### Author Topic: quiz5 tut0401  (Read 2119 times)

#### baixiaox

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##### quiz5 tut0401
« on: November 01, 2019, 01:33:15 PM »
Solve $y'' + 4y = 3csc2t$, $0 < t<\frac{\pi}{2}$

For solution to homogeneous equation $y'' + 4y = 0$, the characteristic polynomial is
\begin{align*}
r^2 + 4 &= 0\\
r_1 &= 2i\quad r_2 = -2i
\end{align*}
therefore, solution to homogeneous equation
\begin{align*}
y_c &= C_1e^{\lambda t}cos\mu t + C_2e^{\lambda t }sin\mu t\\
&= C_1cos2t + C_2sin2t
\end{align*}
For solution to $y'' + 4y = 3csc(2t)$
, since $p(t) = 0$, $q(t) = 4$, $g(t) = 3csc(2t)$ are both continuous on $0< t< \frac{\pi}{2}$
\begin{equation*}
W = \det \begin{bmatrix}
y_1 & y_1' \\ y_2 & y_2'
\end{bmatrix} = \begin{bmatrix}
cos2t & sin2t' \\ -2sin2t & 2cos2t'
\end{bmatrix} = 2
\end{equation*}
Thus, $y_1(t)$ and $y_2(t)$ form a fundamental set of solutions.

Therefore
\begin{align*}
u_1(t) &= -\int \frac{y_2(t)g(t)}{W[y_1, y_2](t)}\\
&= -\int\frac{(sin2t)(3csc2t)}{2}dt\\
&= -\int\frac{3}{2}dt\\
&= -\frac{3}{2}t
\end{align*}

\begin{align*}
u_2(t) &= \int\frac{y_1(t)g(t)}{W[y_1, y_2](t)}\\
&= \int\frac{(cos2t)(3csc2t)}{2}dt\\
&= \frac{3}{2}\int\frac{\cos{2t}}{\sin{2t}}dt\\
&= \frac{3}{2}\int \cot{2t}dt\\
&= \frac{3}{4}\ln{|\sin{2t}|}
\end{align*}
Therefore, the particular solution is $y_p(t) = u_1(t)y_1(t) + u_2(t)y_2(t)$, we get
\begin{align*}
y_p &= \cos{2t}(-\frac{3}{2}t) + \sin{2t}(\frac{3}{4}\ln{|\sin{2t}|})\\
&= \frac{3}{4}\sin{2t}\ln{|\sin{2t}|} - \frac{3}{2}t\cos{2t}
\end{align*}
Therefore the general solution is
\begin{equation*}
C_1cos2t + C_2sin2t + \frac{3}{4}\sin{2t}\ln{|\sin{2t}|} - \frac{3}{2}t\cos{2t}
\end{equation*}