### Author Topic: Lec5101 quiz5  (Read 1041 times)

#### xuanzhong

• Jr. Member
• Posts: 12
• Karma: 1
##### Lec5101 quiz5
« on: November 01, 2019, 01:47:05 PM »
Find the general solution of the given differential equation.
$$y^{\prime\prime}+4y=3csc(2t), 0<t<pi/2$$
For homogeneous equation: $r^2+4=0$
we get:
$$r_{1}=2i, r_{2}=-2i$$
$$y_{c}(t)=c_{1}cos2t+c_{2}sin2t$$

For non-homogeneous equation:
$$W[y_{1},y_{2}](t) = \begin{vmatrix} cos2t & sin2t \\ -2sin2t & 2cos2t \\ \end{vmatrix} = 2$$

Therefore,
$$u_{1}(t)=-(\int\frac{sin2t*3csc2t}{2}dt)$$
$$=-(\int\frac{3}{2}dt)$$
$$=-\frac{3}{2}t$$
$$u_{2}(t)=-(\int\frac{cos2t*3csc2t}{2}dt)$$
$$=\frac{3}{2}(\int\frac{cos2t}{sin2t}dt)$$
$$=\frac{3}{2}(\int{cot2t}dt)$$
$$=\frac{3}{4}ln|sin2t|$$

Hence, the particular solution is $y_{p}(t)=u_{1}(t)y_{1}(t)+u_{2}(t)y_{2}(t)$
$$y_{p}(t)=cos2t\cdot(-\frac{3}{2}t)+sin2t\cdot(\frac{3}{4}ln|sin2t|)$$
$$=\frac{3}{4}sin2tln|sin2t|-\frac{3}{2}tcos2t$$

Therefore, the general solution is:
$$y(t)=y_{c}(t)+y_{p}(t)$$
$$=c_{1}cos2t+c_{2}sin2t+\frac{3}{4}sin2tln|sin2t|-\frac{3}{2}tcos2t$$