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### Messages - Shang Wu

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1
##### Quiz-4 / TUT5103 Quiz4
« on: October 18, 2019, 02:21:01 PM »
Find the solution of the given initial value problem.
\begin{align*}
y''+y&=0, \\
y(\pi /3)&=2, \\
y'(\pi /3)&=-4
\end{align*}
Solution:
First we solve the characteristic equation:
\begin{align*}
r^2+1&=0\\
r&=\pm i
\end{align*}
so here $\lambda = 0, \mu = 1$, the general solution for the homogeneous equation is
\begin{align*}
y_1=c_1\cos(t)+c_2\sin(t)
\end{align*}
We can check the Wronskian is $1\neq 0$.
Differentiate y, we get
\begin{align*}
y'=-c_1\sin(t)+c_2\cos(t)
\end{align*}
Now substitute the initial values:
\begin{align*}
y(\pi /3)&=2 \\
c_1\cos(\pi /3) +c_2\sin(\pi/3) &=2\\
\frac{1}{2}c_1+\frac{\sqrt{3}}{2}c_2 &=2\\
y'(\pi /3) &=-4\\
-c_1\sin(\pi/3)+c_2\cos(\pi/3)&=-4\\
-\frac{\sqrt{3}}{2}c_1+\frac{1}{2}c_2 &=-4
\end{align*}
Solve for $c_1, c_2$, we get
\begin{align*}
c_1 = 1+2\sqrt{3}, c_2 = \sqrt{3}-2
\end{align*}
Then the solution of the initial value problem is
\begin{align*}
y = (1+2\sqrt{3})\cos(t) + (\sqrt{3}-2)\sin(t)
\end{align*}

2
##### Quiz-3 / TUT5103 Quiz3
« on: October 11, 2019, 01:56:28 PM »
Find the general solution of the given differential equation.
\begin{align*}
2y''-3y'+y=0
\end{align*}
Solution:
First we find the the characteristic equation, that is
\begin{align*}
2r^2-3r+1=0
\end{align*}
then we solve for the roots $r_1, r_2$
\begin{align*}
r&=\frac{-b\pm \sqrt{b^2-4ac}}{2a} \\
&=\frac{-(-3)\pm\sqrt{(-3)^2-4(2)(1)}}{2(2)} \\
&= \frac{3 \pm 1}{4}
\end{align*}
So
\begin{align*}
r_1 &=\frac{3 + 1}{4}=1,\\
r_2 &=\frac{3 - 1}{4}=\frac{1}{2}
\end{align*}
and
\begin{align*}
y_1 &= e^t, \\
y_2 &=e^{\frac{1}{2}t}
\end{align*}
Therefore we get the general solution
\begin{align*}
y&=c_1y_1+c_2y_2\\
&=c_1e^t + c_2e^{\frac{1}{2}t}
\end{align*}

3
##### Quiz-2 / TUT5103 Quiz2
« on: October 04, 2019, 02:33:50 PM »
Show that the given equation is not exact but becomes exact when multiplied by the given integrating factor. Then solve the equation.
\begin{align*}
(\frac{\sin(y)}{y}- 2e^{-x}\sin(x))+(\frac{\cos(y)+2e^{-x}\cos(x)}{y})y' = 0, \mu(x,y) = ye^{x}.
\end{align*}
We first need to check if $M_y=N_x$ to decide if the equation is exact.
Notice
\begin{align*}
M &= \frac{\sin(y)}{y}- 2e^{-x}\sin{(x)} \\
N &=\frac{\cos(y)+2e^{-x}\cos(x)}{y}
\end{align*}
Then
\begin{align*}
M_y &= \frac{\cos(y)y-\sin(y)}{y^2} \\
N_x &= (\frac{\cos(y)}{y}+\frac{2e^{-x}\cos(x)}{y})_x \\
&= \frac{-2e^{-2}\cos(x) - 2e^{-x}\sin(x)}{y}
\end{align*}
We showed the given equation is not exact since $M_y \neq N_x$.
Then we try to multiply the equation by the integrating factor $\mu(x,y) = ye^x$. Then we get
\begin{align*}
(\sin(y)e^{x}- 2y\sin{(x)})+(e^{x}\cos(y)+2\cos(x))y' = 0
\end{align*}
Now we check $M_y$ and $N_x$ of the new equation.
Notice
\begin{align*}
M &= \sin(y)e^{x}- 2y\sin{(x)}\\
N &= e^{x}\cos(y)+2\cos(x)\\
\end{align*}
Then
\begin{align*}
M_y &= e^x \cos(y)-2\sin(x) \\
N_x &= e^x\cos(y)-2\sin(x)
\end{align*}
We verified that $M_y = N_x$, so the new equation is exact.
Now we integrate $M$ to solve for the general equation.
\begin{align*}
\psi = \int (\sin(y)e^{x}- 2y\sin{(x)})dx = \sin(y)e^{x}+ 2y\cos{(x)}+h(y)
\end{align*}
where $h(y)$ is a function of just y and we can find it by partial differentiate $\psi$ with respect to $y$. So
\begin{align*}
\psi_y = e^x\cos(y)+2\cos(x) + h'(y) = N(x,y) = e^{x}\cos(y)+2\cos(x)
\end{align*}
so we get $h'(y)=o$ and $h(y)=c, c$ is a constant. Then
\begin{align*}
\psi = \sin(y)e^{x}+ 2y\cos{(x)} +c
\end{align*}
Therefore we get the general solution
\begin{align*}
\sin(y)e^{x}+ 2y\cos{(x)} = C
\end{align*}

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