Author Topic: LEC0101 quiz5  (Read 715 times)

Linqian Shen

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LEC0101 quiz5
« on: November 01, 2019, 03:02:29 PM »
Find the general solution of the given differential equation.
$$
y^{\prime \prime}+y=\tan (t)
$$


$$
\begin{array}{c}{r^{2}+1=0} \\ {r=\pm i} \\ {y=c_{1} \cos t+c_{2} \sin t}\end{array}
$$
$$
\begin{aligned}
w&=\left|\begin{array}{cc}{\cos t} & {\sin t} \\ {-\sin t} & {\cos t}\end{array}\right|\\
&=\cos ^{2} t+\sin ^{2} t=1\\
w_{1}&=\left|\begin{array}{ll}{0} & {\sin t} \\ {1} & {\cos t}\end{array}\right|=-\sin t\\
w_{2}&=\left|\begin{array}{cc}{\cos t} & {0} \\ {-\sin t} & {1}\end{array}\right|=\cos t
\end{aligned}
$$
$$
\begin{aligned}
y p(t)&=\cos t \int \frac{-s i n t-\tan s}{1} d s+s \sin t \int \frac{\cos s-t a n s}{1} d s\\
&=\cos t \int-\sin s \frac{\sin s}{\cos s} d s+\sin t \int \cos s \frac{\sin s}{\cos s} d s\\
&=-\cos t \int \frac{1-\cos ^{2} s}{\cos s} d s+\sin t \int \sin s d s\\
&=-\cos t \int \sec -\cos s d s+\sin t(-\cos s)\\
&=-\cos \ln (\sec t+\tan t)+\cos t \sin t-\sin t \cos t
\end{aligned}
$$
$$
y(t)=c_{1} \cos t+c_{2} \sin t-\cos t \ln (\sec t+\tan t)
$$