Author Topic: TT2B-P1  (Read 861 times)

Victor Ivrii

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TT2B-P1
« on: November 20, 2018, 05:54:05 AM »
(a) Find the general solution of
\begin{equation*}
y''-5y'+6y= \frac{6e^{4t}}{e^{2t}+1}.
\end{equation*}

(b) Find solution, such that $y(0)=0$, $y'(0)=0$.

Guanyao Liang

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Re: TT2B-P1
« Reply #1 on: November 20, 2018, 06:37:17 AM »
This is my answer.

Nick Callow

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Re: TT2B-P1
« Reply #2 on: November 20, 2018, 06:41:50 AM »
Part A

1. Find the complementary solution by considering the homogeneous equation $y'' - 5y' + 6y = 0$.

Try a solution of the form $y(t) = e^{rt}$. Doing so, we get the characteristic equation:
\begin{align}
r^2 - 5r + 6 = 0 \\
(r-2)(r-3) = 0
\end{align}

Therefore, $r = 2, 3$. We have solutions $y_1(t) = e^{2t}$ and $y_2(t) = e^{3t}$. Therefore, the general solution of the homogeneous equation is given by
\begin{align}
y_c(t) = c_1e^{2t} + c_2e^{3t}
\end{align}

2. Using the method of variation of parameters, we can find a particular solution. Let $g(t) = \frac{6e^{4t}}{e^{2t} + 1}$.

Find the Wronskian, Wr[$y_1(t), y_2(t)$] = $e^{2t}3e^{3t} - e^{3t}2e^{2t} = 3e^{5t} - 2e^{5t} = e^{5t}$

\begin{align}
y_p(t) & = -y_1(t) \int_{t_0}^{t}\frac{g(s)y_2(s)}{W(s)}ds + y_2(t) \int_{t_0}^{t}\frac{g(s)y_1(s)}{W(s)}ds \\
y & = -e^{2t} \int_{t_0}^{t}\frac{\frac{6e^{4s}}{e^{2s}+1}e^{3s}}{e^{5s}}ds + e^{3t} \int_{t_0}^{t}\frac{\frac{6e^{4s}}{e^{2s}+1}e^{2s}}{e^{5s}}ds \\
& = -e^{2t} \int_{t_0}^{t}\frac{6e^{2s}}{e^{2s}+1}ds + e^{3t}\int_{t_0}^{t}\frac{6e^s}{e^{2s}+1}ds \\
& = -6e^{2t} \int_{t_0}^{t}\frac{e^{2s}}{e^{2s} + 1}ds + 6e^{3t}\int_{t_0}^{t}\frac{e^s}{e^{2s} + 1}ds \\
& = -3e^{2t}\ln({e^{2t} + 1}) + 6e^{3t}\arctan({e^t})
\end{align}

3. Find the general solution by combining complementary and particular solutions.

\begin{align}
y(t) & = y_c(t) + y_p(t) \\
& = c_1e^{2t} + c_2e^{3t} -3e^{2t}\ln({e^{2t} + 1}) + 6e^{3t}\arctan({e^t})
\end{align}

Jingyi Wang

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Re: TT2B-P1
« Reply #3 on: November 20, 2018, 09:34:09 AM »
For c, I think the final answer should be: