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### Messages - Ruodan Chen

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##### Term Test 2 / Re: Problem 4 (main sitting)
« on: November 19, 2019, 10:41:19 AM »
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

2
##### Term Test 2 / Re: Problem 2 (main sitting)
« on: November 19, 2019, 10:32:48 AM »
2) $y'''+4y''+y'-6y=24e^{t}$

a)

Since the coefficient of y'' is 4,

Then, Wronskain is

$w=ce^{-\int4dt}=ce^{-4t}$

b)

Use homogeneous equation to find fundamental solutions,

$y'''+4y''+y'-6y=24e^{t}$

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

Then $r=1, r=-2, r=-3$

Then the solution is $y_{c}(t)=c_{1}e^{t}+c_{2}e^{-2t}+c_{3}e^{-3t}$

$w = \begin{array}{ccc} e^{t} & e^{-2t} & e^{-3t}\\ e^{t} & -2e^{-2t} & -3e^{-3t}\\ e^{t} & 4e^{-2t} & 9e^{-3t} \end{array}=-12e^{-4t}$

$w = -12e^{-4t} =ce^{-4t}$

This is consistent with what we get in part (a) with $c=-12$

c)

Use under determined coefficients method to find the general equation for y(t)

$y'''+4y''+y'-6y=24e^{t}$

$y_{p}(t)=Ate^{t}$

$y'_{p}(t)=Ae^{t}+Ate^{t}$

$y''_{p}(t)=2Ae^{t}+Ate^{t}$

$y'''_{p}(t)=3Ae^{t}+Ate^{t}$

Plug into the equation,

$3Ae^{t}+Ate^{t}+4(2Ae^{t}+Ate^{t})+Ae^{t}+Ate^{t}-6Ate^{t}=12Ae^{t}=24e^{t}$

We get A=2

$y_{p}(t)=2te^{t}$

Thus, $y(t)=c_{1}e^{t}+c_{2}e^{-2t}+c_{3}e^{-3t}+2te^{t}$

3
##### Term Test 2 / Re: Problem 1 (main sitting)
« on: November 19, 2019, 10:28:21 AM »
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, $y(t) = cos2t(ln(cost)) + tsin2t -\frac{1}{2}sin2ttant$

4
##### Quiz-5 / TUT5103 Quiz5
« on: November 01, 2019, 03:41:59 PM »
Find the general solution of the given differential equation.

y'' + 4y = 3csc(2t), 0 < t < $\frac{\pi}{2}$

$r^{2}$ + 4 = 0

$r^{2}$ = $-4$

r = $\pm2i$

$y_{c}$(t) = $c_{1}$cos2t + $c_{2}$sin2t

$y_{1}$ = cos2t

$y_{2}$ = sin2t

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

Y(t) = cos2t$\int$$\frac{-sin2s3csc2s}{2}ds + sin\int\frac{cos2s3csc2s}{2}ds = -cos2t(\frac{3}{2})\intsin2scsc2sds + sin2t(\frac{3}{2})\int$$\frac{cos2s}{sin2s}$ds

= -cos2t($\frac{3}{2}$)t + sin2t($\frac{3}{2}$)ln|sin2t|

y(t) = $y_{c}$(t) + Y(t) = $c_{1}$cos2t + $c_{2}$sin2t - cos2t($\frac{3}{2}$)t + sin2t($\frac{3}{2}$)ln|sin2t|

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