Author Topic: TUT 0602 Quiz 5  (Read 30 times)

Yichen Ji

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TUT 0602 Quiz 5
« on: November 05, 2019, 03:12:52 AM »
Question: find the general solution of the differential equation $4y''+y=2sec(\frac{t}{2})$ where $-\pi<t<\pi$

Solution:
By the method of variation of parameters, first we find the solution of the homogeneous equation
\begin{equation*}
    4y''+y=0
\end{equation*}
The characteristic polynomial equation
\begin{equation*}
    4r^2+1=0
\end{equation*}
We get $r_1=\frac{1}{2}i$ and $r_2=-\frac{1}{2}i$
Then the homogeneous solution
\begin{equation*}
    y_h(t)=c_1cos(\frac{t}{2})+c_2sin(\frac{t}{2})
\end{equation*}
Then, substitute the two constants$c_1$ and $c_2$ for $u_1$ and $u_2$, we get
\begin{equation}
    y=u_1cos(\frac{t}{2})+u_2sin(\frac{t}{2})
\end{equation}
Take differentiation, we get
\begin{equation*}
    y'=u_1'cos(\frac{t}{2})+u_2'sin(\frac{t}{2})-\frac{1}{2}u_1sin(\frac{t}{2})+\frac{1}{2}u_2cos(\frac{t}{2})
\end{equation*}
Set
\begin{equation}
    u_1'cos(\frac{t}{2})+u_2'sin(\frac{t}{2})=0
\end{equation}
So
\begin{equation*}
  y'=-\frac{1}{2}u_1sin(\frac{t}{2})+\frac{1}{2}u_2cos(\frac{t}{2})
\end{equation*}
Differentiate again,we get
\begin{equation}
      y''=-\frac{1}{2}u_1'sin(\frac{t}{2})+\frac{1}{2}u_2'cos(\frac{t}{2})-\frac{1}{4}u_1cos(\frac{t}{2})-\frac{1}{4}u_2sin(\frac{t}{2})
\end{equation}
Substitute (1)(3) back to the original equation and use (2) for substitution,we get
\begin{equation*}
    u_2'\frac{sin(\frac{t}{2})}{cos(\frac{t}{2})}sin(\frac{t}{2})+u_2'cos(\frac{t}{2})=\frac{1}{cos(\frac{t}{2})}
\end{equation*}
Reorganizing, we get
\begin{align*}
    u_2' &=1\\
    u_2 &=t+c_2\\
    u_1' &=-tan(\frac{t}{2})\\
    u_1 &=2ln|cos(\frac{t}{2})|+c_1
\end{align*}
Finally we get the general solution
\begin{equation*}
    y=c_1cos(\frac{t}{2})+c_2sin(\frac{t}{2})+2ln|cos(\frac{t}{2})|cos(\frac{t}{2})+tsin(\frac{t}{2})
\end{equation*}
where$-\pi<t<\pi$