Author Topic: Q5 TUT 0401  (Read 206 times)

Victor Ivrii

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Q5 TUT 0401
« on: November 02, 2018, 03:15:49 PM »
Use the method of variation of parameters (without reducing the order) to determine the general solution of the given differential equation:
$$y''' + y' = \sec (t),\qquad -\pi /2 < t < \pi /2.$$

Tianfangtong Zhang

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Re: Q5 TUT 0401
« Reply #1 on: November 02, 2018, 03:19:43 PM »

The homogeneous equation is $y^{'''} + y^{'} = 0$

$r^3 + r = 0$

$r = 0, \pm i$

Thus the roots are $r = 0, \pm i$

Thus $y_{c} = c_1 + c_2\cos(t) + c_3\sin(t)$

\begin{equation*}
W(t) =
W(1, \cos (t), \sin (t))
= \begin{bmatrix}
1 & \cos(t) & \sin(t) \\
0 & -\sin(t) & \cos(t) \\
0 & -\cos(t) &-\sin(t)
\end{bmatrix}
= \sin^{2}(t) + \cos^{2} (t)
= 1
\end{equation*}

\begin{equation*}
W_{1}(t) =
\begin{bmatrix}
0 & \cos(t) & \sin(t) \\
0 & -\sin(t) & \cos(t) \\
1 & -\cos(t) &-\sin(t)
\end{bmatrix}
= \sin^{2}(t) + \cos^{2} (t)
= 1
\end{equation*}

\begin{equation*}
W_{2}(t) =
\begin{bmatrix}
1 & 0 & \sin(t) \\
0 & 0 & \cos(t) \\
1 & 1 &-\sin(t)
\end{bmatrix}
= -\cos (t)
\end{equation*}

\begin{equation*}
W_{3}(t) =
\begin{bmatrix}
1 & \cos (t) & 0 \\
0 & -\sin(t) & 0 \\
0 & -\cos(t) & 1
\end{bmatrix}
= -\sin (t)
\end{equation*}

\begin{align*}
y_{p}(t) &= y_{1}\int \frac{g(t)W_{1}(t)}{W(t)} dt + y_{2}\int \frac{g(t)W_{2}(t)}{W(t)} dt + y_{3}\int \frac{g(t)W_{3}(t)}{W(t)} dt \\
\\
&= 1\int \frac{\sec(t) 1}{1} dt + \cos (t)\int \frac{\sec (t) (-\cos(t))}{1} dt + \sin(t)\int \frac{\sec(t)(-\sin(t))}{1} dt\\
\\
&= \ln (\sec(t)+\tan(t))-t\cos(t) + \sin(t)\ln(\cos(t))
\end{align*}

Thus, the general solution is

$y(t) = c_1 + c_2\cos(t) + c_3\sin(t) + \ln(\sec(t)+\tan(t))-t\cos(t)+\sin(t)\ln(\cos(t))$

Samarth Agarwal

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Re: Q5 TUT 0401
« Reply #2 on: November 04, 2018, 01:58:22 AM »
First, we must change to homogeneous equation
$$ y'''(t) + y'(t) = 0 $$
Then we must find the characteristic equation
$$ r^{3} + r = 0 $$
$$ r ( r^{2} + 1) = 0 $$
$$ \mbox{Therefore, } r = 0 \mbox{ or } r = i \mbox{ or } r = -i $$
$$ \mbox{Therefore, } y_1(t) = 1 \mbox{ and } y_2(t) = \cos(t) \mbox{ and } y_3(t) = \sin(t) $$
$$ \mbox{Therefore, } y_c(t) = c_1 + c_2\cos(t) + c_3\sin(t) $$

$$ \mbox{Therefore, the Wronskian = W} (y_1, y_2, y_3)(t) =
\begin{vmatrix}
1&\cos(t)&\sin(t)\\
0&-\sin(t)&\cos(t)\\
0&-\cos(t)&-\sin(t)\\
\end{vmatrix} = \sin^{2}(t) + \cos^{2}(t) = 1
$$

$$ \mbox{Now we must find } W_1(y_1, y_2, y_3) $$
$$ \mbox{Therefore, } W_1(y_1, y_2, y_3)(t) =
\begin{vmatrix}
0&\cos(t)&\sin(t)\\
0&-\sin(t)&\cos(t)\\
1&-\cos(t)&-\sin(t)\\
\end{vmatrix} = \cos^{2}(t) + \sin^{2}(t) = 1
$$

$$ \mbox{Now we must find } W_2(y_1, y_2, y_3) $$
$$ \mbox{Therefore, } W_2(y_1, y_2, y_3)(t) =
\begin{vmatrix}
1&0&\sin(t)\\
0&0&\cos(t)\\
0&1&-\sin(t)\\
\end{vmatrix} = -\cos(t)
$$

$$ \mbox{Now we must find } W_3(y_1, y_2, y_3) $$
$$ \mbox{Therefore, } W_3(y_1, y_2, y_3)(t) =
\begin{vmatrix}
1&\cos(t)&0\\
0&-\sin(t)&0\\
0&-\cos(t)&1\\
\end{vmatrix} = -\sin(t)
$$

$$ \mbox{Therefore, }u_1' = \sec(t) $$
$$ \mbox{Therefore, }u_2' = \sec(t)(-\cos(t)) = -1 $$
$$ \mbox{Therefore, }u_3' = \sec(t)(-sin(t)) = \frac{1}{\cos(t)}(-\sin(t)) = -\tan(t) $$

$$ \mbox{Therefore, } u_1 = \int_{t_0}^{t} \sec(t) dt = \log(\sec(t) + \tan(t)) $$
$$ \mbox{Therefore, } u_2 = \int_{t_0}^{t} -1 dt = -t $$
$$ \mbox{Therefore, } u_3 = \int_{t_0}^{t} -\tan(t) dt = \log(\cos(t)) $$
$$ \mbox{Therefore, } y_p(t) =  \log(\sec(t) + \tan(t)) + (-t)(\cos(t)) + \log(\cos(t))(\sin(t)) $$

$$ \mbox{We know that } y(t) = y_c(t) + y_p(t) $$
$$ y(t) = c_1 + c_2\cos(t) + c_3\sin(t) +  \log\bigl(\sec(t) + \tan(t)\bigr) + (-t)(\cos(t)) + \log(\cos(t))(\sin(t)) $$
« Last Edit: November 04, 2018, 08:04:31 PM by Victor Ivrii »

Victor Ivrii

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Re: Q5 TUT 0401
« Reply #3 on: November 04, 2018, 08:06:49 PM »
Samarth, avoid \mbox, it is evi (not really, but there is a proper command \text{    }