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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}

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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 $$W = ce^{t + 2ln(t)} = ct^2e^t$$

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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.

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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.

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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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MAT244--Lectures & Home Assignments / Re: Problem 10 in ex 2.1

« on: September 28, 2018, 12:03:38 AM »

I believe the problem your talking about is finding the general solution of: $$ty'(t)-y(t) = t^2e^{-t}, t > 0$$
In that case, this differential is first order but not separable, therefore we will use the method of integrating factors.
Be more specific, what kind of equation is this one.
The first problem that arises is this is not in the standard form we use for integrating factors where $$y'(t) + a(t)y(t) = b(t)$$
So we need to fix that. We can do that by diving every term by $t$ to isolate the derivative term. This produces the equation $$y'(t) - \frac{1}{t}y(t) = te^{-t}$$
Now, we can find the integrating factor which we will denote by $\mu(t)$. $$\mu(t) = e^{\int\frac{-1}{t}dt} = e^{\ln(\frac{1}{t})} = \frac{1}{t}$$
Now we multiply each side of the differential by $\mu(t)$ and isolate for $y(t)$ as follows:
\begin{gather*}
\frac{1}{t}y'(t) - \frac{1}{t^2}y(t) = e^{-t}\implies\\
\frac{d}{dt} (\frac{1}{t}y(t)) = e^{-t}\implies\\
\int \frac{d}{dt} (\frac{1}{t}y(t))dt = \int e^{-t}dt\implies\\
 \qquad\qquad\qquad\frac{1}{t}y(t) + c_1 = -e^{-t} + c_2\implies\qquad\qquad\qquad\color{red}{\checkmark}\\
 \frac{1}{t}y(t) = -e^{-t} + k\implies\\
 y(t) = -te^{-t} + kt.
\end{gather*}
No need to introduce $c_1$ and $c_2$ at all: marked line is excessive

Therefore, the general solution of the above differential is given by: $$ y(t) = -te^{-t} + kt $$

Note that this is only valid when $t > 0$. You can see from above if $t = 0$ this makes no sense as we had to divide by $t$ several times.

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MAT244--Lectures & Home Assignments / Re: Describing Direction Fields

« on: September 23, 2018, 01:18:13 AM »

A possible analytical approach.

1. If we want to describe the behaviour of $y(t)$ as $t \rightarrow \infty$ then we need to actually find out what $y(t)$ is. We can do this through the method of integrating factors.

Given $y'(t) - y(t) = e^{-t}$,  let $\mu(t)$ represent the integrating factor. $\mu(t) = e^{\int a(t)dt}$, where $a(t) = -1$. Therefore, 
$$ \mu(t) = e^{\int -1dt}, \qquad \mu(t)  = e^{-t}, \qquad y'(t) - y(t) = e^{-t}$$
$e^{-t}(y'(t) - y(t)) = e^{-2t}$
$\frac{d}{dt}(e^{-t}y(t)) = e^{-2t}$
$\int \frac{d}{dt}(e^{-t}y(t))dt = \int e^{-2t}dt$
$e^{-t}y(t) + c_1 = -\frac{1}{2}e^{-2t} + c_2$
$e^{-t}y(t) = -\frac{1}{2}e^{-2t} + k$
$$y(t) = \frac{-\frac{1}{2}e^{-2t} + k}{e^{-t}} = -\frac{1}{2}e^{-t} + k e^t$$

Now that we have an equation for $y(t)$, we can describe its behaviour.

2. Zhihong I think this is what you were talking about where we could find the limit:
$$
\lim_{t \rightarrow \infty} \frac{-\frac{1}{2}e^{-2t} + k}{e^{-t}}
$$
which will depend on what we select $k$ to be. You can then see for what values of $k$ how the function $y(t)$ behaves and this should of course agree with the direction field but be more analytical and accurate.

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MAT244--Lectures & Home Assignments / Re: question about the acid water problem

« on: September 22, 2018, 11:59:39 PM »

He posted the link to Quercus so that's where I found it. You are right though, for some questions the problem is not listed and only the visual solutions are there. But I have seen a few where the equation is listed. Also, he hasn't posted the question you are talking about yet. I'm sure it will be up soon.

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MAT244--Lectures & Home Assignments / Re: question about the acid water problem

« on: September 21, 2018, 11:42:44 PM »

You can access the problems in class at this link: http://www.math.toronto.edu/afenyes/teaching/mat244-fall2018/