Author Topic: Lec5101 quiz5  (Read 490 times)

xuanzhong

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

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