Show Posts

This section allows you to view all posts made by this member. Note that you can only see posts made in areas you currently have access to.

Messages - yangyiq5

Pages: [1]

Term Test 1 / Re: Problem 3 (afternoon)

« on: October 23, 2019, 07:30:41 AM »

Rewrite the question:
y’’ -5y’ + 6y = 52 cos(2x)
1. $$r^{2} – 5r +6 = 0$$
$$(r-2)(r-3) = 0$$
$$r_{1} = 2$$
$$r_{2} = 3$$
$$y_{c}(t) = C_{1}e^{2t} + C_{2}e^{3t}$$
2. y’’ -5y’ + 6y = 52 cos(2x)
$$y_{p}(t) = A cos(2x) +B sin(2x)$$
$$y_{p}’(t) = -2Asin(2x) +2B cos(2x)$$
$$y_{p}’’(t) = -4Acos(2x) -4Bsin(2x)$$
$$(-4A-10B+6A)cos(2x)+(-4B+10A+6B)sin(2x) = (2A -10B) cos(2x)+(2B+10A)sin(2x)$$
$$2A-10B = 52$$
$$2B+10A = 0$$
$$A =1$$
$$B = -5$$
$$y_{p}(t) = cos(2x) – 5sin(2x)$$
$$y = C_{1}e^{2t} + C_{2}e^{3t}+ cos(2x) – 5sin(2x)$$
$$y’ = 2C_{1}e^{2t} +3C_{2}e^{3t}-10 cos(2x) – 2sin(2x)$$
$$when x=0,y = 0, y’=0$$
$$C_{1} = -13$$
$$C_{2} = 12$$
$$y = -13e^{2t} + 12e^{3t}+ cos(2x) – 5sin(2x)$$

Term Test 1 / Re: Problem 1 (afternoon)

« on: October 23, 2019, 07:30:09 AM »

$$M_{y} = -2ycos(xy) + xy^{2}sin(xy)$$
$$N_{x} = -ycos(xy) + xy^{2}sin(xy) - 2ycos(xy)$$
$$R_{2} = \frac{M_{y}-N_{x}}{M} = \frac{-2ycos(xy) + xy^{2}sin(xy)-(ycos(xy) + xy^{2}sin(xy) -2ycos(xy))}{-y^{2}cos(xy)}$$
$$R_{2} = -\frac{1}{y}$$
$$\mu = e^-{\int -\frac{1}{y}} = y$$
$$multiple at both sides$$
$$-y^{3}cos(xy)+(-xy^{2}cos(xy)-2ysin(xy)+3y^{2})y’ = 0$$
$$\exists \varphi$$
$$\varphi _{x} = M $$
$$\varphi _{y} =N$$
$$\varphi = \int -y^{3}cos(xy) dx = -y^{2}sin(xy) + h(y)$$
$$\varphi_{y} = -2ysin(xy)-xy^{2}cos(xy) + h’(y)= -xy^{2}cos(xy)- 2ysin(xy)+3y^{2}$$
$$h’(y) = 3y^{2}$$
$$h(y) = y^{3} + C$$
$$\varphi = -y^{2}sin(xy) + y^{3} + C$$
$$when x = n/3 y = 1 $$
$$C = 3/2$$

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

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$