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### Topics - shangluy

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##### Quiz-4 / TUt0402 question
« on: October 18, 2019, 02:00:36 PM »
Solve $y'' - y' - 2y = cosh2t$

Firstly, consider homogeneous equation $y'' - y' - 2y = 0$, the characteristic polynomial is
\begin{align*}
r^2 - r - 2 &= 0\\
(r + 1)(r - 2) &= 0
\end{align*}

Therefore, the solution to homogeneous equation is $c_1 e^{-t} + c_2 e^{2t}$

Then need to solve $y'' - y' - 2y = cosh2t$, since
\begin{equation*}
cosh(2t) = \frac{e^{2t} + e^{-2t}}{2}
\end{equation*}

we come to guess $y = Ate^{2t} + Be^{-2t}$ is a solution.
\begin{align*}
y' &= Ae^{2t} + 2Ate^2t - 2Be^{-2t}\\
y'' &= 2Ae^{2t} + 2Ae^{2t} + 4Ate^{2t} + 4Be^{-2t}\\
&= (4A + 4At)e^{2t} + 4Be^{-2t}
\end{align*}
then plug $y$, $y'$ and $y''$ back in $y'' - y' - 2y$, we get

\begin{align*}
y'' - y' - 2y =& (4A + 4At)e^{2t} + 4Be^{-2t} - (Ae^{2t} + 2Ate^2t - 2Be^{-2t}) - 2(Ate^{2t} + Be^{-2t}) \\
&= (4A - A)e^{2t} + (4A - 2A - 2A)te^{2t} + (4B + 2B - 2B)e^{-2t}\\\
&= 3Ae^{2t} + 4Be^{-2t}
\end{align*}
since $y'' - y' - 2y = cosh2t$
\begin{align*}
&3Ae^{2t} + 4Be^{-2t} = \frac{e^{2t} + e^{-2t}}{2}\\
\end{align*}
Therefore the solution to $y'' - y' - 2y = cosh2t$ is $y = \frac{1}{6}te^{2t} + \frac{1}{8}e^{-2t}$
Then, the general solution is
\begin{equation*}
y = c_1 e^{-t} + c_2 e^{2t} + \frac{1}{6}te^{2t} + \frac{1}{8}e^{-2t}
\end{equation*}

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##### Quiz-3 / TUT0402 question
« on: October 11, 2019, 02:00:04 PM »
Find the Wronskian of $cos^2\theta$ and $1 + cos2\theta$

Let $f = cos^2\theta$ and $g = 1 + cos2\theta$, then

since $W = fg' - f'g$, we get
\begin{align*}
W &= cos^2\theta(-2sin2\theta) - (-sin2\theta)(1 + cos2\theta)\\
&= -2cos^2\theta sin2\theta + sin2\theta(1 + cos2\theta)\\
&= sin2\theta(-cos2\theta + cos2\theta)\\
&= 0
\end{align*}
Therefore the Wronskian of $cos^2\theta$ and $1 + cos2\theta$ is $0$

3
##### Quiz-2 / tut0402 question
« on: October 04, 2019, 02:00:04 PM »
Solve \begin{align*}
\frac{x dx}{(x^2 + y^2)^\frac{3}{2}} + \frac{y dy}{(x^2 + y^2)^\frac{3}{2}} = 0
\end{align*}

We can multiple both side by $(x^2 + y^2)^\frac{3}{2}$, then we have

xdx + ydy = 0

Let $M = x$ and $N = y$

Since $M_y = N_x$,  the equation is exact so $\exists \psi(x, y)\quad s.t \quad \frac{\partial \psi}{\partial x} = M$ and $\frac{\partial \psi}{\partial y} = N$
Therefore
\begin{align*}
\psi(x, y) &= \int M dx \\
&= \int x dx\\
&= \frac{1}{2}x^2 + h(y)\\
\end{align*}
Since
\begin{align*}
\frac{\partial \psi}{\psi y} &= h'(y)\\
\implies h'(y) &= y\\
\implies h(y) &= \frac{1}{2}y^2 + C
\end{align*}
Therefore, the solution is $\frac{1}{2}x^2 + \frac{1}{2}y^2 = C$, rearrange we get $x^2 + y^2 = C$, where $x \neq 0$ and $y \neq 0$

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##### Chapter 1 / TUT0402 question
« on: October 04, 2019, 12:21:24 AM »
Solve $\frac{xdx}{(x^2 + y^2)^\frac{3}{2}}$

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##### Quiz-1 / n
« on: October 04, 2019, 12:08:50 AM »
n

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##### Quiz-1 / TUT0402 question
« on: September 27, 2019, 02:20:15 PM »
TUT0402 question

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