Author Topic: Q5 TUT 5102  (Read 305 times)

Victor Ivrii

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Q5 TUT 5102
« on: November 02, 2018, 03:18:54 PM »
Use the method of variation of parameters (without reducing the order) to determine the general solution of the given differential equation:
$$y''' - y' = \coth(t),\qquad t>0$$

Tianfangtong Zhang

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Re: Q5 TUT 5102
« Reply #1 on: November 02, 2018, 03:20:04 PM »
homogeneous equation is
$y^{'''} - y^{'} = 0$

then $r^3 - r = 0$

$r = 0, \pm 1$

thus, $y_{c}(t) = c_{1} + c_{2}e^t + c_{3}e^{-t}$

\begin{align*}
W(t) &=

\begin{bmatrix}
1 & e^t & e^{-t} \\
0 & e^t & -e^{-t}\\
0 & e^t & e^{-t}
\end{bmatrix} \\
\\

&= 1*(-1)^{1+1}
\begin{bmatrix}
e^t & -e^{-t}\\
e^t & e^{-t}
\end{bmatrix}
+ e^t(-1)^{2+1}
\begin{bmatrix}
0 & -e^{-t}\\
0 & e^{-t}
\end{bmatrix}
+ e^{-t}(-1)^{3+1}
\begin{bmatrix}
0 & e^t\\
0 & e^t
\end{bmatrix}\\
\\

&= e^t*e^t - (-e^{-t})e^t + 0 + 0 \\
\\

&= 1 + 1 \\
\\

&= 2
\end{align*}

\begin{align*}
W_{1}(t) &=
\begin{bmatrix}
0 & e^t & e^{-t}\\
0 & e^t & -e^{-t}\\
1 & e^t & e^{-t}
\end{bmatrix}\\
\\

&=1*(-1)^{1+3}
\begin{bmatrix}
e^t & e^{-t}\\
e^t & -e^{-t}
\end{bmatrix}\\
\\

&= -1-1 \\
\\

&= -2
\end{align*}

\begin{align*}
W_{2}(t) &=
\begin{bmatrix}
1 & 0 & e^{-t}\\
0 & 0 & -e^{-t}\\
0 & 1 & e^{-t}
\end{bmatrix}\\
\\

&=1*(-1)^{1+1}
\begin{bmatrix}
0 & -e^{-t}\\
1 & e^{-t}
\end{bmatrix}\\
\\

&= e^{-t}
\end{align*}
\\

\begin{align*}
W_{3}(t) &=
\begin{bmatrix}
1 & e^t & 0\\
0 & e^t & 0\\
0 & e^t & 1
\end{bmatrix}\\
\\

&=1*(-1)^{1+1}
\begin{bmatrix}
e^{t} & 0\\
e^{t} & 1
\end{bmatrix}\\
\\

&= e^{t}
\end{align*}

\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{\cosh(t)(-2)}{2)} dt + e^t\int \frac{\cosh(t)e^{-t}}{2} dt + e^{-t}\int \frac{\cosh(t)e^t}{2} dt \\
\\

&= -1\sinh(t) + \frac{1}{2}e^t\int \cosh(t)e^{-t} dt + \frac{1}{2}e^{-t}\int \cosh(t) e^t dt \\
\\

&= -\sinh(t) + \frac{1}{2}e^t \int \cosh(t)e^{-t}dt + \frac{1}{2}e^{-t}\int \cosh(t)e^t dt \\
\\

&= -\sinh(t) + \frac{1}{2}e^t(\frac{t}{2} - \frac{e^{-2t}}{4}) + \frac{1}{2}e^{-t}(\frac{t}{2}+ \frac{e^{2t}}{4})
\end{align*}

Therefore $y(t) = c_{1} + c_{2}e^t + c_{3}e^{-t} -\sinh(t) + \frac{1}{2}e^t(\frac{t}{2} - \frac{e^{-2t}}{4}) + \frac{1}{2}e^{-t}(\frac{t}{2}+ \frac{e^{2t}}{4})$
« Last Edit: November 03, 2018, 01:32:16 AM by Tianfangtong Zhang »

Michael Poon

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Re: Q5 TUT 5102
« Reply #2 on: November 02, 2018, 04:36:21 PM »
Tianfangtong Zhang I think you made a small typo,

$W_1(t) = -2$

not 2.

I think you fixed it in your final answer though.

Tianfangtong Zhang

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Re: Q5 TUT 5102
« Reply #3 on: November 03, 2018, 01:33:22 AM »
thank you, fix it

Michael Poon

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Re: Q5 TUT 5102
« Reply #4 on: November 10, 2018, 01:42:01 PM »
As mentioned before in the forum, $g(t)$ should be $\coth(t)$ not $\cosh(t)$

I recalculated:

$$
W(t) = 2e^{2t}\\

W_1(t) = -2e^{2t}\\

W_2(t) = e^t\\

W_3(t) = e^t

$$

This gives

$$
\begin{align}

y_p(t) &= \int{\frac{\coth(t)(-2e^{2t})dt}{2e^{2t}}} + e^t\int{\frac{\coth(t)e^t}{2e^{2t}}dt} + e^t\int{\frac{\coth(t)e^t}{2e^{2t}}dt}\\

y_p(t) &= -\int{\coth(t)dt} + e^t\int{\coth(t)e^{-t}dt}\\

y_p(t) &= -\ln|\sinh(t)| + e^t(e^{-t} + \ln|1-e^{-t}| - \ln|e^{-t} + 1|)\\

y(t) &= c_1+c_2e^t+c_3e^{-t} -\ln|\sinh(t)| + e^t(e^{-t} + \ln|1-e^{-t}| - \ln|e^{-t} + 1|)
\end{align}


$$