Author Topic: TUT0303 Quiz5  (Read 471 times)

Yiyang Huang

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TUT0303 Quiz5
« on: November 01, 2019, 02:02:22 PM »
Find the general solution of the given differential equation.
$$
y^{\prime \prime}+4 y=3 \csc (2 t), \quad 0<t<\pi / 2
$$

$$
\begin{aligned} r^{2}+4 &=0 \\ r^{2} &=-4 \\ r &=\pm 2 i \\ y_{c(t)}=& c_{1} \cos 2 t+c_{2} \sin 2 t \end{aligned}
$$
$$
w=\left|\begin{array}{cc}{\cos 2 t} & {\sin 2 t} \\ {-2 \sin 2 t} & {2 \cos 2 t}\end{array}\right|=2 \cos ^{2} 2 t+2 \sin ^{2} 2 t=2
$$
$$
\begin{array}{l}{w_{1}=\left|\begin{array}{cc}{0} & {\sin 2 t} \\ {1} & {2 \cos 2 t}\end{array}\right|=-\sin 2 t} \\ {w_{2}=\left|\begin{array}{cc}{\cos 2 t} & {0} \\ {-2 \sin 2 t} & {1}\end{array}\right|=\cos 2 t}\end{array}
$$
$$
\begin{aligned} y_{p}(t) &=\cos 2 t \int \frac{(-\sin 2 t)(3 \csc (2 t))}{2} d t+\sin 2 t \int \frac{(\cos 2 t)(3 \csc (2 t))}{2}d t \\ &=\cos 2 t \int-\frac{3}{2} d t+\sin 2 t \int \frac{3}{2} \cot (2 t) d t \\ &=(\cos 2 t)\left(-\frac{3}{2} t\right)+\frac{3}{4} \sin 2 t \ln |\sin (2 t)| \\ &=-\frac{3}{2} t \cos 2 t+\frac{3}{4} \sin 2 t \ln |\sin (2 t)| \end{aligned}
$$
$$
y(t)=c_{1} \cos 2 t+c_{2} \sin 2 t+\frac{3}{4} \sin 2 t \ln |\sin (2 t)|-\frac{3}{2} t \cos 2 t
$$