MAT244--2019F > Term Test 2

Problem 1 (main sitting)

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Victor Ivrii:
(a) Find the general solution of
$$y''+4y=\frac{1}{\cos^2(t)},\qquad -\frac{\pi}{2}<t<\frac{\pi}{2}.$$

(b) Find solution satisfying
$$y(0)=y'(0)=0.$$

Yiheng Bian:
(a):
$$r^2+4=0\\ r_1=2i,r_2=-2i$$
So we can get:
$$y_c(t)=c_1cos2t+c_2sin2t$$
$$W=\begin{vmatrix} cos2t & sin2t \\ -2sin2t & 2cos2t \end{vmatrix}=2$$
$$W_1=\begin{vmatrix} 0 & sin2t \\ 1 & 2cos2t \end{vmatrix}=-sin2t$$
$$W_2=\begin{vmatrix} cos2t & 0 \\ -2sin2t & 1 \end{vmatrix}=cos2t$$
Therefore:
$$Y(t)=cos2t\int{\frac{-sin2s*\frac{1}{(cos(s))^2}}{2}}ds+sin2t\int{\frac{cos2s*\frac{1}{(cos(s))^2}}{2}}ds\\ Y(t)=cos2t\int{-\frac{sins}{coss}}ds+sin2t\int{\frac{2(cos(s))^2-1}{2(cos(s)^2)}}ds\\ Y(t)=cost2t*lncost+sin2t*(t-0.5tant)$$
So general solution is :
$$y(t)=c_1cos2t+c_2sin2t+cost2t*lncost+sin2t*(t-0.5tant)$$

(b):
$$y'(t)=-2c_1sin2t+2c_2cos2t+cos2t*\frac{-sint}{cost} - 2sin2t*lncost+sin2t(1-0.5(sect)^2)+2cos2t(t-0.5tant)$$
$$\text{Take } y(0)=y'(0)=0\\ \text{We can get that: }c_1=c_2=0$$
So finally:
$$y=cost2t*lncost+sin2t*(t-0.5tant)$$

OK, except LaTeX sucks:

1) * IS NOT a sign of multiplication
2)  "operators" should be escaped: \cos, \sin, \tan, \ln
$$\boxed{ y= \ln (\cos(t)) \cos(2t) + \Bigl(t-\frac{1}{2}\tan (t)\Bigr)\sin(2t). }$$

Anyue Huang:
Solution see attachment.

Yuying Chen:
$\text(a)\\$
$r^2+4=0\\$
$\qquad r=\pm2i\\$
$\therefore y(t)=c_1 \cos{2t}+c_2 \sin{2t}\\$
$W=\begin{vmatrix} \cos2t & \sin2t \\ -2\sin2t & 2\cos2t \\ \end{vmatrix}=2\\$
$W_1=\begin{vmatrix} 0 & \sin2t \\ 1 & 2\cos2t \\ \end{vmatrix}=-\sin2t\\$
$W_2=\begin{vmatrix} \cos2t & 0 \\ -2\sin2t & 1 \\ \end{vmatrix}=\cos2t\\$
$y_c(t)=\cos2t \int \frac{-\sin2s·\frac{1}{\cos^2{s}}}{2}ds+\sin2t \int \frac{\cos2s·\frac{1}{cos^2{s}}}{2}ds\\$
$\qquad = \frac{1}{2}\cos2t \int \frac{-2sins·coss}{cos^2{s}}ds+\frac{1}{2}\sin2t\int\frac{2cos^2{s}-1}{cos^2{s}}ds\\$
$\qquad =-\cos2t \int tans ds+ \frac{1}{2}\sin2t \int (2-\sec^2{s})ds\\$
$\qquad = \cos2t·\ln (\cos t) + \frac{1}{2}(2t-\tan t)\\$
$y(t)=c_1 \cos{2t}+c_2 \sin{2t}+ \cos2t·\ln (\cos t) + \frac{1}{2}(2t-\tan t)\\$
$\text(b)\\$
$y^{\prime}(t)=-2c_1\sin2t =2c_2\cos2t-2\sin2t·\ln\cos(t)-\cos2t·tant+\cos2t(2t- tant)+\frac{1}{2}\sin2t(2-\sec^2{t})\\$
$y(0)=c_1+\ln1=0 \implies c_1=0\\$
$y^{\prime}(0)=2c_2+(0-\tan0)=0 \implies c_2=0\\$
$\therefore y(t)=\cos2t·\ln (\cos t)+\frac{1}{2}\sin2t(2t-tant)$

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