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Topics - yuhan cheng

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1
Quiz-4 / tut0402 quiz4
« on: October 18, 2019, 02:01:58 PM »
22.$$
y^{\prime \prime}+2 y^{\prime}+2 y=0, y(\pi / 4)=2, y^{\prime}(\pi / 4)=-2
$$
$$
\begin{array}{c}{y^{\prime \prime}+2 y^{\prime}+2 y=0} \\ {r^{2}+2 r+2=0} \\ {r=-1 \pm i} \\ {y=c_{1} e^{-t} \cos t+c_{2} e^{-t} \sin t}\end{array}
$$

We know that general solution is $y=c_{1} e^{-t} \cos t+c_{2} e^{-t} \sin t$



Initial condition $y(\pi / 4)=2$
$$
\begin{array}{l}{c_{1} e^{-\pi / 4} \cdot \frac{\sqrt{2}}{2}+c_{2} e^{-\pi / 4} \cdot \frac{\sqrt{2}}{2}=2} \\ {c_{1}+c_{2}=2 \sqrt{2} e^{\pi / 4}}\end{array}
$$
$$
\begin{array}{l}{y^{\prime}=-c_{1} e^{-t} \cos t-c_{1} e^{-t} \sin t-c_{2} e^{-t} \sin t+c_{2} e^{-t} \cos t} \\ {y^{\prime}(\pi / 4)=-2} \\ {-c_{1} e^{-\pi / 4} \frac{\sqrt{2}}{2}-c_{1} e^{-\pi / 4} \frac{\sqrt{2}}{2}-c_{2} e^{-\pi / 4} \frac{\sqrt{2}}{2}+c_{2} e^{-\pi / 4} \frac{\sqrt{2}}{2}=-2} \\ {-\sqrt{2} c_{1}=-2 e^{\pi / 4}} \\ {c_{1}=\sqrt{2} e^{\pi / 4}} \\ {c_{2}=2 \sqrt{2} e^{\pi / 4}-\sqrt{2} e^{\pi / 4}=\sqrt{2} e^{\pi / 4}}\end{array}
$$

Solution: $y=\sqrt{2} e^{-t+\pi / 4} \cos t+\sqrt{2} e^{-t+\pi / 4} \sin t$











2
Quiz-3 / TUT 5102 QUIZ3
« on: October 11, 2019, 02:00:45 PM »
\[
y''-2y'-2y=0
\]


\[
y''-2y'-2y=0
\]
We assume that $y=e^{rt}$, and then it follows that $r$ must be a root of characteristic equation
\[
r^2-2r-2=0
\]
We use the quadratic formula which is
\[
r=\frac{-b\pm\sqrt{b^2-4ac}}{2a}
\]
Hence,
\[
\left\{
\begin{array}{c}
r_1=1+\sqrt3\\
r_2=1-\sqrt3\\
\end{array}
\right.
\]
Since the general solution has the form of
\[
y=c_1e^{r_1t}+c_2e^{r_2t}
\]
Therefore, the general solution of the given differential equation is
\[
y=c_1e^{(1+\sqrt3)t}+c_2e^{(1-\sqrt3)t}
\]

3
Quiz-3 / TUT 5102 QUIZ3
« on: October 11, 2019, 02:00:00 PM »
12. $y^{\prime \prime}+3 y^{\prime}=0, y(0)=-2, y^{\prime}(0)=3$

Find roots of characteristic equation:
$$
\begin{array}{l}{r^{2}+3 r=0} \\ {r(r+3)=0} \\ {r_{1}=0} \\ {r_{2}=-3}\end{array}
$$

General solution of equation is:
$$
y=c_{1}+c_{2} e^{-3 t}
$$

Initial terms:
$$
\begin{array}{l}{y(0)=-2} \\ {-2=c_{1}+c_{2} e^{0}} \\ {-2=c_{1}+c_{2}}\end{array}
$$
$$
\begin{array}{l}{y^{\prime}(0)=3} \\ {-3 c_{2} e^{-3 t}=3} \\ {-3 c_{2} e^{0}=3} \\ {-3 c_{2}=3}\end{array}
$$
$$
\begin{array}{l}{c_{2}=-1} \\ {c_{1}+c_{2}=-2} \\ {c_{1}=-2+1=-1}\end{array}
$$

Solution of problem: $$y=-1-e^{-3 t}$$

4
Quiz-2 / TUT0402 quiz2
« on: October 04, 2019, 02:04:45 PM »
$$
1+(\frac{x}{y}-\sin(y))y^{\prime}=0.
$$

\noindent Let
$$
M(x,y)=1\quad \text{and}\quad N(x,y) =(\frac{x}{y}-\sin(y))
$$
Then
$$
\frac{\partial}{\partial y}M(x,y)=0\quad\text{and}\quad \frac{\partial}{\partial x}N(x,y)=\frac{1}{y}
$$
We can see that this equation is not exact, however, note that
$$
\frac{d\mu}{d y}=(\frac{N_x-M_y}{M})\mu=\frac{\mu}{y}\quad\rightarrow\quad \mu=y
$$
Multiplying our original euqation by$\mu(y)$, we have
$$
y+(x-y\sin(y))y^{\prime}=0
$$
We can see that this equaion is exact, since
$$
\frac{\partial}{\partial y}(y)=1=\frac{\partial}{\partial x}(x-y\sin(y))
$$
Thus, there exists a function $\psi(x,y)$ such that
\begin{align}
\psi_x (x,y)&=y\\
\psi_y (x,y)&=x-y\sin(y)
\end{align}
Integating (1) with respect to x, we get
$$
\psi(x,y)=xy+h(y)
$$
for some arbitary function $h$ of $y$. Next, differentiating with respect to $y$, and equating with (2), we get
$$
\psi_y(x,y)=x+h^{\prime}(y)
$$
Therefore,
$$
h^{\prime}(y)=-y\sin(y)\quad\rightarrow\quad h(y)=s\int y\sin(y)dy=y\cos(y)-\sin(y)
$$
and we have
$$
\psi(x,y)=xy+y\cos(y)-\sin(y)
$$
Thus, the solutions of the differential equation are given implicity by
$$
xy+y\cos(y)-\sin(y)=C
$$

5
Quiz-2 / TUT0402 quiz2
« on: October 04, 2019, 02:00:32 PM »
tut 0402 quiz2

6
Quiz-1 / TUT0402 QUIZ1
« on: September 27, 2019, 02:00:00 PM »
tut 0402 quiz 1

Pages: [1]