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Messages - Nick Callow

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1
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}

2
Quiz-3 / Re: Q3 TUT 0801
« on: October 12, 2018, 06:35:23 PM »
To find the Wronskian of the equation without solving we can apply Abel's Theorem. However, we must first isolate the second derivative term in $t^2y''(t) - t(t+2)y'(t) + (t+2)y(t) = 0$. We can do this by dividing all terms by $t^2$. Doing so yields the equation $$y'(t) - \frac{t+2}{t}y'(t) + \frac{t+2}{t^2} = 0$$ Now we will compute the Wronskian $$W = ce^{-\int p(t)dt }$$ where $p(t) = -\frac{t+2}{t}$. Aside: $- \int -\frac{t+2}{t}dt = t + 2ln(t)$.

Therefore, we get that $$W = ce^{t + 2ln(t)} = ct^2e^t$$

3
Quiz-3 / Re: Q3 TUT 0501
« on: October 12, 2018, 06:16:17 PM »
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.

4
Quiz-3 / Re: Q3 TUT 0601
« on: October 12, 2018, 06:06:11 PM »
Find a differential equation with general solution $c_1e^{2t} + c_2e^{-3t}$.

Given that $r = -3, 2$ we know the characteristic equation of this general solution will be of the form $(r-2)(r+3)$. Expanding this, we get that the differential equation has the form $r^2 + r - 6$. Therefore, a differential equation with the above general solution will be $y''(t) + y'(t) - 6y(t) = 0$.

We can check this by working in reverse. Suppose we have a differential equation $y''(t) + y'(t) - 6y(t) = 0$. Then we can try 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 have, $r^2e^{rt} + re^{rt} - 6e^{rt} = 0$. We can factor out an $e^{rt}$ and the characteristic equation becomes $r^2 + r - 6$. Factoring this, we have $(r-2)(r+3)$. Since our $r = -3,2$ we know two particular solutions are $y_1(t) = e^{2t}$ and $y_2(t) = e^{-3t}$. The general form of this will be $c_1e^{2t} + c_2e^{-3t}$. This matches what the question was asking, so we are finished.

5
Quiz-1 / Re: Q1: TUT 0701
« on: September 28, 2018, 06:00:48 PM »
My solution to this quiz can be found on this attachment.

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