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MAT244--Lectures & Home Assignments / Sec 7.1 Q4
« on: October 30, 2018, 12:05:54 PM »
What does $u^{(4)}$ stand for?
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Term Test 1 / Re: TT1 Problem 2 (main)
« on: October 16, 2018, 07:29:31 AM »
(a) reformat equation into
$$y'' + p(x)y' + q(x)$$
where
$$p(x) = -\frac{x}{x-1}$$
by equation of wronskian
\begin{gather*}
W(x) = \exp(-\int p(x) dx)\\
W(x) = \exp(\int \frac{x}{x-1} dx) = \exp(\int \frac{u+1}{u}du) = c_0ue^u = c_0(xe^{x-1} - e^{x-1})
\end{gather*}
choose $c_0 = e$
$$W(x) = xe^x - e^x$$
(b) Plug in $y = x, y' = 1, y'' = 0$, equation becomes
$$-x + x = 0$$
therefore $y=x$ is a solution, let second solution be $y_2$ and let $y_1 = x$ since
$$W(x) = y_1y_2' - y_1'y_2$$
Plug in $y_1 = x, y_1' = 1$
$$xe^x - e^x = xy_2' - y_2$$
By inspection
$$y_2 = e^x$$
Hence, general solution is
$$y(x) = c_1x + c_2e^x$$
c) Set $y(0) = 1$
$$c_2 = 1$$
Set $y'(0) = 0, y'(0) = c_1 + c_2e^x$
$$c_1 + c_2 = 0, c_1 = -c_2 = -1$$
Hence the solution is
$$y(x) = -x + e^x$$
$$y'' + p(x)y' + q(x)$$
where
$$p(x) = -\frac{x}{x-1}$$
by equation of wronskian
\begin{gather*}
W(x) = \exp(-\int p(x) dx)\\
W(x) = \exp(\int \frac{x}{x-1} dx) = \exp(\int \frac{u+1}{u}du) = c_0ue^u = c_0(xe^{x-1} - e^{x-1})
\end{gather*}
choose $c_0 = e$
$$W(x) = xe^x - e^x$$
(b) Plug in $y = x, y' = 1, y'' = 0$, equation becomes
$$-x + x = 0$$
therefore $y=x$ is a solution, let second solution be $y_2$ and let $y_1 = x$ since
$$W(x) = y_1y_2' - y_1'y_2$$
Plug in $y_1 = x, y_1' = 1$
$$xe^x - e^x = xy_2' - y_2$$
By inspection
$$y_2 = e^x$$
Hence, general solution is
$$y(x) = c_1x + c_2e^x$$
c) Set $y(0) = 1$
$$c_2 = 1$$
Set $y'(0) = 0, y'(0) = c_1 + c_2e^x$
$$c_1 + c_2 = 0, c_1 = -c_2 = -1$$
Hence the solution is
$$y(x) = -x + e^x$$
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MAT244--Lectures & Home Assignments / Re: Typo in assignment 3.4
« on: October 15, 2018, 06:41:36 PM »
Also, there is an earlier one in Sec2.4 27, where $y'$ should be replaced by $z'$. Not sure if a separate post is appropriate.
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MAT244--Lectures & Home Assignments / Typo in assignment 3.4
« on: October 15, 2018, 04:45:56 PM »
In Method of Reduction, part a, after wronskian is expanded $y_1$ should be replaced by $y_1'$
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Quiz-3 / Re: Q3 TUT 0701
« on: October 12, 2018, 06:09:14 PM »
Reformat equation into form $y'' + p(t)y' + q(t) = 0$
We can see that
$p(t) = \frac{\sin(t)}{\cos(t)}$
By the equation for wronskian
$$W(t) = \exp \bigl(\int p(t) dt\bigr) = \exp \bigl(\int \frac{\sin(t)}{\cos(t)} dt\bigr) = \exp (-\ln(\cos(t))+\ln (c_0)) = \frac{c_0}{\cos(t)}$$
where $c_0$ depends on choice of $y_2$ and $y_2$
We can see that
$p(t) = \frac{\sin(t)}{\cos(t)}$
By the equation for wronskian
$$W(t) = \exp \bigl(\int p(t) dt\bigr) = \exp \bigl(\int \frac{\sin(t)}{\cos(t)} dt\bigr) = \exp (-\ln(\cos(t))+\ln (c_0)) = \frac{c_0}{\cos(t)}$$
where $c_0$ depends on choice of $y_2$ and $y_2$
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Thanksgiving Bonus / Re: Thanksgiving bonus 3
« on: October 05, 2018, 06:35:26 PM »
$y_1 = x^3, y_1' = 3x^2, y_1'' = 6x$
$y_2 = y_2' = y_2'' = e^x$
Expand W(y, y_1, y_2)
$$y(y_1'y_2'' - y_2'y_1'') - y'(y_1y_2'' - y_2y_1'') + y''(y_1y_2' - y_2y_1') = 0.$$
Plug in $y_1$, $y_2$ and their derivatives, equation becomes
$$(x^3 - 3x^2)e^xy'' + (6x - x^3)e^xy' + (3x^2 - 6x)e^xy' = 0$$
or
$$ (x^2 - 3x)y'' + (6 - x^2)y' + (3x - 6)y' = 0$$
Btw, I think typo is still there. Two $y_2''$. FIXED
$y_2 = y_2' = y_2'' = e^x$
Expand W(y, y_1, y_2)
$$y(y_1'y_2'' - y_2'y_1'') - y'(y_1y_2'' - y_2y_1'') + y''(y_1y_2' - y_2y_1') = 0.$$
Plug in $y_1$, $y_2$ and their derivatives, equation becomes
$$(x^3 - 3x^2)e^xy'' + (6x - x^3)e^xy' + (3x^2 - 6x)e^xy' = 0$$
or
$$ (x^2 - 3x)y'' + (6 - x^2)y' + (3x - 6)y' = 0$$
Btw, I think typo is still there. Two $y_2''$. FIXED
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Quiz-2 / Re: Q2 TUT 0701
« on: October 05, 2018, 05:42:24 PM »
\begin{gather*}
N_x = \frac{2x}{y} - \frac{3y}{x^2}, M_y = -\frac{6}{y^2}, xM = 3x^2 + \frac{6x}{y}, yN = x^2 + 3\frac{y^2}{x}, \\\frac{N_x - M_y}{xM - yN} = \frac{\frac{2x}{y} - \frac{3y}{x^2} + \frac{6}{y^2}}{2x^2 - \frac{3y^2}{x} + \frac{6x}{y}} = R = \frac{1}{xy}
\end{gather*}
From assignment
\begin{gather*}
\mu (z) = \exp(\int \frac{1}{z} d(z)) = z,\\
\mu M = 3x^2y + 6x, \mu N = x^3 + 3y^2,\\
\int \mu M dx = x^3y + 3x^2 + f(y) + c_0,\\
\int \mu N dy = x^3y + y^3 + g(x) + c_1.
\end{gather*}
Combine the previous two result gives
$$ \phi(x, y) = x^3y + 3x^2 + y^3 = c,$$
where
$\frac{\partial\phi}{\partial x} = uM, \frac{\partial\phi}{\partial x} = uN$ We do not need this
N_x = \frac{2x}{y} - \frac{3y}{x^2}, M_y = -\frac{6}{y^2}, xM = 3x^2 + \frac{6x}{y}, yN = x^2 + 3\frac{y^2}{x}, \\\frac{N_x - M_y}{xM - yN} = \frac{\frac{2x}{y} - \frac{3y}{x^2} + \frac{6}{y^2}}{2x^2 - \frac{3y^2}{x} + \frac{6x}{y}} = R = \frac{1}{xy}
\end{gather*}
From assignment
\begin{gather*}
\mu (z) = \exp(\int \frac{1}{z} d(z)) = z,\\
\mu M = 3x^2y + 6x, \mu N = x^3 + 3y^2,\\
\int \mu M dx = x^3y + 3x^2 + f(y) + c_0,\\
\int \mu N dy = x^3y + y^3 + g(x) + c_1.
\end{gather*}
Combine the previous two result gives
$$ \phi(x, y) = x^3y + 3x^2 + y^3 = c,$$
$\frac{\partial\phi}{\partial x} = uM, \frac{\partial\phi}{\partial x} = uN$
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MAT244--Lectures & Home Assignments / Exercise 2.4
« on: October 01, 2018, 03:48:04 PM »
I believe there is a typo in exercise 2.4 question 27b.
I was checking the textbook and seems like the substitution should be $v = y^{1-n}$ though what I see from course website is $v = y^{1\hat{a}'n}$
I was checking the textbook and seems like the substitution should be $v = y^{1-n}$ though what I see from course website is $v = y^{1\hat{a}'n}$
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Quiz-1 / Re: Q1: TUT 0201, TUT 5101 and TUT 5102
« on: September 29, 2018, 09:23:56 PM »
Rephrase equation
$y' + \frac{2}{t}y = \frac{sin(t)}{t}$
Find integrating factor
$u(t) = e^{\int \frac{2}{t}} = t^2$
the constant from integration is chosen to be zero. Now
$y = \frac{1}{u(t)}\int u(t)\frac{sin(t)}{t}$
Use int by parts,
$y = -\frac{cos(t)}{t} + \frac{sin(t)}{t^2} + \frac{c_1}{t^2}$
Since $t$ is in denominator and the nominators are bounded, as $t \to \infty$, $y \to 0$
$y' + \frac{2}{t}y = \frac{sin(t)}{t}$
Find integrating factor
$u(t) = e^{\int \frac{2}{t}} = t^2$
the constant from integration is chosen to be zero. Now
$y = \frac{1}{u(t)}\int u(t)\frac{sin(t)}{t}$
Use int by parts,
$y = -\frac{cos(t)}{t} + \frac{sin(t)}{t^2} + \frac{c_1}{t^2}$
Since $t$ is in denominator and the nominators are bounded, as $t \to \infty$, $y \to 0$
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Quiz-1 / Re: Q1: TUT 0701
« on: September 28, 2018, 06:00:02 PM »
First divide by $t^3$ on both side of the equation, we get
$$y' + \frac{4}{t}y = \frac{e^{-t}}{t^3}$$
Using the method of integrating factor we have equation for $u(t)$
$$u(t) = e^{\int \frac{4}{t}dt} = e^{4\ln(t) + c} = t^4$$
where constant $c$ is arbitrary, it's chosen to be 0 here. Then
$$\bigl(y u(t)\bigr)' = u(t)\frac{e^{-t}}{t^3}$$
rearranging gives equation
$$y = \frac{1}{u(t)}\int u(t)\frac{e^{-t}}{t^3}$$
substitute in $u(t) = t^4$
$$y = \frac{1}{t^4}\int te^{-t}$$
use integration by parts
$$y = -\frac{e^{-t}}{t^3} - \frac{e^{-t}}{t^4} + \frac{c_1}{t^4}$$
to check $c_1$, plug in condition $y(-1) = 0$
$$y(-1) = e - e + c_1 = c_1= 0$$
Plug in $c_1 = 0$ gets
$$y = -\frac{e^{-t}}{t^3} - \frac{e^{-t}}{t^4} $$
$$y' + \frac{4}{t}y = \frac{e^{-t}}{t^3}$$
Using the method of integrating factor we have equation for $u(t)$
$$u(t) = e^{\int \frac{4}{t}dt} = e^{4\ln(t) + c} = t^4$$
where constant $c$ is arbitrary, it's chosen to be 0 here. Then
$$\bigl(y u(t)\bigr)' = u(t)\frac{e^{-t}}{t^3}$$
rearranging gives equation
$$y = \frac{1}{u(t)}\int u(t)\frac{e^{-t}}{t^3}$$
substitute in $u(t) = t^4$
$$y = \frac{1}{t^4}\int te^{-t}$$
use integration by parts
$$y = -\frac{e^{-t}}{t^3} - \frac{e^{-t}}{t^4} + \frac{c_1}{t^4}$$
to check $c_1$, plug in condition $y(-1) = 0$
$$y(-1) = e - e + c_1 = c_1= 0$$
Plug in $c_1 = 0$ gets
$$y = -\frac{e^{-t}}{t^3} - \frac{e^{-t}}{t^4} $$
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MAT244--Misc / Re: quiz
« on: September 21, 2018, 09:45:58 PM »
Check the arrow on quiz 1. http://www.math.toronto.edu/courses/mat244h1/20189/homeassignments-fall.html
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MAT244--Lectures & Home Assignments / Re: linear differential equations
« on: September 18, 2018, 10:23:08 PM »
Maybe in that website independent variable is $t$ and $x$, $y$ are the dependent variables. That could be why it's said to be nonlinear.
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MAT244--Misc / Re: Tutorial Pacing
« on: September 18, 2018, 02:03:06 PM »
Will the quizzes cover materials based on lectures, tutorials, or assignments due before the date of the quiz?
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MAT244--Lectures & Home Assignments / Re: Directional field
« on: September 17, 2018, 07:51:31 PM »
When your equation for $y'$ only has dependence on $y$, directional field has same value for same $y$.
But $y'$ could have dependence on other variables like $t$. $y' = y + t$ for example, even with $y$ fixed, could yield different slope at different $t$.
But $y'$ could have dependence on other variables like $t$. $y' = y + t$ for example, even with $y$ fixed, could yield different slope at different $t$.
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MAT244--Lectures & Home Assignments / Re: substitution in integral
« on: September 12, 2018, 06:08:56 AM »
dt can exist alone. Look up the technique called separation of variable.
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