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

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##### 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 / 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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