Problem 1 (main sitting)

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

[ZYR]

For the third solution, I believe there are some typo in Question a). For the last two equations, $y_2$ is missing, it should be $\frac{1}{2} \sin2t( 2t - tant)$, or simplify $\sin2t(t - \frac{1}{2}tant)$.

[Vy Nguyen]

(Almost perfect typing appreciated). V.I.

a)
Solve the homogeneous equation: $y''+4y = 0$
$$r^2+4=0$$
$$r=\pm2i$$
fundamental set of solutions:
$$y_1 (t)=\cos(2t)$$
$$y_2 (t)=\sin(2t)$$
homogeneous solution:
$$y_h (t)=c_1 \cos(2t) + c_2 \sin(2t)$$
Solve for a particular solution using variation of parameters:
Find the Wronskian:
$$W=\begin{vmatrix}
    \cos(2t) & \sin(2t) \\
    -2\sin(2t) & 2\cos(2t)\\
    \end{vmatrix}$$
$$=2\cos^2 (2t) + 2\sin^2 (2t) = 2$$
Solve for $\mu_1 (t)$:
$$\mu_1 (t)=-\int \frac {y_2 (t) g(t)}{W} dt$$
$$=\int \frac {-\sin(2t)}{2\cos^2 t} dt$$

use identity $2\cos^2 t = \cos(2t) +1$

$$=\int \frac {-\sin(2t)}{\cos(2t)+1} dt$$

let $u=\cos(2t)+1$ then $\frac{1}{2}du=-\sin(2t) dt$

$$=\frac{1}{2} \int \frac {1}{u} du$$
$$=\frac{1}{2} \ln(u)$$
$$=\frac{1}{2} \ln(\cos(2t)+1)$$
Solve for $\mu_2 (t)$:
$$\mu_2 (t)=\int \frac {y_1 (t) g(t)}{W} dt$$
$$=\int \frac {\cos(2t)}{2\cos^2 t} dt$$

use identity $\cos(2t)=2\cos^2 (t)-1$

$$=\int \frac {2\cos^2 t-1}{2\cos^2 t} dt$$
$$=\int 1 - \frac {1}{2\cos^2 t} dt$$
$$=\int 1 - \frac {1}{2}\sec^2 t \, dt$$
$$=t - \frac{1}{2}\tan t$$
particular solution:
$$y_p(t)=\frac{1}{2}\cos(2t)\ln [\cos(2t)+1] + t\sin(2t) - \frac{1}{2}\sin(2t)\tan t$$
Therefore, general solution to ODE is:
$$y(t) = y_h(t) + y_p(t)$$
$$y(t) = c_1\cos(2t) + c_2\sin(2t) + \frac{1}{2}\cos(2t)\ln [\cos(2t)+1] + t\sin(2t) - \frac{1}{2}\sin(2t)\tan t$$

b)
Take the derivative of the general solution:
$$y'(t)=-2c_1\sin(2t)+2c_2\cos(2t)-\sin(2t)\ln[\cos(2t)+1] - \frac{(\cos(2t))(\sin(2t))}{\cos(2t)+1} + \sin(2t) + 2t\cos(2t) - \cos(2t)\tan t - \frac{1}{2}\sin(2t)\sec^2t$$
Use the initial conditions to solve for constants $c_1$ and $c_2$:
$$y(0)=0=c_1+\frac{\ln2}{2}$$
$$c_1=-\frac{\ln2}{2}$$
$$y'(0)=0=2c_2$$
$$c_2=0$$
Therefore the solution that satisfies the initial conditions is:
$$y(t)=-\frac{\ln2}{2}\cos(2t)+\frac{1}{2}\cos(2t)\ln[\cos(2t)+1]+t\sin(2t)-\frac{1}{2}\sin(2t)\tan t$$

[Mingdi Xie]

This is my solutions for problem1.