### 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 - xuanzhong

Pages: [1]
1
##### Quiz 3 / quiz 3 lec5101 - 5b
« on: October 08, 2020, 10:16:37 PM »
Compute line integral: $\int_{r} Re(z) \,dz$, where r is line segmemt from 1 to i
$$f(z) = Re(z)$$
$$r(t) = (1-t)z_0 + tz_1 = (1-t)*1 + ti = 1-t+it , 0<= t <= 1$$
$$r^{'}(t) = i-1$$
$$f(r(t)) = Re(r(t)) = 1-t$$
$$\int_{r} f(z) \,dz = \int_{0}^{1}f(r(t))r^{'}(t) \,dt$$
$$=\int_{0}^{1}(1-t)(i-1) \,dt$$
$$=(i-1)*(t-\tfrac{1}{2} t^2)\Big|_0^1$$
$$=(i-1)(1-\tfrac{1}{2}-0)$$
$$=\tfrac{1}{2}(i-1)$$

2
##### Quiz 1 / quiz 1 5101
« on: September 25, 2020, 07:44:43 PM »
Find the locus of points for:
$$Re(z^{2}) = 4$$
$$Let z=x+iy$$
$$Re((x+iy)^{2}) = 4$$
$$Re(x^{2}+2ixy-y^{2}) = 4$$
$$x^{2} - y^{2} = 4$$
$$\frac{x^{2}}{4} - \frac{y^{2}}{4} = 1$$
Therefore, the locus of z is a hyperbolic curve with $\frac{x^{2}}{4} - \frac{y^{2}}{4} = 1$

3
##### Term Test 2 / Re: Problem 3 (noon)
« on: November 19, 2019, 06:00:45 AM »
Here's the solution for sketching.

4
##### Term Test 2 / Re: Problem 3 (morning)
« on: November 19, 2019, 05:54:32 AM »
HERE'S THE SOLUTION FOR SKETCHING.

5
##### Term Test 2 / Re: Problem 4 (noon)
« on: November 19, 2019, 05:47:01 AM »
here's the solution including sketching.

6
##### Term Test 2 / Re: Problem 3 (main sitting)
« on: November 19, 2019, 05:19:08 AM »
Here's the solution including sketching.

7
##### Term Test 2 / Re: Problem 2 (main sitting)
« on: November 19, 2019, 05:03:39 AM »
a)Write equation for Wronskian of y1,y2,y3.
$$W=ce^{-\int p(t)dt}=ce^{-\int 4dt}=ce^{-4t}$$
b)Find fundamental system of solutions for honogenuous equation, and find Wronskian. Comapare with (a).
$$r^3+4r^2+r-6=0$$
$$(r-1)(r+2)(r+3)=0$$
$$r=1, r=-2, r=-3$$
$$y_c(t)=c_1e^t+c_2e^{-2t}+c_3e^{-3t}$$
$$W=\begin{vmatrix} e^t & e^{-2t} & e^{-3t}\\ e^t & -2e^{-2t} & -3e^{-3t}\\ e^t & 4e^{-2t} & 9e^{-3t}\\ \end{vmatrix} = 12e^{-4t}$$
similar solution to (a), but $c=12$

(c)Find the general solution.
$$y_p(t)=Ate^t$$
$$y^\prime=Ae^t+Ate^t=Ae^t(1+t)$$
$$y^{\prime\prime}=2Ae^t+Ate^t=Ae^t(2+t)$$
$$y^{\prime\prime\prime}=3Ae^t+Ate^t=Ae^t(3+t)$$
$$y^{\prime\prime\prime}+4y^{\prime\prime}+y^\prime-6y=24e^t$$
$$Ae^t(3+t+8+4t+1+t-6t)=24e^t$$
$$12Ae^t=24e^t$$
$$A=2$$
Hence, $y(t)=c_1e^t+c_2e^{-2t}+c_3e^{-3t}+2te^t$

8
##### Quiz-5 / Lec5101 quiz5
« on: November 01, 2019, 01:47:05 PM »
Find the general solution of the given differential equation.
$$y^{\prime\prime}+4y=3csc(2t), 0<t<pi/2$$
For homogeneous equation: $r^2+4=0$
we get:
$$r_{1}=2i, r_{2}=-2i$$
$$y_{c}(t)=c_{1}cos2t+c_{2}sin2t$$

For non-homogeneous equation:
$$W[y_{1},y_{2}](t) = \begin{vmatrix} cos2t & sin2t \\ -2sin2t & 2cos2t \\ \end{vmatrix} = 2$$

Therefore,
$$u_{1}(t)=-(\int\frac{sin2t*3csc2t}{2}dt)$$
$$=-(\int\frac{3}{2}dt)$$
$$=-\frac{3}{2}t$$
$$u_{2}(t)=-(\int\frac{cos2t*3csc2t}{2}dt)$$
$$=\frac{3}{2}(\int\frac{cos2t}{sin2t}dt)$$
$$=\frac{3}{2}(\int{cot2t}dt)$$
$$=\frac{3}{4}ln|sin2t|$$

Hence, the particular solution is $y_{p}(t)=u_{1}(t)y_{1}(t)+u_{2}(t)y_{2}(t)$
$$y_{p}(t)=cos2t\cdot(-\frac{3}{2}t)+sin2t\cdot(\frac{3}{4}ln|sin2t|)$$
$$=\frac{3}{4}sin2tln|sin2t|-\frac{3}{2}tcos2t$$

Therefore, the general solution is:
$$y(t)=y_{c}(t)+y_{p}(t)$$
$$=c_{1}cos2t+c_{2}sin2t+\frac{3}{4}sin2tln|sin2t|-\frac{3}{2}tcos2t$$

9
##### Term Test 1 / Re: Problem 1 (main sitting)
« on: October 23, 2019, 02:29:17 PM »
$$M_{y}=1+6ye^{3x}$$
$$N_{x}=4ye^{2x}$$
$M_{y} ≠N_{x}$,it is not exact
$$R_{2} =\frac{ M_{y} -N_{x}}{N}=\frac{1+2ye^{2x} }{1+2ye^{2x}}=1$$
$$μ=e^{∫R_{2}dx} =e^{∫1 dx} =e^x$$
Multiplying both sides by $\mu$, we get
$$ye^x+3y^2e^{3x} +(e^x+2ye^{3x}) y^\prime=0$$
$$M_{y}^\prime=e^x+6ye^{3x}$$
$$N_{x}^\prime=e^x+6ye^{3x}$$
$M_{y}^\prime=N_{x}^\prime$,it is exact
$$∃φ(x,y) such that\ φ_{x} =M^\prime,φ_{y} =N^\prime$$
$$φ(x,y)=∫{M^\prime dx}=∫{ye^x+3y^{2}e^{3x}dx}=ye^x+y^{2}e^{3x} +h(y)$$
$$φ_{y} =e^x+2ye^{3x} +h(y)^\prime=e^x+2ye^{3x}$$
Then $h(y)^\prime=0$

Hence h(y)=c
$$φ(x,y)=ye^x+y^{2}e^{3x} =c$$
Since y(0)=1
$$1⋅e^0+1^2⋅e^0=2=c$$
$$φ(x,y)=ye^x+y^{2}e^{3x} =2$$

10
##### Quiz-3 / TUT0601 quiz3
« on: October 11, 2019, 06:29:27 PM »
Find the Wronskian of two solutions of the given differential equation without solving the equation.
$$cos(t)y^{\prime\prime}+sin(t)y^\prime-ty=0$$

First, we divide both sides of the equation by cos(t):
$$y^{\prime\prime}+\frac{sin(t)}{cos(t)}y^\prime-\frac{1}{cos(t)}y=0$$

Now the given second-order differential equation has the form:
$$L[y]= y^{\prime\prime}+p(t)y^\prime-\ q(t)y=0$$

Then by Abel’s Theorem: the Wronskian W[y1,y2](t) is given by
$$W[y_1,y_2\ ](t)=cexp(-\int p(t)dt)$$
$$=cexp(-\int\frac{sin(t)}/{cos(t)}dt〗)$$
$$=cexp(\int\frac{1}{cos(t)}\ d(cos(t)〗)$$
$$=ce^{ln|cos(t)|}\$$
$$=ccos(t)$$

11
##### Quiz-2 / TUT0601 quiz 2
« on: October 04, 2019, 07:25:01 PM »
Show that the given equation is not exact but becomes exact when multiplied by the given integrating factor. Then solve the equation.
$$\frac{x^2}{y^3}+x(1+y^2\ )\frac{dy}{dx} =0,\mu(x,y)=\frac{1}{xy^3}$$
First, define $M(x,y)=\ x^2\ y^3$,$N(x,y)=x(1+y^2)$
$$M_y=\partial y/\partial x\ (x^2\ y^3\ )=3x^2\ y^2$$
$$N_y=\partial y/\partial x[x(1+y^2\ )]=1+y^2$$

Since $M_y\neq N_y$, this implies the given equation is not exact.
Now, multiplying both sides by the integrating factor $\mu(x,y)$, that is
$$\frac{1}{xy^3}x^2y^3+\frac{1}{xy^3}x(1+y^2\ )\frac{dy}{dx}=x+(\frac{1}{y^3}+\frac{1}{y}) \frac{dy}{dx}=0$$

Define $M^\prime\ (x,y)=\ x$,$N^\prime\ (x,y)=\frac{1}{y^3}+\frac{1}{y}$
$$M_y=\partial y/\partial x\ (x)=0$$
$$N_y=\partial y/\partial x[\frac{1}{y^3}+\frac{1}{y}]=0$$

Since $M_y$=$N_y$, this implies the given equation is exact.
Thus, we know there exists a function $\phi(x,y)$ such that $\partial\phi/\partial x=\ M^\prime\ (x,y)=\ x$
Then $\phi(x,y)=\int M^\prime\ dx=\frac{1}{2} x^2+h(y)$
Take derivative on both sides with respect to y we get
$$\partial\phi/\partial y=h^\prime\ (y)=N^\prime\ (x,y)=\frac{1}{y^3}+\frac{1}{y}$$

Integrating with respect to y we have
$$h(y)=\frac{-1}{2}\frac{1}{y^2}+ln|y|+C$$

Hence, $\phi(x,y)=\frac{1}{2} x^2-\frac{1}{2} \frac{1}{y^2}+ln|y|=C$ is the general solution to the given DE.
Besides, notice that the constant function y(x)=0\ \forall x is also a solution to the given DE.

12
##### Quiz-1 / TUT0601 quiz 1
« on: September 27, 2019, 03:06:48 PM »
Find the solution of the given initial value problem.
$$\frac{dy}{dx}-2y=e^{2t},y(0)=2$$
$$\mu(t)=e^{\int-2dt}=e^{-2t}$$

Multiplying both sides by $\mu(t)$:
$$e^{-2t}*\frac{dy}{dx}\ -\ 2e^{-2t}y=e^{-2t}{*e}^{2t}$$
$$\frac{d}{dx}(e^{-2t}y)=1$$

By integrating both sides:
$$e^{-2t}y=t+c$$
$$y=\frac{t+c}{e^{-2t}}$$
$$y=(t+c)e^{2t}$$

since $y\left(0\right)=2$, then $2=\left(0+c\right)\ast e^0=c\ast1=c$

Hence the solution to the initial value problem is $y=(t+2)e^{2t}$

Pages: [1]