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##### Quiz-6 / LEC5101 QUIZ6
« on: November 15, 2019, 01:07:58 PM »
Question:

a. Find the general solution of the given system of equations and describe the behavior of the solution as $t \rightarrow \infty$.

b. Draw a direction field and plot a few trajectories of the system.

${x}^{\prime}=\left(\begin{array}{ll}{3} & {-2} \\ {2} & {-2}\end{array}\right) {x}$

$A=\left|\begin{array}{cc}{3} & {-2} \\ {2} & {-2}\end{array}\right|$

det$(A-r I)={\rm det}\left|\begin{array}{cc}{3-r} & {-2} \\ {2} & {-2-r}\end{array}\right|=(r+2)(r-3)+4=0$

$(r+2)(r-3)+4=0$

$r^{2}-3 r+2 r-6+4=0$

$r^{2}-r-2=0$

$(r-2) \cdot(r+1)=0$

$r_{1}=2 \quad, \quad r_{2}=-1$

For $r_{1}=2$

$A-r_{1} I=\left|\begin{array}{cc}{3-2} & {-2} \\ {2} & {-2-2}\end{array}\right|=\left|\begin{array}{cc}{1} & {-2} \\ {2} & {-4}\end{array}\right|$

$\left|\begin{array}{cc:cc}{1} & {-2} & {0} \\ {2} & {-4} & {0}\end{array}\right|$ ${R_{2}-2 R_{1}}$ $\left|\begin{array}{cc:cc}{1} & {-2} & {0} \\ {0} & {0} & {0}\end{array}\right|$

$x_{2}=t \quad x_{1}-2 x_{2}=0$

$x_{1}=2 t$

$\left|\begin{array}{l}{x_{1}} \\ {x_{2}}\end{array}\right|=\left|\begin{array}{l}{2} \\ {1}\end{array}\right|t$

For $r_{2}=-1$

$A-r I=\left|\begin{array}{cc}{3-(-1)} & {2} \\ {2} & {-2-(-1)}\end{array}\right|=\left|\begin{array}{cc}{4} & {-2} \\ {2} & {-1}\end{array}\right|$

$\left|\begin{array}{cc:cc}{4} & {-2} & {0} \\ {2} & {-1} & {0}\end{array}\right|$ ${R_{2} \times 2-R_{1}}$ $\left|\begin{array}{cc:cc}{4} & {-2} & {0} \\ {0} & {0} & {0}\end{array}\right|$

$x_{2}=t \quad 4 x_{1}-2 x_{2}=0$

$x_{1}=\frac{1}{2} t$

$\left|\begin{array}{l}{x_{1}} \\ {x_{2}}\end{array}\right|=\left|\begin{array}{c}{\frac{1}{2}} \\ {1}\end{array}\right|t$$=\left|\begin{array}{l}{1} \\ {2}\end{array}\right|t$

The general solution is

$x=c_{1} e^{-t}\left|\begin{array}{l}{1} \\ {2}\end{array}\right|+c_{2} e^{2 t}\left|\begin{array}{l}{2} \\ {1}\end{array}\right|$

$\left\{\begin{array}{l}{x_{1}(t)=c_{1} e^{-t}+2 c_{2} e^{2 t}} \\ {x_{2}(t)=2 c_{1} e^{-t}+c_{2} e^{2 t}}\end{array}\right.$

when $t \rightarrow+\infty$

when $c_{2}=0,\lim \limits_{t \rightarrow+\infty} x_{1}(t)=0$ and $\lim \limits_{t \rightarrow +\infty} x_{2}(t)=0$

Then $x$ approaches to zero as $t \rightarrow+\infty$

when $c_{2} \neq 0,\lim \limits_{t \rightarrow+\infty} x_{1}(t)=+\infty$ and $\lim \limits_{t \rightarrow+\infty} x_{2}(t)=+\infty$

Then $x$ approaches to $+\infty$ as $+\rightarrow +\infty$

$(b) \quad r_{1}=2>0 \quad\quad r_{2}=-1<0$
The graph is in the attachment

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##### Quiz-5 / LEC5201 QUIZ5
« on: October 31, 2019, 10:52:18 PM »
Question:Find the general solution of the given differential equation.
y′′+9y=9sec2(3t) 0<t<π/6

$r^{2}+9=0 \quad r^{2}=-9 \quad r=\pm 3 i$

$y_{C}(t)=C_{1} \cos (3 t)+C_{2} \sin (3 t)$

$y^{\prime \prime}+9 y=9 \sec ^{2}(3 t)$

$w\left(y_1, y_{2}\right)=\left|\begin{array}{ll}{y_{1}(t)} & {y_{2}(t)} \\ {y_{1}^{\prime}(t)} & {y_2^{\prime}(t)}\end{array}\right|=\left|\begin{array}{cc}{\cos (3 t)} & {\sin (3 t)} \\ {-3 \sin (3 t)} & {3 \cos (3 t)}\end{array}\right|=3 \cos ^{2}(3 t)+3 \sin ^{2}(3 t)=3$

$w_{1}=\left|\begin{array}{ll}{0} & {\sin (3t)} \\ {1} & {3 \cos (3t)}\end{array}\right|=-\sin (3t) \quad \quad \quad w_{2}=\left|\begin{array}{cc}{\cos (3t)} & {0} \\ {-3 \sin (3t)} & {1}\end{array}\right|=\cos (3t)$

$y_{p}(t)=y_{1}(t) \underbrace{\int \frac{q(s)w_1(s)}{w(s)} d s}_{u_{1}(s)}+y_{2}(t) \underbrace{\frac{q(s) \cdot w_{2}(s)}{w(s)} d s}_{u_{2}(s)}$

\begin{aligned} u_{1}(s) &=-\int \frac{9 \sec ^{2}(3 s) \sin (3 s)}{3} d s \\ &=-3 \int \frac{\sin (3 s)}{\cos ^{2}(3 s)} d s \\ &=-3 \int \sec (3 s)\tan(3s) d s \\ &=-\sec 3 s \end{aligned}

\begin{aligned} u_{2}(s) &=\int \frac{9 \cos (3 s) \cdot \sec ^{2}(3s)}{3} d s \\ &=\int 3 \frac{\cos (3s)}{\cos ^{2}(3s)} d s \\ &=\int 3 \sec (3 s) d s \\ &=\ln|\sec (3s)+\tan (3s)|\end{aligned}

\begin{aligned} y_{p}(t) &=\cos (3 t)(-\sec (3t))+\sin (3t) \cdot \ln |\sec (3 t)+\tan (3 t)| \\ &=\sin (3 t)[\ln |\sec (3 t)+\tan (3 t)|]-1 \end{aligned}

$y(t)=y_{p}(t)+y_{C}(t)= C_{1} \cos (3 t)+C_{2} \sin (3 t)+ \sin (3 t)[\ln |\sec (3 t)+\tan (3 t)|]-1$

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