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### Topics - yangyiq5

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##### Quiz-3 / QUIZ3 TUT5301
« on: October 11, 2019, 02:00:10 PM »
Question : find the Wronskian of the given pair of function

$$X ;Xe^{x}$$

w=$$\left[ \begin{matrix} x & xe^{x} \\ 1 & e^{x}+xe^{x} \end{matrix} \right] \tag{3}$$
$$=x(1+x)e^{x}-xe^{x}$$
$$=x^{2}e^{x}$$

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##### Quiz-2 / quiz 2 TUT 5103
« on: October 04, 2019, 02:01:32 PM »
Question:
$x^{2}y^{3}+x(1+y^{2}){y}'=0$
$\mu \left ( x,y \right )=\frac{1}{xy^{3}}$
Solution:
$M = x^{2}y^{3}$
$My = 3x^{2}y^{2}$
$N= x(1+y^{2})$
$Nx = 1+y^{2}$
$since My \neq Nx$
$this equation is not exact$
$multiple \mu (x,y) at both sides$
$x +\frac{1+y^{2}}{y^{3}}{y}' = 0$
$now My = Nx = 0, this equation is exact$
$\exists \varphi (x,y) s.t. \varphi x= M and\varphi y = N$
$\varphi = \int x dx = \frac{1}{2}x^{2} + h(y)$
$\varphi y = h'(y) =\frac{1+y^{2}}{y^{3}} = \frac{1}{y^{3}}+\frac{1}{y}$
$h(y) = -1/2y^{-2} + ln\left | y \right | + c$
$so\varphi = \frac{1}{2}x^{2} -1/2y^{-2} + ln\left | y \right | + c$
$\frac{1}{2}x^{2} -1/2y^{-2} + ln\left | y \right | = c$

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##### Quiz-1 / tut5103 quiz1
« on: September 27, 2019, 02:36:46 PM »
${y}' = \frac{x^{2}+3x^2}{2xy}=\frac{y}{2x}+\frac{3y}{2x} = \frac{1}{2}\left ( \frac{y}{x} \right )^{-1}+\frac{3}{2}\left ( \frac{y}{x} \right )$

$this equation is homogenous$

$u = \frac{y}{x}$
$y = ux$
${y}'={u}'x+u$
${u}'x+u = \frac{1}{2}u^{-1}+\frac{3}{2}u$
seperable
$\left ( \frac{2u}{1+u^{2}} \right )du=\frac{1}{x}dx$
$ln\left |1+u^{2} \right | = ln\left | x \right |$
$1+u^{2}=cx$
$1+\left ( \frac{y}{x} \right )^{2}=cx$
$x^2+y^2= cx^3$

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##### Quiz-1 / tut5103 quiz1
« on: September 27, 2019, 02:03:56 PM »
y’ = (x^2+3x^2)/(2xy)

sol:
y’ = (x^2+3x^2)/(2xy) = x/2y + 3y/2x = 1/2(y/x) ^ (-1) +3/2(y/x)
this equation is homogenous
u = y/x   y = ux
y’ = du/dx*x +u
du/dx*x +u = 1/2u^ (-1) +3/2u
separable
(2u/(1+u^2)) du = 1/x dx
Integral both sides
ln|1+u^2| = ln|x|
1+u^2 = cx
y^2/x^2 +1 = cx
y^2+x^2-cx^3=0

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