Author Topic: TUT5103 Quiz5  (Read 761 times)

Ruodan Chen

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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($\frac{3}{2}$)$\int$sin2scsc2sds + 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|