrt{2}t-\sqrt{2}sin\sqrt{2}t\\
-2cos\sqrt{2}t
\end{array})+ie^{t}(\begin{array}{c}
2isin\sqrt{2}t+\sqrt{2}icos\sqrt{2}t\\
-2isin\sqrt{2}t
\end{array})$

$x(t)=c_{1}e{}^{t}(\begin{array}{c}
2cos\sqrt{2}t-\sqrt{2}sin\sqrt{2}t\\
-2cos\sqrt{2}t
\end{array})+c_{2}e{}^{t}(\begin{array}{c}
2isin\sqrt{2}t+\sqrt{2}icos\sqrt{2}t\\
-2isin\sqrt{2}t
\end{array})$
You should consider real solutions

1)

a)

First, solve the homogeneous equation $y''' + y'' + 4y = 0$

Then, $r^{2}+4=0$

Then, $r=\pm2i$

Therefore, the homogeneous solution is $y_{c}(t)=c_{1}cos2t+c_{2}sin2t$

So, we have $w, w_{1}, w_{2}$ by plugging in $y_{1}(t) = cos2t$, $y_{2}(t) = sin2t$

$w=\begin{array}{cc}
cos2t & sin2t\\
-2sin2t & 2cos2t
\end{array}=2cos^{2}2t+2sin^{2}2t=2 $

$w_{1}=\begin{array}{cc}
0 & sin2t\\
1 & 2cos2t
\end{array}=-sin2t$

$w_{2}=\begin{array}{cc}
cos2t & 0\\
-2sin2t & 1
\end{array}=cos2t$

Then we use $w$, $w_{1}$, $w_{2}$ to solve the non_homo part

$y_{p}(t) = cos2t\int\frac{-sin2s(\frac{1}{cos^{2}s})}{2}ds + sin2t\int\frac{cos2s(\frac{1}{cos^{2}s})}{2}ds$

$y_{p}(t) = -cos2t\int\frac{2cosssins(\frac{1}{cos^{2}s})}{2}ds + sin2t\int\frac{(2cos^{2}s-1)(\frac{1}{cos^{2}s})}{2}ds$

$y_{p}(t) = -cos2t\int\frac{sins}{coss}ds + \frac{1}{2}sin2t\int\frac{2cos^{2}s-1}{cos^{2}s}ds $

$y_{p}(t) = -cos2t\int(tans)ds + \frac{1}{2}sin2t\int2 - sec^{2}sds $

$y_{p}(t) = -cos2t(-ln(cost)) + \frac{1}{2}sin2t(2t-tant) $

$y_{p}(t) = cos2t(ln(cost)) + tsin2t -\frac{1}{2}sin2ttant$

Then, we have the general solution for $y(t)$,

$y(t) = y_{c}(t) + y_{p}(t) = c_{1}cos2t+c_{2}sin2t + cos2t(ln(cost)) + tsin2t -\frac{1}{2}sin2ttant$

(b)

Derivativing y(t) to get y'(t),

$y'(t)=-2sin2t(ln(cost))+cos2t\frac{-sint}{cost}+sin2t+2tcost-cos2ttant-\frac{1}{2}sin2tsec^{2}t$

Plug in $y(0) = y'(0) = 0$

Get $c_{1}=c_{2}=0$

Therefore, $y(t) = cos2t(ln(cost)) + tsin2t -\frac{1}{2}sin2ttant$