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

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##### 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

y^{\prime \prime}+2 y^{\prime}+y=2 e^{-t}

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.

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##### 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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##### Quiz-1 / TUT0303 Quiz1
« on: September 27, 2019, 02:00:01 PM »
Find the general solution of the equation
$$y^{\prime}-y=2 t e^{2 t}, \quad y(0)=1$$
$$p(t)=-1$$
\begin{aligned} \mu &=e^{\int p(t) d t} \\ &=e^{\int-1 d t} \\ &=e^{-t} \end{aligned}
\begin{aligned} e^{-t} y^{\prime}-e^{-t} y &=2 t e^{t} \\\left(e^{-t} y\right)^{\prime} &=2 t e^{t} \\ e^{-t} y &=\int 2 t e^{t} d t \end{aligned}
$$\begin{array}{ll}{u=2 t} & {v=\int d v=\int e^{t} d t=e^{t}} \\ {d u=2} & {d v=e^{t}}\end{array}$$
$$e^{-t} y=2 t e^{t}-\int e^{t} \cdot 2 d t\text{(integration by parts)}$$
\begin{aligned} e^{-t} y &=2 t e^{t}-2 e^t+c \\ &=2 e^{t}(t-1)+c \\ y &=2 e^{2 t}(t-1)+C e^{t} \end{aligned}
$$\text{substitute}\quad y(0)=1$$
\begin{aligned} 1 &=2 e^{0}(0-1)+c e^{0} \\ 1 &=-2+c \\ & c=3 \end{aligned}
$$y=2 e^{2 t}(t-1)+3 e^{t}$$

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