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Messages - Yiyang Huang

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1
Quiz-6 / TUT0303 Quiz6
« on: November 15, 2019, 02:00:40 PM »
a. Express the general solution of the given system of equations in terms of real-value functions.
$$
x^{\prime}=\left[\begin{array}{rr}{-2} & {1} \\ {1} & {-2}\end{array}\right] x
$$
$$
\begin{aligned} \operatorname{det}(A-\lambda I) &=\left|\begin{array}{cc}{-2-\lambda} & {1} \\ {1} & {-2-\lambda}\end{array}\right| \\ &=(-2-\lambda)(-2-\lambda)-1 \\ &=3+4 \lambda+\lambda^{2} \\ &=(\lambda+3)(\lambda+1) \end{aligned}
$$
$$
\lambda=-3, \lambda=-1
$$

when $\lambda=-3$, $(A-(-3) I) x=0$
$$
\left[\begin{array}{ll|l}{1} & {1} & {0} \\ {1} & {1} & {0}\end{array}\right] \sim\left[\begin{array}{ll|l}{1} & {1} & {0} \\ {0} & {0} & {0}\end{array}\right]
$$
$$
\begin{array}{c}{x_{2}=t, \quad x_{1}=-t} \\ {x=t\left[\begin{array}{c}{-1} \\ {1}\end{array}\right]}\end{array}
$$

when $\lambda=-1$, $(A-(-1) I) x=0$
$$
\left[\begin{array}{rr|r}{-1} & {1} & {0} \\ {1} & {-1} & {0}\end{array}\right] \sim\left[\begin{array}{rr|r}{1} & {-1} & {0} \\ {0} & {0} & {0}\end{array}\right]
$$
$$
\begin{aligned} x_{2} =t &\quad x_{1}=t \\ x=& t\left[\begin{array}{l}{1} \\ {1}\end{array}\right] \end{aligned}
$$

Hence, $y=c_{1} e^{-t}\left[\begin{array}{l}{1} \\ {1}\end{array}\right]+c_{2} e^{-3 t}\left[\begin{array}{c}{-1} \\ {1}\end{array}\right]$

2
Quiz-5 / TUT0303 Quiz5
« on: November 01, 2019, 02:02:22 PM »
Find the general solution of the given differential equation.
$$
y^{\prime \prime}+4 y=3 \csc (2 t), \quad 0<t<\pi / 2
$$

$$
\begin{aligned} r^{2}+4 &=0 \\ r^{2} &=-4 \\ r &=\pm 2 i \\ y_{c(t)}=& c_{1} \cos 2 t+c_{2} \sin 2 t \end{aligned}
$$
$$
w=\left|\begin{array}{cc}{\cos 2 t} & {\sin 2 t} \\ {-2 \sin 2 t} & {2 \cos 2 t}\end{array}\right|=2 \cos ^{2} 2 t+2 \sin ^{2} 2 t=2
$$
$$
\begin{array}{l}{w_{1}=\left|\begin{array}{cc}{0} & {\sin 2 t} \\ {1} & {2 \cos 2 t}\end{array}\right|=-\sin 2 t} \\ {w_{2}=\left|\begin{array}{cc}{\cos 2 t} & {0} \\ {-2 \sin 2 t} & {1}\end{array}\right|=\cos 2 t}\end{array}
$$
$$
\begin{aligned} y_{p}(t) &=\cos 2 t \int \frac{(-\sin 2 t)(3 \csc (2 t))}{2} d t+\sin 2 t \int \frac{(\cos 2 t)(3 \csc (2 t))}{2}d t \\ &=\cos 2 t \int-\frac{3}{2} d t+\sin 2 t \int \frac{3}{2} \cot (2 t) d t \\ &=(\cos 2 t)\left(-\frac{3}{2} t\right)+\frac{3}{4} \sin 2 t \ln |\sin (2 t)| \\ &=-\frac{3}{2} t \cos 2 t+\frac{3}{4} \sin 2 t \ln |\sin (2 t)| \end{aligned}
$$
$$
y(t)=c_{1} \cos 2 t+c_{2} \sin 2 t+\frac{3}{4} \sin 2 t \ln |\sin (2 t)|-\frac{3}{2} t \cos 2 t
$$

3
Term Test 1 / Re: Problem 3 (morning)
« on: October 23, 2019, 08:05:12 AM »
Find the general solution for the equation
$$
y^{\prime \prime}-6 y^{\prime}+8 y=48 \sinh (2 x)
$$
$$
\begin{array}{c}{r^{2}-6 r+8=0} \\ {(r-4)(r-2)=0} \\ {r=4 \quad r=2} \\ {y_c=c_{1} e^{4 x}+c_{2} e^{2 x}}\end{array}
$$
$$
y^{\prime \prime}-6 y^{\prime}+8 y=48 \sinh (2 x)
$$
$$
\begin{array}{l}{\sinh (x)=\frac{e^{x}-e^{-x}}{2}} \\ {\sinh (2 x)=\frac{e^{2 x}-e^{-2 x}}{2}} \\ {48 \sinh (2 x)=24 e^{2 x}-24 e^{-2 x}}\end{array}
$$
$$
y^{\prime \prime}-6 y^{\prime}+8 y=24 e^{2 x}
$$

let $y_{p}(t)=A x e^{2 x}$
$$
y^{\prime}=A e^{2 x}+2 A x e^{2 x} \quad y^{\prime \prime}=4 A e^{2 x}+4 A x e^{2 x}
$$
$$
\begin{aligned} 4 A e^{2 x}+4 A x e^{2 x}-6 A e^{2 x}-12 A x e^{2 x}+8 A x e^{2 x} &=24 e^{2 x} \\ 4 A-6 A &=24 \\ A &=-12 \\ y_{P}(t) &=-12 \times 2^{2 x} \end{aligned}
$$
$$
y^{\prime \prime}-6 y^{\prime}+8 y=-24 e^{-2 x}
$$
Let $y_p(t)=A e^{-2 x} \quad y^{\prime}=-2 A e^{-2 x} \quad y^{\prime \prime}=4 A e^{-2 x}$
$$
\begin{aligned} 44 e^{-2 x}+12 A e^{-2 x}+8 A e^{-2 x} &=-24 e^{-2 x} \\ A &=-1 \\ y_{p} &=-e^{-2 x} \end{aligned}
$$
$$
y=c_{1} e^{4 x}+c_{2} e^{2 x}-12 x e^{2 x}-e^{-2 x}
$$

4
Quiz-4 / TUT0303 Quiz4
« on: October 18, 2019, 02:00:05 PM »
Find the general solution for equation
$$
y^{\prime \prime}+2 y^{\prime}+y=2 e^{-t}
$$

It's a non-homogeneous DE. To find complimentry solution,
$$
y^{\prime \prime}+2 y^{\prime}+y=0
$$

Assume that $y=e^{r t}$ is the solution of this eq
Then the characteristic equation is

$$
\begin{array}{l}{r^{2}+2 r+1=0} \\ {r_{1}=r_{2}=-1 \quad \text { (repeated roots) }}\end{array}
$$

Hence, $y_{c}(t)=c_{1} e^{-t}+c_{2} t e^{-t}$

To find the particular solution
\begin{equation}
y^{\prime \prime}+2 y^{\prime}+y=2 e^{-t}
\end{equation}

we assume
$$
y_{p}(t)=A t^{2} e^{-t}
$$

Then,
$$
\begin{array}{l}{y_{p}^{\prime}(t)=2 A t e^{-t}-A t^{2} e^{-t}} \\ {y_{p}^{\prime \prime}(t)=2 A e^{-t}-4 A t e^{-t}+A t^{2} e^{-t}}\end{array}
$$

substitute $y_{p}, y_{p}^{\prime}, y_{p}^{\prime \prime}$ into equation ( 1)

we get
$$
\begin{aligned} 2 A e^{-t} &=2 e^{-t} \\ A &=1 \end{aligned}
$$

so $y_{p}(t)=t^{2} e^{-t}$

Hence the general solution is
$$
y=y_{c}(t)+y_{p}(t)=c_{1} e^{-t}+c_{2} t e^{-t}+t^{2} e^{-t}
$$

5
Quiz-3 / TUT0303 Quiz3
« on: October 11, 2019, 02:00:00 PM »
Verify that the functions $y_1$ and $y_2$ are solutions of the given differential equation. Do they constitute a fundamental set of solutions?
$$
y^{\prime \prime}+4 y=0, y(t)=\cos (2 t), y_{2}(t)=\sin (2 t)
$$
$$
\begin{aligned}
w=\left|\begin{array}{cc}{y_{1}(t)} & {y_{2}(t)} \\ {y_{1}^{\prime}(t)} & {y_{2}^{\prime}(t)}\end{array}\right|&=\left|\begin{array}{cc}{\cos (2 t)} & {\sin (2 t)} \\ {-2 \sin (2 t)} & {2 \cos (2 t)}\end{array}\right|\\
&=2 \cos ^{2}(2 t)+2 \sin ^{2}(2 t) \neq 0
\end{aligned}
$$
$$
\begin{aligned} y_{1}=\cos (2 t) \quad& y_{1}^{\prime}=-2 \sin 2 t \quad y_{1}^{\prime \prime}=-4 \cos 2 t \\ y_{1}^{\prime \prime}+4 y_{1}=&(-4 \cos 2 t)+4 \cos (2 t)=0 \\ y_{2}=\sin (2 t)\quad & y_{2}^{\prime}=2 \cos (2 t) \quad y_{2}^{\prime \prime}=-4 \sin (2 t) \\ y_{2}^{\prime \prime}+4 y_{2} &=-4 \sin (2 t)+4 \sin (2 t)=0 \end{aligned}
$$

Hence, they constitute a fundamental set of solutions.

6
Quiz-2 / TUT0303 Quiz2
« on: October 04, 2019, 02:00:57 PM »
1.Show that the given equation is not exact but becomes exact when multiplied by the given integrating factor.Then solve the equation.
$$
x^2y^3+x(1+y^2)y^{\prime}=0,\qquad\mu(x,y)=1/xy^3.
$$

$$
\begin{aligned}
   M&=x^2y^3& N=x+xy^2\\
   My&=3x^2y^2& Nx=1+y^2 \text{not exact.}
   \end{aligned}
$$
$$\left. \begin{array}{l}{ \displaystyle\frac { x ^ { 2 } y ^ { 3 } } { x y ^ { 3 } } + ( \frac { x } { x y ^ { 3 } } + \frac { x y ^ { 2 } } { x y ^ { 3 } } ) y ^ { \prime } = 0 }\\{ \displaystyle x + ( \frac { 1 } { y ^ { 3 } } + \frac { 1 } { y } ) y ^ { \prime } = 0 }\\{ M_y = N_x = 0 }\end{array} \right.$$

Hence$$\begin{aligned}
\exists \phi_{(x,y)} s.t. \phi _ { x } &= M \\ \phi &= \int x d x \\ &= \frac { 1 } { 2 } x ^ { 2 } + h ( y ) \\ \Phi _ { y } &= \frac { 1 } { y ^ { 3 } } + \frac { 1 } { y } \\ h ^ { \prime } ( y ) &= - \frac { 1 } { 2 } y ^ { - 2 } + \ln | y | + c \\\phi_{(x,y)}&=\frac{1}{2}x^2-\frac{1}{2}y^{-2}+\ln | y |+c
\end{aligned}
$$

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