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Topics - Yingyingz

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1
Quiz-5 / LEC5101 Quiz 5
« on: November 01, 2019, 02:00:00 PM »
Verify that the given functions y 1and y 2 satisfy the corresponding homogeneous equation;
then find a particular solution of the given nonhomogeneous equation.
$$
\begin{array}{l}{(1-t) y^{\prime \prime}+t y^{\prime}-y=2(t-1)^{2} e^{-t}, 0<t<1} \\ {y_{1}(t)=e^{t}, y_{2}(t)=t}\end{array}
$$

$$
y^{\prime \prime}+\frac{t}{1-t} y^{\prime}-\frac{1}{1-t} y=-2(t-1) e^{-t}
$$
$$
\begin{array}{l}{w=\left|\begin{array}{ll}{e^{t}} & {t} \\ {e^{t}} & {1}\end{array}\right|=e^{t}-t e^{t}} \\ {w_{1}=\left|\begin{array}{ll}{0} & {t} \\ {1} & {1}\end{array}\right|=-t \quad w_{2}=\left|\begin{array}{ll}{e^{t}} & {0} \\ {e^{t}} & {1}\end{array}\right|=e^{\tau}}\end{array}
$$
$$
\begin{aligned} Y(t) &=e^{t} \int \frac{-t \cdot\left(-2(t-1) e^{-t}\right)}{e^{t}-t e^{t}} d t+t \int \frac{e^{t}\left(-2(t-1) e^{-t}\right)}{e^{t}-t e^{t}} d t \\ &=e^{t} \int \frac{-t \cdot\left(t^{2}(t-1) e^{-t}\right)}{e^{t}(1-t)}+t \int \frac{e^{t}\left(t+2(t-1) e^{-t}\right)}{e^{t}(1-t)} d t \\ &=e^{t} \int-2 t e^{-2 t} d t+\int 2 e^{-t} d t \\ &=-2 e^{t} \int t e^{-2 t} d t-2 t e^{-t} \end{aligned}
$$
$$
\begin{array}{ll}{u=t} & {d v=e^{-2 t}} \\ {d u=d t} & {v=-\frac{1}{2} e^{-2 t}}\end{array}
$$
$$
\begin{array}{l}{\displaystyle =-2 e^{t}\left(-\frac{1}{2} e^{-2 t} \cdot t+\int \frac{1}{2} e^{-2 t} d t\right)+t \int 2 e^{-t} d t} \\ {\displaystyle =-2 e^{t}\left(-\frac{1}{2} e^{-2 t} t-\frac{1}{4} e^{-2 t}\right)-2 t e^{-t}} \\ {\displaystyle =t e^{-t}+\frac{1}{2} e^{-t}-2 t e^{-t}} \\ {\displaystyle =\left(\frac{1}{2}-t\right) e^{-t}}\end{array}
$$

$\therefore$ the particular solution of the given non-homogeneous equation is
$$
Y(t)=\left(\frac{1}{2}-t\right) e^{-t}
$$

2
Quiz-4 / tUT5103 Quiz4
« on: October 18, 2019, 02:00:01 PM »
$
\text{14.}\quad y^{\prime \prime}+4 y^{\prime}+4 y=0, y(-1)=2, y^{\prime}(-1)=1
$

The characteristic equation is,
$$
\begin{aligned} r^{2}+4 r+4 &=0 \\(r+2)^{2} &=0 \\ r_{1}=r_{2} &=-2 \end{aligned}
$$

since it has repeated roots, the general solution is:
$$
\begin{array}{l}{y(t)=c_{1} e^{-2 t}+c_{2} t e^{-2 t}} \\ {y^{\prime}(t)=-2 c_{1} e^{-2 t}+c_{2} e^{-2 t}-2 c_{2} t e^{-2 t}}\end{array}
$$

$\operatorname{plug}$ in $y(-1)=2, \quad y^{\prime}(-1)=1$

we get $$\left\{\begin{array}{l}{c_{1} e^{2}-c_{2} e^{2}=2} \\ {-2 c_{1} e^{2}+c_{2} e^{2}+2 c_{2} e^{2}=1}\end{array}\right.$$

$$
\therefore\left\{\begin{array}{l}{c_{1}=7 e^{-2}} \\ {c_{2}=5 e^{-2}}\end{array}\right.
$$

substitute $C_{2}$ and $C_{2}$ into $y(t)$

$$
=7 e^{-2} e^{-2 t}+5 e^{-2} t e^{-2 t}
$$

$\therefore$ The general solution is $y(t)=7 e^{-2(t+1)}+5 t e^{-2(t+1)}$

3
Quiz-3 / TUT5103 Quiz3
« on: October 11, 2019, 02:00:00 PM »
Find the solution of the given initial problem.
$$
y^{\prime \prime}+y^{\prime}-2 y=0, y(0)=1, y^{\prime}(0)=1
$$

We assume $y=e^{r t}$, then $r$ must be a root of the characteristic equation:
$$
\begin{array}{l}{r^{2}+r-2=0} \\ \therefore (r-1)(r+2)=0 \\ {r_{1}=1 \quad r_{2}=-2}\end{array}
$$

$\therefore$ the general solution is $y=c_{1} e^{t}+c_{2} e^{-2 t}$
$$
\begin{array}{l}{y^{\prime}=c_{1} e^{t}-2 c_{2} e^{-2 t}} \\ {\operatorname{plug} \text { in } y(0)=1, y^{\prime}(0)=1}\end{array}
$$
$$
\left\{\begin{array}{l}{C_{1}+C_{2}=1\qquad {{\small 1}}} \\ {C_{1}-2 C_{2}=1\quad~~ {{\small 2}}}\end{array}\right.
$$

${{\small 1}}-{{\small 2}}$ , we get
$$
\begin{array}{r}{3 C_{2}=0} \\ {C_{2}=0}\end{array}
$$
$$
\therefore C_{1}=1
$$

$\therefore$ the solution is $y=e^{t}$

4
Quiz-2 / TUT5103 Quiz2
« on: October 04, 2019, 02:00:04 PM »
7.Find an integrating factor and solve the given equation
$$
\underbrace{1}_{M}+(\underbrace{\frac{x}{y}-\sin(y)}_{N})y^{\prime}=0.
$$


$M_y=0$$N_x=\frac{1}{y}$

$\because$ $M_y\neq N_x$

$\therefore$ not  exact

$R_1=\underbrace{M_y-N_x}_{M'}=\underbrace{0-\frac{1}{y}}_{N'}=-\frac{1}{y}$

$$\left. \begin{array} { l } { \mu = e ^ { - \int R_1 d r } = e ^ { - \int \frac { 1 } { y } } = e ^ { \operatorname { ln } | y | } = y } \\ { \therefore y + ( x - y \operatorname { sin } ( y ) ) y ^ { \prime } = 0 \quad \because M'_y = N'_x = 1 \quad \therefore \text{exact function}} \\{\therefore \exists\quad\varphi(x,y)\qquad s.t \,\varphi_x=M'=y}\\{ \psi = \varphi y d x = y x + h ( y ) }\\{ \varphi y = x + h ^ { \prime } ( y ) = x - y \operatorname { sin } ( y ) }\\{ h ^ { \prime } ( y ) = - y \operatorname { sin } ( y ) }\\{ h ( y ) = - \int y \operatorname { sin } ( y ) }\\{ = y \operatorname { cos } ( y ) - \operatorname { sin } ( y ) }\end{array} \right.$$
$$
\varphi=yx+y\cos(y)-\sin(y)=c
$$

5
Quiz-1 / TUT5103 Quiz1
« on: September 27, 2019, 02:00:07 PM »
Find the general solution of the equation and prove it is a homogereous equation
$$
\because\left(x^{2}+3 y^{2}\right) d x=(2 x y) d y
$$
$$
\begin{array}{l}{\therefore\text { the equation is homogeneous. }} \\ {\quad y^{\prime}=\frac{1+3 \frac{y^{2}}{x^{2}}}{2 \frac{y}{x}}} \\ {\text { let } u=\frac{y}{x}}\end{array}
$$
$$
\begin{array}{l}{\therefore y=u x} \\ {\therefore \frac{d y}{d x}=\frac{d(u x)}{d x}=\frac{d u}{d x} x+u} \\ {\therefore \frac{d u}{d x} x=\frac{1+3 u^{2}}{2 u}-\frac{2 u^{2}}{2 u}=\frac{1+u^{2}}{2 u}}\end{array}
$$
$$
\begin{aligned} \int \frac{2 u}{1+u^{2}} d u &=\int \frac{1}{x} d x \\ \ln \left|1+u^{2}\right| &=\ln |x|+C \\ e^{\ln \left|1+u^{2}\right|} &=\ln |x|+c \\ e^{\ln \left|1+u^{2}\right|} &=e^{\ln |x|+c} \\ e^{\ln | | t+u^{2} |} &=e^{\ln |x|} \cdot e^{c} \\ 1+u^{2} &=c x \\\left(\frac{y}{x}\right)^{2} &=c x-1 \\ \frac{y^{2}}{x^{2}} &=c x-1 \\ y^{2} &=c x^{3}-x^{2} \end{aligned}
$$

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