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MAT244--2019F => MAT244--Test & Quizzes => Quiz-5 => Topic started by: Wang Jingyao on October 31, 2019, 06:31:02 PM

Title: LEC5101 Quiz5
Post by: Wang Jingyao on October 31, 2019, 06:31:02 PM
Find the general solution of the given differential equation.

$y’’+4y=3\csc(2t)$,$\quad 0<t<\dfrac{\pi}{2}$

For homogeneous equation:

$y’’+4y=0$

Characteristic equation:

$r^2+4=0$ $\Longrightarrow$ $\left\{\begin{array}{l}r_1=2i\\r_2=-2i \end{array}\right.$

Complementary solution:

$y_c(t)=c_1\cos(2t)+c_2\sin(2t)$

For nonhomogeneous equation $y’’+4y=3\csc(2t)$, we have:

$p(t)=0,\quad q(t)=4,\quad g(t)=3\csc(2t)$

Then,

$W[y_1(t),y_2(t)]=$$\left | \begin{matrix} y_1(t) & y_2(t) \\ y_1’(t) & y_2’(t) \end{matrix} \right |$$ =$$\left | \begin{matrix} \cos(2t) & \sin(2t) \\ -2\sin(2t) & 2\cos(2t) \end{matrix} \right |$$ =2$

Thus, $y_1(t)$ and $y_2(t)$ form a fundamental set of solutions. Therefore,

\begin{align} u_1(t)&=-\int\dfrac{y_2(t)g(t)}{ W[y_1(t),y_2(t)]}dt\\ \notag \\ &=-\int\dfrac{\sin(2t) \cdot 3\csc(2t)}{2}dt\\ \notag \\ &=-\int\dfrac{3}{2}dt\\ \notag \\ &=-\dfrac{3}{2}t\\ \end{align}

\begin{align} u_2(t)&=\int\dfrac{y_1(t)g(t)}{ W[y_1(t),y_2(t)]}dt\\ \notag \\ &=\int\dfrac{\cos(2t) \cdot 3\csc(2t)}{2}dt\\ \notag \\ &=\dfrac{3}{2}\int\dfrac{\cos(2t)}{\sin(2t)}dt\\ \notag \\ &=\dfrac{3}{2}\int \cot(2t)dt\\ \notag \\ &=\dfrac{3}{4}\ln|\sin(2t)|\\ \end{align}

Since,

$Y(t)=u_1(t)y_1(t)+u_2(t)y_2(t)$

Therefore,

\begin{align} Y(t)&= \cos(2t) \cdot (-\dfrac{3}{2}t)+\sin(2t) \cdot (\dfrac{3}{4}\ln|\sin(2t)|) \notag \\ \notag \\ &=\dfrac{3}{4}\sin(2t)\ln|\sin(2t)|-\dfrac{3}{2}t\cos(2t) \notag \\ \end{align}

Thus, the general solution is,

\begin{align} y(t)&=y_c(t)+Y(t) \notag \\ \notag \\ y(t)&= c_1\cos(2t)+c_2\sin(2t)+ \dfrac{3}{4}\sin(2t)\ln|\sin(2t)|-\dfrac{3}{2}t\cos(2t) \notag \\ \end{align}