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### Messages - Yuefan Wang

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##### Quiz-6 / Lec5101
« on: November 15, 2019, 02:00:01 PM »
Find the general solution
$$\begin{array}{c}{x^{\prime}=\left(\begin{array}{cc}{4} & {-3} \\ {8} & {-6}\end{array}\right) x} \\ {\left(\begin{array}{cc}{4-\lambda} & {-3} \\ {8} & {-6-x}\end{array}\right)\left(\begin{array}{l}{x_{1}} \\ {x_{2}}\end{array}\right)=\left(\begin{array}{c}{0} \\ {0}\end{array}\right)} \\ {\left(\begin{array}{cc}{4-\lambda} & {-3} \\ {8} & {-6-\lambda}\end{array}\right)=0} \\ {r(r+2)=0} \\ {r_{1}=0, \quad r_{2}=-2}\end{array}$$

when $r_{1}=0$
$$\begin{array}{c}{\left(\begin{array}{cc}{4} & {-3} \\ {8} & {-6}\end{array}\right)\left(\begin{array}{l}{x_{1}} \\ {x_{2}}\end{array}\right)=\left(\begin{array}{l}{0} \\ {0}\end{array}\right)} \\ {x_{1}=\left(\begin{array}{l}{3} \\ {4}\end{array}\right)}\end{array}$$
$$\begin{array}{c}{\left(\begin{array}{cc}{6} & {-3} \\ {8} & {-4}\end{array}\right)\left(\begin{array}{l}{x_{1}} \\ {x_{2}}\end{array}\right)=\left(\begin{array}{c}{0} \\ {0}\end{array}\right)} \\ {\lambda_{2}=\left(\begin{array}{l}{1} \\ {2}\end{array}\right)}\end{array}$$
$$x^{\prime}(t)=\left(\begin{array}{l}{3} \\ {4}\end{array}\right) \quad x^{2}(t)=\left(\begin{array}{l}{1} \\ {2}\end{array}\right)t^{-2}$$

general solution
$$x=c_{1}\left(\begin{array}{l}{3} \\ {4}\end{array}\right)+c_{2}\left(\begin{array}{l}{1} \\ {2}\end{array}\right) t^{-2}$$

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##### Quiz-5 / LEC5101 Quiz 5
« on: November 01, 2019, 02:00:00 PM »
Find the general solution of $y^{\prime \prime}+4 y^{\prime}+4 y=t^{-2} e^{-2 t}, t>0$

$$\begin{array}{l}{\text { homo: } y^{\prime \prime}+4 y^{\prime}+4 y=0} \\ {\qquad \begin{array}{c}{(r+2)^{2}=0} \\ {r_{1}=-2, r_{2}=-2} \\ {\therefore y_{c}(t)=c_{1} e^{-2 t}+c_{2} t e^{-2 t}}\end{array}}\end{array}$$
\begin{array}{l}{\text { non-homo: }} \\ {\qquad \begin{aligned} w\left(y_{1}, y_{2}\right)=\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}{e^{-2 t}} & {te^{-2 t}} \\ {-2 e^{-2 t}} & {-2 t^{-2t}+e^{-2 t} }\end{array}\right| \\ &=-2 t e^{-4 t}+e^{-4 t}+t e^{-4 t}=e^{-4t} \end{aligned}}\end{array}
$$\begin{array}{l}{w_{1}\left(y_{1}, y_{2}\right)=\left|\begin{array}{cc}{0} & {e^{-2 t}} \\ {1} & {-2 t e^{-2 t}+e^{-2 t} |}\end{array}\right|=-t e^{-2 t}} \\ {w_{2}\left(y_{1}, y_{2}\right)=\left|\begin{array}{cc}{e^{-2 t}} & {0} \\ {-2 e^{-2 t}} & {1}\end{array}\right|=e^{-2 t}}\end{array}$$
\begin{aligned} Y(t)&=e^{-2 t} \int \frac{\left(-s e^{-2 s}\right) \cdot\left(s^{-2} e^{-2 s}\right)}{e^{-2 s}} d s+t e^{-2 t} \int \frac{\left(e^{-2 s}\right) \cdot\left(s^{-2} e^{-2 s}\right)}{e^{-2 s}} d s\\ &=e^{-2 t} \int -t^{-1} d t+t e^{-2 t} \int t^{-2} d t\\&=-e^{-2 t} \ln t-e^{-2 t}\\ y(t) &=y_{c}(t)+Y(t) \\ &=c_{1} e^{-2 t}+c_{2} t e^{-2 t}-e^{-2 t} \ln t-e^{-2 t} \\ &=c_{1} e^{-2 t}+c_{2} t e^{-2 t}-e^{-2 t} \ln t \end{aligned}

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##### Term Test 1 / Re: Problem 3 (afternoon)
« on: October 23, 2019, 07:52:33 AM »
Question:

find a general solution of $y^{\prime \prime}-5 y^{\prime}+6 y=52 \cos (2 x)$
and satisfy $y(0)=0, y^{\prime}(0)=0$
$$\begin{array}{c}{\because y^{\prime \prime}-5 y^{\prime}+b y=0} \\ {r^{2}-5 r+b=0} \\ {(r-2)(r-3)=0}\end{array}$$
$$\begin{array}{l}{\therefore r_{1}=2, r_{2}=3} \\ {\therefore y_{c}(t)=c_{1} e^{2 t}+c_{2} e^{3 t}}\end{array}$$
$$\begin{array}{l}{\because y^{\prime \prime}-5 y^{\prime}+6 y=52 \cos (2 x)} \\ {\therefore y_{p}(t)=A \cos (2 x)+B \sin (2 x)} \\ {y^{\prime} p(t)=-2 A \sin (2 x)+2 B \cos (2 x)} \\ {y^{\prime \prime} p(t)=-4 A \cos (2 x)-4B \sin (2 x)}\end{array}$$
$$\begin{array}{rl}{-4 A \cos (2 x)-4 B \sin (2 x)+10 A \sin (2 t)-10} & {B \cos (2 x)+6 A \cos (2 x)+6 B \sin (2 x)} \\ {} & {=52 \cos (2 x)}\end{array}$$
$$(-4 A-10 B+6 A) \cos (2 x)+(-4 B+10 A+6 B) \sin (2 x)=52 \cos (2 x)$$
$$\left\{\begin{array}{l}{2 A-10 B=52} \\ {2 B+10 A=0}\end{array} \quad \therefore\left\{\begin{array}{l}{A=1} \\ {B=-5}\end{array}\right.\right.$$
$$\begin{array}{l}{\therefore y_p(x)=\cos (2 x)-5 \sin (2 x)} \\ {\therefore y(x)=y_c(x)+y_p(x)=c_{1} e^{2 t}+c_{2} e^{3 t}+\cos (2 x)-5 \sin (2 x)}\\{\therefore y^{\prime}(x)=2 c_1 e^{2 t}+3 c_{2} e^{3 t}-2 \sin (2 x)-10 \cos (2 x)} \\ {\therefore y(0)=0, y^{\prime}(0)=0}\\{\therefore\left\{\begin{array}{l}{c_{1}+c_{2}=-1} \\ {2 c_{1}+3 c_{2}=10}\end{array}\therefore \left\{\begin{array}{l}{c_{1}=-13} \\ {c_{2}=12}\end{array}\right.\right.}\\{\therefore y(x)=-13 e^{2 t}+12 e^{3 t}+\cos (2 x)-5 \sin (2 x)} \end{array}$$

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##### Quiz-4 / QUIZ4 TUT5103
« on: October 18, 2019, 02:00:11 PM »
$Question: y^{\prime \prime}+2 y^{\prime}=3+4 \sin (2 t)$
$$\begin{array}{l}{r^{2}+2 r=0} \\ {r(r+2)=0} \\ {r_{1}=0, r_{2}=-2}\end{array}$$
$$y_{c}(t)=c_{1}+c_{2} e^{-2 t}$$
$$y^{\prime \prime}+2 y^{\prime}=3 \quad y_p(t)=3 a+b \quad y^{\prime} _p(t)=a \quad y^{\prime \prime}_p(t)=0$$
$$\begin{array}{r}{2 a=3} \\ {a=\frac{3}{2}}\end{array}$$
$$\therefore y_{p(t)}=\frac{3}{2} t$$
$$\begin{array}{l}{y^{\prime \prime}+2 y^{\prime}=4 \sin (2 t)} \\ {y_{P}(t)=A \cos (2 t)+B \sin (2 t)} \\ {y^{\prime} p(t)=-2 A \sin (2 t)+2 B \cos (2 t)} \\ {y^{\prime \prime} p(t)=-4 A \cos (2 t)-4 B \sin (x t)}\end{array}$$
$$-4 A \cos (2 t)-4 B \sin (2 t)-4 A \sin (2 t)+4 B \cos (2 t)=4 \sin 2 t$$
$$\left\{\begin{array}{ll}{-4 A+4 B=0}\\{-4 B-4 A=4}\end{array}\right.\qquad{\left\{\begin{array}{l}{A=-\frac{1}{2}} \\ {B=-\frac{1}{2}}\end{array}\right.}$$
$$y_{p}(t)=\frac{1}{2} \cos (2 t)-\frac{1}{2} \sin (2 t)$$
$$\therefore y(t)=y_{c}(t)+y_{p}(t)=c_{1}+c_{2} e^{-2 t}+\frac{3}{2} t-\frac{1}{2} \cos (2 t)-\frac{1}{2} \sin (2 t)$$

5
##### Quiz-3 / TUT5103
« on: October 11, 2019, 02:00:35 PM »
Find the Wronskian of two solutions of the given differential equation without solving the
equation.
$$\left(1-x^{2}\right) y^{\prime \prime}-2 x y^{\prime}+\alpha(\alpha+1) y=0$$

$$\begin{array}{l}{y^{\prime \prime}-\displaystyle\frac{2 x}{1-x^{2}} y^{\prime}+\frac{a(a+1)}{1-x^{2}} y=0} \\ {P(t)=-\displaystyle\frac{2 x}{1-x^{2}}}\end{array}$$

\begin{aligned} w=c e^{\int-p(t) d t} &=c e^{\displaystyle\int {\frac{2 x}{1-x^{2}} d x}} \qquad u=1-x^{2}, d u=-2 x d x\\ &=\operatorname{ce}^{\displaystyle \int-\frac{1}{u}d u} \\ &=\operatorname{ce}^{-\ln(u)}\\&=\operatorname{ce}^{-\ln \left(1-x^{2}\right)} \\ &=c{(\left.\ln \left(1-x^{2}\right)\right)^{-1}} \\ &=\frac{c}{1-x^{2}} \end{aligned}

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##### Quiz-2 / Re: TUT5103
« on: October 04, 2019, 02:05:32 PM »
$${ ( x + 2 ) \operatorname { sin } ( y ) \ +\operatorname { cos } ( y )'(y)=0 }\mu=x e ^ { x }$$
$$\left. \begin{array} { l } { M = ( x + 2 ) \operatorname { sin } ( y ) \quad N = x \operatorname { cos } ( y ) } \\ { M _ { y } = ( x + 2 ) \operatorname { cos } ( y ) \quad N x = \operatorname { cos } y } \\ { \because M y \neq N x \quad \therefore \text { hot exact } } \\ { x e ^ { x } ( x + 2 ) \operatorname { sin } ( y ) + x ^ { 2 } e ^ { x } \operatorname { cos } ( y ) y }\\{\therefore \text{now the function is exact}} \end{array} \right.$$

$$\left. \begin{array} { l }{\exists~\varphi(x,y)~s.t.\quad\varphi = \int M d x }\\{ = \int x e ^ { x } ( x + 2 ) \operatorname { sin } ( y ) d x }\\{ = \int ( x ^ { 2 } e ^ { x } + 2 x e ^ { x } ) \operatorname { sin } ( y ) }\\{ = ( x ^ { 2 } e ^ { x } ) \operatorname { sin } ( y ) + h ( y ) }\\{\varphi_y = ( x ^ { 2 } e ^ { x } ) \operatorname { cos } ( y ) + h ^ { \prime } ( y ) + h ^ { \prime } ( y ) + h ^ { \prime } ( y ) }\\{h^{\prime}(y)=0}\\{h(y)=\text{constant}}\\{\varphi=(x^2e^x)\sin y=C.}\end{array} \right.$$

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##### Quiz-2 / TUT5103
« on: October 04, 2019, 02:01:18 PM »
$$\left. \begin{array} { l } { M = ( x + 2 ) \operatorname { sin } ( y ) \quad N = x \operatorname { cos } ( y ) } \\ { M _ { y } = ( x + 2 ) \operatorname { cos } ( y ) \quad N x = \operatorname { cos } y } \\ { \because M y \neq N x \quad \therefore \text { not exact } } \\ { x e ^ { x } ( x + 2 ) \operatorname { sin } ( y ) + x ^ { 2 } e ^ { x } \operatorname { cos } ( y ) y }\\{\therefore \text{now the function is exact}} \end{array} \right.$$

$$\left. \begin{array} { l }{\exists~\varphi(x,y)~s.t.\quad\varphi = \int M d x }\\{ = \int x e ^ { x } ( x + 2 ) \operatorname { sin } ( y ) d x }\\{ = \int ( x ^ { 2 } e ^ { x } + 2 x e ^ { x } ) \operatorname { sin } ( y ) }\\{ = ( x ^ { 2 } e ^ { x } ) \operatorname { sin } ( y ) + h ( y ) }\\{\varphi_y = ( x ^ { 2 } e ^ { x } ) \operatorname { cos } ( y ) + h ^ { \prime } ( y ) + h ^ { \prime } ( y ) + h ^ { \prime } ( y ) }\\{h^{\prime}(y)=0}\\{h(y)=\text{constant}}\\{\varphi=(x^2e^x)\sin y=C.}\end{array} \right.$$

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