Author Topic: Question 8, Sec 3.6  (Read 272 times)

Yifei Gu

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Question 8, Sec 3.6
« on: November 29, 2018, 09:22:21 PM »
Having a bit trouble finding solutions, can anyone take a look.
Question: $y'' + 4y = 3\csc 2t, 0 < t <\frac{\pi}{2}$

xin xiong

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Re: Question 8, Sec 3.6
« Reply #1 on: November 29, 2018, 09:45:35 PM »
\begin{equation*}
\begin{aligned}
&p(t) = 0, q(t) = 4, g(t) = 3\csc(2t)\\
& y'' + 4y = 0\\
& r^2 + 4 = 0 \implies r = \pm2i\\
&y_c(t) = c_1 \cos 2t + c_2 \sin 2t \implies \text{homogeneous sol'n}\\
\\
&W(y_1,y_2)(t) = \begin{vmatrix}\cos2t & \sin2t\\  -2\sin2t & 2\cos2t\end{vmatrix}=2\cos^22t+2\sin^2 2t = 2\\
\\
\\
&y(t) = u_1(t)y_1(t)+u_2(t)y_2(t)\\
&u_1(t) = -\int\frac{y_2(t)g(t)}{W(y_1,y_2)(t)}dt,\ \ u_2(t) = \int\frac{y_1(t)g(t)}{W(y_1,y_2)(t)}dt \\
\\
\\
&u_1(t) = -\int\frac{\sin2t(3\csc2t)}{2}dt=\frac{3}{2}\int\sin2t(\csc2t)dt = -\frac{3}{2}\int 1 dt = -\frac{3}{2}t\\
\\
& u_2(t) = \int\frac{\cos2t(3\csc2t)}{2}dt = \frac{3}{2}\int \cos 2t \left(\frac{1}{\sin 2t}\right)dt= \frac{3}{2}\left(\frac{\ln \vert \sin 2t \vert}{2}\right) = \frac{3}{4}\ln\vert\sin2t\vert\\
\\
\\
&Y(t) = u_1(t)y_1(t)+u_2(t)y_2(t) = -\frac{3}{2}t\cos2t+\frac{3}{4}\ln\vert\sin 2t\vert(\sin2t)\\
&y = y_c(t)+ Y(t)= c_1\cos 2t + c_2 \sin2t - \frac{3}{2}t\cos2t + \frac{3}{4}\ln|\sin 2t|(\sin 2t)
\end{aligned}