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### Messages - suyichen

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1
##### Quiz-6 / TUT0801
« on: November 15, 2019, 02:02:17 PM »
a. Find the general solution of given system of equations and describe the behavior as $t \rightarrow+\infty$

b. Draw a direction field and plot a few trajectories of the system

$$x^{\prime}=\left(\begin{array}{cc}{2} & {-1} \\ {3} & {-2}\end{array}\right) x$$
\begin{aligned} \operatorname{det}(A-\lambda I)=&\left|\begin{array}{cc}{2-\lambda} & {-1} \\ {3} & {-2-\lambda}\end{array}\right| =0 \\(2-\lambda)(-2-&\lambda)-(-3) =0 \\-4+\lambda^{2}&+3 =0 \\ \lambda=1 &\quad \lambda=-1 \end{aligned}

$\lambda=1$
$$\left[\begin{array}{rr|r}{1} & {-1} & {0} \\ {3} & {-3} & {0}\end{array}\right] \sim\left[\begin{array}{rr|r}{1} & {-1} & {0} \\ {0} & {0} & {0}\end{array}\right]$$
$$x_{2}=t \quad x_{1}-t=0 \quad \Rightarrow \quad x_{1}=t$$
$$\left[\begin{array}{l}{x_{1}} \\ {x_{2}}\end{array}\right]=\left[\begin{array}{l}{1} \\ {1}\end{array}\right]t$$

$\therefore$ eigenvector $\left(\begin{array}{l}{1} \\ {1}\end{array}\right)$ corresponding to eigenvalue $\lambda=1$

$\lambda=-1$
$$\left[\begin{array}{rr|r}{3} & {-1} & {0} \\ {3} & {-1} & {0}\end{array}\right] \sim\left[\begin{array}{rr|r}{3} & {-1} & {0} \\ {0} & {0} & {0}\end{array}\right]$$
$$x_{2}=t \quad 3 x_{1}-t=0 \quad \Rightarrow \quad x_{1}=\frac{t}{3}$$
$$\left[\begin{array}{l}{x_{1}} \\ {x_{2}}\end{array}\right]=\left[\begin{array}{l}{1} \\ {3}\end{array}\right] t$$

$\therefore$ eigeuvector $\left(\begin{array}{l}{1} \\ {3}\end{array}\right)$ correspouding to eigenvalue $\lambda=-1$

$\therefore$ general solution is

\begin{aligned} x&=c_{1} e^{t}\left(\begin{array}{l}{1} \\ {1}\end{array}\right)+c_{2} e^{-t}\left(\begin{array}{l}{1} \\ {3}\end{array}\right) \\ \lambda &=-1 \end{aligned}

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##### Quiz-5 / TUT0801
« on: November 01, 2019, 10:20:25 AM »
Find the general solution of the given differential equation.
$$y^{\prime \prime}-2 y^{\prime}+y=e^{t} /\left(1+t^{2}\right)$$

$$\begin{array}{c}{r^{2}-2 r+1=0} \\ {(r-1)(r-1)=0} \\ {r=1,1} \\ {y_{c}=c_{1} e^{t}+c_{2} t e^{t}}\end{array}$$
\begin{aligned} w &=\left|\begin{array}{cc}{e^{t}} & {t e^{t}} \\ {e^{t}} & {e^{t}+t e^{t}}\end{array}\right| \\ &=e^{2 t}+t e^{2 t}-\left(t e^{2 t}\right) \\ &=e^{2 t} \\ w_{1}&=\left|\begin{array}{cc}{0} & {t e^{t}} \\ {1} & {e^{t}+t e^{t}}\end{array}\right|=-t e^{t}\\ w_{2}&=\left|\begin{array}{cc}{e^{t}} & {0} \\ {e^{t}} & {1}\end{array}\right|=e^{t} \end{aligned}
$$Y(t)=e^{t} \int \frac{-t e^{t} \cdot \frac{e^{t}}{1+t^{2}}}{e^{2} t} d t+t e^{t} \int \frac{e^{t} \cdot \frac{e^{t}}{1+t^{2}}}{e^{2} t}$$
\begin{aligned}-\int \frac{t e^{2 t}}{1+t^{2}} \cdot \frac{1}{e^{2} t} &=-\int \frac{t}{1+t^{2}} d t\qquad\qquad \int \frac{e^{2 t}}{1+t^{2}} \cdot \frac{1}{e^{2 t}}&=\int \frac{1}{1+t^{2}} d t \\ &=-\frac{1}{2} \ln \left(1+t^{2}\right) &=\arctan t\end{aligned}
$$Y(t)=-\frac{1}{2} e^{t} \ln \left(1+t^{2}\right)+t e^{t} \arctan t$$

The general solution is
$$y(t)=c_{1} e^{t}+c_{2} t e^{t}-\frac{1}{2} \ln \left(1+t^{2}\right) e^{t}+t e^{t} \arctan t$$

3
##### Term Test 1 / Re: Problem 4 (afternoon)
« on: October 23, 2019, 08:45:31 AM »
Find the general soluton

(a) $y^{\prime \prime}+2 y^{\prime}+17 y=40 e^{x}+130 \sin (4 x)$

(b) And the solution with $y(0)=0 \quad y^{\prime}(0)=0$

(a)
$$\begin{array}{l}{y^{\prime \prime}+2 y^{\prime}+17 y=0} \\ {\Rightarrow r^{2}+2 r+17=0}\end{array}$$
\begin{aligned} r=\frac{-2 \pm \sqrt{2^{2}-4(1)(17)}}{2(1)} &=\frac{-2 \pm \sqrt{-64}}{2}=\frac{-2 \pm 8 i}{2}=-1 \pm 4 i \\ \lambda=-1 \quad \mu=4 \end{aligned}
$$y_{p}(x)=c_{1} e^{-x} \cos (4 x)+c_{2} e^{-x} \sin (4 x)$$

$$\begin{array}{l}{y^{\prime \prime}+2 y^{\prime}+17 y=40 e^{x}} \\ {y_{1}(x)=A e^{x}, \quad y_{1}^{\prime}(x)=A e^{x}, y_{1}^{\prime \prime}(x)=A e^{x}}\end{array}$$
$$\begin{array}{c}{A e^{x}+2 A e^{x}+17 A e^{x}=40 e^{x}} \\ {20 A e^{x}=40 e^{x}} \\ {A=2}\end{array}$$
$$\therefore y_{1}(x)=2 e^{x}$$
$$\begin{array}{l}{y^{n}+2 y^{\prime}+17 y=130 \sin (4 x)} \\ {y_{2}(x)=B \sin (4 x)+c \cos (4 x)} \\ {y_{2}^{\prime}(x)=4 B \cos (4 x)-4 c \sin (4 x)} \\ {y_{2}^{\prime \prime}(x)=-16 B \sin (4 x)-16 c \cos (4 x)}\end{array}$$

$-16B \sin (4 x)-16 c \cos (4 x)+8B \cos (4 x)-8 c \sin (4 x)+17 B \sin (4 x)+17 c \cos (4 x)=130 \sin (4 x)$

$(-16 B-8 c+17 B) \sin (4 x)+(-16 c+8 B+17 c) \cos (4 x)=130 \sin (4 x)$

$\left\{\begin{array}{l}{B-8 c=130} \\ {8 B+c=0}\end{array}\right.$$\Rightarrow\left\{\begin{array}{l}{B-8 C=130} \\ {64 B+8 C=0}\end{array}\right.$$\begin{array}{rl}{65 B=130} & {B=2} \\ {} & {c=-16}\end{array}$

$$y_{2}(x)=2 \sin (4 x)-16 \cos (4 x)$$

$\therefore$ Tue general solution is
$$y(x)=c_{1} e^{-x} \cos (4 x)+c_{2} e^{-x} \sin (4 x)+2 e^{x}+2 \sin (4 x)-16 \cos (4 x)$$

(b)

\begin{aligned} y^{\prime}(x) =&-c_{1} e^{-x} \cos (4 x)+4 c_{1} e^{-x}(-\sin (4 x))+\left(-c_{2} e^{-x} \sin (4 x)+4 c_{2} e^{-x} \cos (4 x)\right) \\ &+2 e^{x}+8 \cos (4 x)+64 \sin (4 x) \end{aligned}
$$\begin{array}{l}{\text { plug } y(0)=0} \\ {\qquad\Rightarrow c_{1}+2-16=0 \quad C_{1}=14}\end{array}$$
$$\begin{array}{rl}{plug } & {y^{\prime}(0)=0} \\ {\Rightarrow} & {-c_{1}+4 c_{2}+2+8=0}\end{array}$$
$$\begin{array}{c}{-14+4 c_{2}+10=0} \\ {c_{2}=1}\end{array}$$

$\therefore$ The solution with Ivc is
$$y(x)=14 e^{-x} \cos (4 x)+e^{-x} \sin (4 x)+2 e^{x}+2 \sin (4 x)-16 \cos (4 x)$$

4
##### Quiz-4 / TUT0801
« on: October 18, 2019, 02:00:01 PM »
Find the solution for $t^{2} y^{\prime \prime}+t y^{\prime}+y=0$

$\displaystyle t^{2} \frac{d y}{d t^{2}}+\alpha t \frac{d y}{d t}+\beta y=0$ is the general form of
Euler equation.

So , we assume $x=\ln t$

$$\begin{array}{l}{\displaystyle\frac{d y}{d t}=\frac{d y}{d x} \cdot \frac{d x}{d t}=\frac{d y}{d x} \cdot \frac{1}{t}} \\ {\displaystyle\frac{d^{2} y}{d t^{2}}=\frac{1}{t^{2}}\left(\frac{d^{2} y}{d x^{2}}-\frac{d y}{d x}\right)}\end{array}$$

$$\begin{array}{l}{\therefore t^{2}\left[\frac{1}{t^{2}}\left(\frac{d^{2} y}{d x^{2}}-\frac{d y}{d x}\right)\right]+t\left(\frac{1}{t} \frac{d y}{d x}\right)+y=0} \\ {\Rightarrow \frac{d^{2} y}{d x^{2}}+y=0}\end{array}$$

$$\therefore \frac{d y}{d x^{2}}+y=0$$

$$\begin{array}{l}{\Rightarrow r^{2}+1=0} \\ {r=\sqrt{-1}} \\ {r=\pm 1 i} \\ {x=0 \quad \mu=1}\end{array}$$

$$\therefore y(x)=c_{1} e^{0} \cos x+c_{2} e^{0} \sin x$$

Then, substitute $x=\ln t$.

The solution for the equation is

$$y(t)=c_{1} \cos (\ln t)+c_{2} \sin (\ln t).$$

5
##### Quiz-3 / TUT0801
« on: October 11, 2019, 02:19:24 PM »
\noindent The wrouskian w of f and g is $3 e^{4 t}$,\\
and if $f(t)=e^{2 t}$, find $g(t)$.
\begin{aligned} W(f, g) &=3 e^{4 t} \\ W(f, g) &=\left|\begin{array}{cc}{f(t)} & {g(t)} \\ {f^{\prime}(t)} & {g^{\prime}(t)}\end{array}\right| \\ &=\left|\begin{array}{cc}{e^{2 t}} & {g(t)} \\ {2 e^{2 t}} & {g^{\prime}(t)}\end{array}\right| \\ &=e^{2 t} g^{\prime}(t)-2 e^{2 t} g(t) \end{aligned}
$$\begin{array}{l}{\therefore e^{2 t} g^{\prime}(t)-2 e^{2 t} g(t)=3 e^{4 t}} \\ {\Rightarrow g^{\prime}(t)-2 g(t)=3 e^{2 t}}\end{array}$$

We have $p(t)=-2$

$$\mu=e^{\int p(t) d t}=e^{\int-2 d t}=e^{-2 t}$$
$$\therefore e^{-2 t} g^{\prime}(t)-2 e^{-2 t} g(t)=3$$
$$\begin{array}{l} {\Rightarrow \left(e^{-2 t} g(t)\right)^{\prime}=3}\\ {\Rightarrow e^{-2 t} g(t)=\int 3 d t\qquad 3 t+c}\\ {\Rightarrow g(t)=e^{2 t}(3 t+c)=3 e^{2 t} t+c e^{2 t}} \end{array}$$

check :
$$g^{\prime}(t)=3 e^{2 t}+6 e^{2 t} t+2 c e^{2 t}$$
\begin{aligned} W(f, g) &=e^{2 t}\left(3 e^{2 t}+6 e^{2 t} t+2 c e^{2 t}\right)-2 e^{2 t}\left(3 e^{2 t} t+c e^{2 t}\right) \\ &=3 e^{4 t} \end{aligned}

Therefore, $g(t)=3 t e^{2 t}+c e^{2 t}$

6
##### Quiz-2 / TUT0801
« on: October 04, 2019, 02:54:51 PM »
Show that the DE is not exact and become exact mutiplied by the integrating factor. Then find the solution for the DE
$$( \frac { \operatorname { sin } ( y ) } { y } - 2 e ^ { - x } \operatorname { sin } ( x ) ) + ( \frac { \operatorname { cos } ( y ) + 2 e ^ { - x } \operatorname { cos } ( x ) } { y } ) y ^ { \prime } = 0 , \quad \mu ( x , y ) = y e ^ { x };$$

Let $M _ { ( x , y ) } = \frac { \operatorname { sin } y } { y } - 2 e ^ { - x } \operatorname { sin } x \quad N _ { ( x , y ) } = \frac { \operatorname { cos } y + 2 e ^ { - x } \operatorname { cos } x } { y }$

\qquad$M_y = \frac { y \operatorname { cos } y - \operatorname { sin } y } { y ^ { 2 } } \quad N_x = - \frac { 2 } { y } e ^ { - x } \operatorname { cos } x - \frac { 2 } { y } e ^ { - x } \operatorname { sin } x$

Since  $M_y\neq N_x$ , which the given DE is not exact.

So,multipliying thr integrating fact $\mu(x,y)=ye^{x}$
$$\left. \begin{array} { l } { ( e ^ { x } \operatorname { sin } y - 2 y \operatorname { sin } x ) + ( e ^ { x } \operatorname { cos } y + 2 \operatorname { cos } x ) y ^ { \prime } = 0 } \\ { M ^ { \prime } ( x , y ) = e ^ { x } \operatorname { sin } y - 2 y \operatorname { sin } x \quad N ^ { \prime } ( x , y ) = e ^ { x } \operatorname { cos } y + 2 \operatorname { cos } x } \\ { M _ { y } ^ { \prime } = e ^ { x } \operatorname { cos } y - 2 \operatorname { sin } x \quad N ^ { \prime }_x = e ^ { x } \operatorname { cos } y - 2 \operatorname { sin } x } \end{array} \right.$$

Since $M^{\prime}y=N^{\prime}x$, which the DE is exact now.

$$\left. \begin{array}{l}{\exists~\psi _ { ( x , y ) } s.t. \psi_y=N^{\prime}=e^x\cos y+2\cos x}\\{ \psi _ { ( x , y ) } = \int N ^ { \prime } d y = \int e ^ { x } \operatorname { cos } y + 2 \operatorname { cos } x d y }\\{ \qquad~= e ^ { x } \operatorname { sin } y + 2 y \operatorname { cos } x + h ( x ) }\\{ \psi _ { x } = \operatorname { sin } y e ^ { x } - 2 y \operatorname { sin } x \cdot \operatorname { th } ^ { \prime } ( x ) = M ^ { \prime } }\\{\therefore h^{\prime}(x)=0\quad h(x)=c}\\{\therefore \psi(x,y)=e^x\sin y+2y\cos x}\end{array} \right.$$

Therefore, the solution of the DE is
$$e^x\sin y+2y\cos y=c$$

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