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**Term Test 2 / Re: TT2B-P1**

« **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}

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}