Toronto Math Forum

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}

Find the solution of the given differential equation $2y''(t) - 3y'(t) + y(t) = 0$.

Since this a second order, linear, homogeneous differential equation, a good guess will be a solution of the form $y(t) = e^{rt}$. Consequently, $y'(t) = re^{rt}$ and $y''(t) = r^2e^{rt}$. Subbing this into the differential equation, we get:
$$2r^2e^{rt} - 3re^{rt} + e^{rt} = 0.$$
 We can then factor out an $e^{rt}$ from the equation to produce
$$e^{rt}(2r^2 - 3r + 1)=0$$
We know that for any choice of $r$, $e^{rt} \neq 0$ therefore, we can discard it. We are left with the characteristic equation $2r^2 - 3r + 1$. Factoring this, we get $(r-1)(2r-1)$.

Since $r = 1$, $\frac{1}{2}$ we now have two particular solutions $y_1(t) = e^t$ and $y_2(t) = e^{\frac{t}{2}}$. The general solution of this will then be:
$$y(t) = c_1e^t + c_2e^{\frac{t}{2}}$$
where $c_1$, $c_2$ are constants.

My solution to this quiz can be found on this attachment.