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##### 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
##### Term Test 1 / Re: Problem 1 (afternoon)
« on: October 23, 2019, 07:45:44 AM »
(a)$$\begin{array}{l}{M=-y^{2} \sin (x y)} \\ {M y=-\left(2 y \sin (x y)+y^{2} \cos (x y) \cdot x\right)} \\ {N=-x y \sin (x y)+2 \cos (x y)+3 y} \\ {N x=-(y \sin (x y)+x y \cos (x y) \cdot y)-2 \sin (x y) y}\end{array}$$
$$M y \neq N_x$$
$$R_{1}=\frac{M y-N x}{M}=\frac{-y \sin (x y)+2 \sin (x y) y}{-y^{2} \sin (x y)}=-\frac{1}{y}$$
$$\therefore \mu=e^{-\int-R_{1} d y}=e^{\int \frac{1}{y} d y}=y$$

Multiply $\mu$ on both sides,
$$-y^{3} \sin (x y)+\left(-x y^{2} \sin (x y)+2 y \cos (x y)+3 y^{2}\right) y^{\prime}=0$$
$$\begin{array}{l}{M^{\prime}=-y^{3} \sin (x y)} \\ {M^{\prime} y=-\left(3 y^{2} \sin (x y)+y^{3} \cos (x y) x\right)} \\ {N^{\prime}=-x y^{2} \sin (x y)+2 y \cos (x y)+3 y^{2}} \\ {N^{\prime} x=-\left(y^{2} \sin (x y)+x y^{2} \cos (x y) y\right)-2 y \sin (x y) y} \\ {\quad=-3 y^{2} \sin (x y)-x y^{3} \cos (x y)}\end{array}$$
$$\therefore M^{\prime} y=N^{\prime} x$$
$\therefore \exists \varphi(x, y), s, t$
$$\begin{array}{l}{\varphi_{x}=M^{\prime}=-y^{3} \sin (x y)} \\ {\varphi=\int M^{\prime} d x=y^{2} \cos (x y)+h(y)} \\ {\varphi_{y}=2 y \cos (x y)-y^{2} \sin (x y) x+h^{\prime}(y)}\end{array}$$
$$\because N^{\prime}=\varphi y=-x y^{2} \sin (x y)+2 y \cos (x y)+3 y^{2}$$
$$\begin{array}{l}{\therefore h^{\prime}(y)=3 y^{2}} \\ {h(y)=\int h^{\prime}(y) d y=y^{3}+C}\end{array}$$
$$\therefore \varphi=y^{2} \cos (x y)+y^{3}=c$$
(b) $$y\left(\frac{\pi}{3}\right)=1$$
\begin{aligned} \therefore \quad x &=\frac{\pi}{3}, \quad y=1 \\ c &=y^{2} \cos (x y)+y^{3} \\ &=\frac{1}{2}+1 \\ &=\frac{3}{2} \end{aligned}
$$\therefore \varphi=y^{2} \cos (x y)+y^{3}=\frac{3}{2}$$

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##### 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 $y(t)=7 e^{-2(t+1)}+5 t e^{-2(t+1)}$

4
##### 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 $y=e^{t}$

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

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