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Messages - Xu Zihan

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1
Final Exam / Re: FE-P2
« on: December 17, 2018, 04:37:59 PM »
I also think the A for At$e^{t}$ should be 10,so the final solution for non home part should be $10te^{t}-e^{-t}+2cost-6sint$ which is same as Jingyi's
You mean $2\cos(t) + 6\sin(t)$ right?
yes, sorry for the typo

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Final Exam / Re: FE-P4
« on: December 14, 2018, 11:26:25 AM »
here is my solution.
$\begin{bmatrix} x'\\ y' \end{bmatrix}=\begin{bmatrix} 1 &-2 \\ 1&-1 \end{bmatrix}+\begin{bmatrix} sec(t)\\0 \end{bmatrix}$
so $det\begin{bmatrix} 1-\lambda &-1 \\ 1& -1-\lambda \end{bmatrix}=0$
$\lambda=i/-i$ and eigenvector for i is $\begin{pmatrix} 1+i\\ 1 \end{pmatrix}$
so $\begin{bmatrix} x'\\ y' \end{bmatrix}=C1e^{it}\begin{bmatrix} 1+i\\ 1 \end{bmatrix}$
$=C1(cost+isint)\binom{1+i}{1}$
$=C1\begin{bmatrix} cost-sint\\ cost \end{bmatrix}+C2i\begin{bmatrix} cost+sint\\ sint \end{bmatrix}$
For non homo:
$\varphi=\begin{bmatrix} cost-sint &cost+sint \\ cost& sint \end{bmatrix}$
Since $\varphi u'=g$
So $\begin{bmatrix} cost-sint &cost+sint \\ cost&sint \end{bmatrix}\begin{bmatrix} u1'\\ u2' \end{bmatrix}=\begin{bmatrix} 1/cost\\ 0 \end{bmatrix}$
so $u1'=-\frac{sint}{cost}$  which means u1=ln(cost)+C2
u2'=1 wihich means u2=t+C1
so the solution is $\begin{bmatrix} x'\\y' \end{bmatrix}=(t+C1)\binom{cost-sint}{cost}+(ln(cost)+C2)\binom{cost+sint}{sint}$

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Final Exam / Re: FE-P2
« on: December 14, 2018, 10:34:57 AM »
$$Homo: r^3 -3r^2+4r-2=0$$
$$r_1=1r_2=1+ir_3=1-i$$
$$\therefore y_c(t)=c_1e^t+c_2e^tcost+c_3e^tsint$$
Non-Homo:
$$y^{'''}-3y{''}+4y{'}-2y=10e^t+10e^{-t} +20cost$$
$$y^{'''}-3y{''}+4y{'}-2y=10e^t$$
$$y_{p1}(t)=Ate^t$$
$$y^{'}=Ate^t+ Ae^t$$
$$y^{''} = Ate^t+ 2Ae^t$$
$$y^{'}= Ate^t+ 3Ae^t$$
substitute above into the function:
$$y^{'''}-3y{''}+4y{'}-2y=10e^t$$
$$we have: A=-10$$
$$\therefore y_{p1}(t)=-10e^t$$
$$y^{'''}-3y{''}+4y{'}-2y=10e^t$$
$$y^{'''}-3y{''}+4y{'}-2y=10e^{-t}$$
$$y_{p2}(t)=Ae^{-t}$$
$$y^{'}= -Ae^{-t}$$
$$y^{''}= Ae^{-t}$$
$$y^{'''}= -Ae^{-t}$$
substitute above into the function:
$$y^{'''}-3y{''}+4y{'}-2y=10e^t$$
we have A=1
$$\therefore y_{p2}(t)=e^{-t}$$
$$y^{'''}-3y{''}+4y{'}-2y=20cost$$
$$y_{p3}(t)=Acost+Bsint$$
$$y^{'}= -Asint+Bcost$$
$$y^{''}= -Acost-Bsint$$
$$y^{'''}= Asint-Bcost$$
substitute above into the function:
we have A=2, B=6
$$\therefore y_{p3}(t)=2cost+6sint$$
$$\therefore y(t)= c_1e^t+c_2e^tcost+c_3e^tsint-10e^t+ e^{-t} + 2cost+6sint$$

I also think the A for At$e^{t}$ should be 10,so the final solution for non home part should be $10te^{t}-e^{-t}+2cost-6sint$ which is same as Jingyi's

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Term Test 2 / Re: TT2A-P1
« on: November 20, 2018, 01:16:03 PM »
Here is my answer, somewhere different from Jiabei's.

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Quiz-5 / Re: Q5 TUT 0501
« on: November 18, 2018, 07:15:42 PM »
after the substitution,

𝑥2 should equal to 𝑒−0.5𝑡(-𝑐1cos(2𝑡)+𝑐2sin(2𝑡))

And C1=-2   C2=2

So the final solution will be:

𝑥1=𝑒−0.5𝑡(−2cos(2𝑡)+2sin(2𝑡))

𝑥2=𝑒−0.5𝑡(2cos(2𝑡)+2sin(2𝑡))

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