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### Topics - Yin Jiekai

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##### Quiz-3 / MAT244 TUT5103 quiz3
« on: October 11, 2019, 02:00:00 PM »
Q: Find the general solution of the given differential equation.
$$y^{\prime\prime}-2y^{\prime}-2y=0$$
A: We assume that $y=e^rt$, and it follows that $r$ must be a root of the characteristic equation
$$r^2 - 2r -2 = 0$$
Then $r= \frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{-(-2)\pm\sqrt{(-2)^2-4\times2}}{2\times1}= 1\pm\sqrt{3}$
Then we find that $r_1 = 1+\sqrt{3}$ and $r_2 = 1-\sqrt{3}$

Since the general solution has the form of$$y = c_1e^{r_1t}+c_2e^{r_2t}$$
Therefore, the general solution of the given differential equation is $$y = c_1e^{(1+\sqrt{3})t}+c_2e^{(1-\sqrt{3})t}$$

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##### Quiz-2 / MAT244 TUT5103 quiz2
« on: October 04, 2019, 02:00:01 PM »
$$( \frac { \operatorname { sin } ( y ) } { y } - 2 e ^ { - x } \operatorname { sin } ( x ) ) + ( \frac { \operatorname { cos } ( y ) + 2 e ^ { - x } \operatorname { cos } ( x ) } { y } ) y ^ { \prime } = 0 , \quad \mu ( x , y ) = y e ^ { x };$$

mutiply both sides by $\mu$:

Then $( \operatorname { sin } ( y ) e ^ { x } - 2 y \operatorname { sin } ( x ) ) + ( \operatorname { cos } ( y ) e ^ { x } + 2 \operatorname { cos } ( x ) ) y ^ { \prime } = 0$

Let $M(x,y)=\sin(y)e^x-2y\sin(x)$ and $N(x,y)=\cos(y)e^x+2\cos(x)$

Then $M_y=\cos(y)e^x-2\sin(x)$ $\quad N_x=\cos(y)e^x-\sin(x)$

Then the equation is exact.

Therefore, there exists a function $\varphi(x,y)$ such that
$$\left. \begin{array} { l } { \varphi _ { x } ( x , y ) = \operatorname { sin } ( y ) e ^ { x } - 2 y \operatorname { sin } ( x ) = M } \\ { \varphi _ { y } ( x , y ) = \operatorname { cos } ( y ) e ^ { x } + 2 \operatorname { cos } ( x ) = N } \end{array} \right.$$

Then $$\left. \begin{array} { l } { \varphi ( x , y ) = \int N d y = \operatorname { sin } ( y ) e ^ { x } + h ( x ) } \\ { \varphi _ { x } ( x , y ) = \operatorname { sin } ( y ) e ^ { x } + h ^ { \prime } ( x ) } \end{array} \right.$$

Then $$\left. \begin{array}{l}{ h ^ { \prime } ( x ) = - 2 y \operatorname { sin } ( x ) }\\{ h ( x ) = \int - 2 y \operatorname { sin } ( x ) }\\{ \quad\,\quad= 2 y \operatorname { cos } ( x ) + c }\end{array} \right.$$

Therefore $\sin(y)e^x+2y\cos(x)=c$

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##### Quiz-1 / MAT244 TUT5103 quiz1
« on: September 27, 2019, 01:59:47 PM »
\newpage
\noindent Find the general solution of the given differential equation, and use it to determine how solutions
behave as $t \rightarrow \infty$
$$t y^{\prime}-y=t^{2} e^{-t}, \quad t>0$$
$$\begin{array}{l}{y^{\prime}-\frac{1}{t} y=t e^{-t}} \\ {p(t)=-\frac{1}{t}} \\ {\mu=e^{\int p(t) d t}=e^{-\ln (t)}=t^{-1}}\end{array}$$
multiply both sides by $\mu$
\begin{aligned} t^{-1} y^{\prime}\cdot t^{-2} y &=e^{-t} \\ \frac{d}{d x}\left(t^{-1} y\right) &=e^{-t} \\ t^{-1} y &=\int e^{-t} \\ t^{-1} y &=-e^{-t}+c \\ y &=-t e^{-t} + c t \end{aligned}

when $t \rightarrow \infty$

case 1: $$\begin{array}{l}{C=0;} \\ {\text { by } L^{\prime} \text { Hopital rule; }} \\ {y \rightarrow 0}\end{array}$$
case 2:$$\begin{array}{l}{C>0;} \\ {y \rightarrow+\infty}\end{array}$$
case 3:$$\begin{array}{l}{C<0;} \\ {y \rightarrow-\infty}\end{array}$$

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