Author Topic: Q4 TUT 5101  (Read 1263 times)

Victor Ivrii

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Q4 TUT 5101
« on: October 26, 2018, 05:45:03 PM »
Find the general solution of the given differential equation.
$$
4y'' + y = 2 \sec(t/2),\qquad -\pi  < t < \pi.
$$

Yulin WANG

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Re: Q4 TUT 5101
« Reply #1 on: October 26, 2018, 06:03:13 PM »
Step1: Find the homogeneous solution $y_{h}$
\begin{align*}
4y'' + y &= 0 \\
4r^{2} + 1 &= 0\\
4r^{2} &= -1\\
r^{2} &= -\frac{1}{4}\\
r &= \pm\frac{1}{2}i\\
\end{align*}
So, $y_{1} = cos(\frac{t}{2})$ and $y_{2} = sin(\frac{t}{2})$

Then, $y_{h} = c_{1}cos(\frac{t}{2}) + c_{2}sin(\frac{t}{2})$

Step2: Find the particular solution $y_{p}$
\begin{align*}
W(y_{1}, y_{2}) = det
\begin{bmatrix}
y_{1} & y_{2} \\
y_{1}' & y_{2}'
\end{bmatrix}
&= y_{1}y_{2}' - y_{2}y_{1}'\\
&= \frac{1}{2}cos(\frac{t}{2})cos(\frac{t}{2}) + \frac{1}{2}sin(\frac{t}{2})sin(\frac{t}{2})\\
&= \frac{1}{2}cos^{2}(\frac{t}{2}) + \frac{1}{2}sin^{2}(\frac{t}{2})\\
&= \frac{1}{2}
\end{align*}
Rewrite the equation: $y'' + \frac{1}{4}y = \frac{1}{2}sec(\frac{t}{2})$

Then g(t) = $\frac{1}{2}sec(\frac{t}{2})$

Since $y_{1} = cos(\frac{t}{2})$, $y_{2} = sin(\frac{t}{2})$, $W(y_{1}, y_{2}) = \frac{1}{2}$, g(t) = $\frac{1}{2}sec(\frac{t}{2})$

So, we have:
\begin{align*}
u_{1} &= -\int \frac{g(t)y_{2}}{W(y_{1}, y_{2})}\\
&= -\int\frac{\frac{1}{2}sec(\frac{t}{2})sin(\frac{t}{2})}{\frac{1}{2}}dt\\
&= -\int tan(\frac{t}{2})dt\\
&= -2ln\lvert sec(\frac{t}{2})\rvert\\
~\\
u_{2} &= \int \frac{g(t)y_{1}}{W(y_{1}, y_{2})}\\
&= \int\frac{\frac{1}{2}sec(\frac{t}{2})cos(\frac{t}{2})}{\frac{1}{2}}dt\\
&= \int1dt\\
&= t
\end{align*}
So, $y_{p} = y_{1}u_{1} + y_{2}u_{2} = -2cos(\frac{t}{2})ln\lvert sec(\frac{t}{2})\rvert + tsin(\frac{t}{2})$

Therefore, $y = y_{h} + y_{p} = c_{1}cos(\frac{t}{2}) + c_{2}sin(\frac{t}{2}) - 2cos(\frac{t}{2})ln\lvert sec(\frac{t}{2})\rvert + tsin(\frac{t}{2})$
« Last Edit: October 26, 2018, 06:06:23 PM by Yulin Wang »