Author Topic: QUIZ4 TUT0401  (Read 491 times)

baixiaox

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QUIZ4 TUT0401
« on: October 18, 2019, 02:37:19 PM »
Solve $y'' + 2y' + 5y = 3sin(2t)$
\begin{align*}
    r^2 + 2r + 5 &= 0\\
    r^2 + 2r + 1 &= -4\\
    (r+1)^2 &= -4\\
    r + 1 &= \pm 2i\\
    r &= -1 \pm 2i
\end{align*}
For solution to homogeneous equation
\begin{align*}
y_c &= C_1e^{\lambda t}cos\mu t + C_2e^{\lambda t }sin\mu t\\
&= C_1e^{-t}cos2t + C_2e^{-t}sin2t
\end{align*}
For solution to $y'' + 2y' + 5y = 3sin(2t)$, guess $y_p = Asin(2t) + Bcos(2t)$
\begin{align*}
   y_p' &= 2Acos(2t) - 2Bsin(2t)\\
   y_p'' &= -4Asin(2t) - 4cos(2t)
\end{align*}
Then
\begin{align*}
    y'' + 2y' + 5y &= -4Asin(2t) - 4cos(2t) + 4Acos(2t) - 4Bsin(2t) + 5Asin(2t) + 5Bcos(2t)\\
    &= (A - 4B)sin(2t) + (B + 4A)cos(2t)\\
    &= 3sin(2t)
\end{align*}
Therefore
\begin{align*}
    &A - 4B = 3 \quad and \quad B +4A = 0\\
    &A = \frac{3}{17}\quad and \quad B = - \frac{12}{17}
\end{align*}
The solution to $y'' + 2y' + 5y = 3sin(2t)$ is $y_p = \frac{3}{17}sin(2t) - \frac{12}{17}cos(2t)$

The general solution is
\begin{equation*}
    y = C_1e^{-t}cos2t + C_2e^{-t}sin2t + \frac{3}{17}sin(2t) - \frac{12}{17}cos(2t)
\end{equation*}