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Final Exam / Re: FE-P6
« on: December 17, 2018, 09:10:17 PM »Question: For $H(x,y)$ what are points $(\pm 4,0)$ and $(0,0)$?From the grape, I don't think any of them is max or min.
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Final Exam / Re: FE-P6
« on: December 17, 2018, 09:09:38 PM »
I think (0,0) is a saddle. why (4,0) and (-4,0) are not centre?
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Final Exam / Re: FE-P4
« on: December 17, 2018, 08:26:55 PM »$\begin{bmatrix} x'\\ y' \end{bmatrix}=\begin{bmatrix} 1 &-2 \\ 1&-1 \end{bmatrix}+\begin{bmatrix} sec(t)\\0 \end{bmatrix}$$\lambda=i$, eigenvector is $\begin{pmatrix} -1\\ i-1 \end{pmatrix}$
so $det\begin{bmatrix} 1-\lambda &-1 \\ 1& -1-\lambda \end{bmatrix}=0$
$\lambda=i/-i$ and eigenvector for i is $\begin{pmatrix} 1+i\\ 1 \end{pmatrix}$
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Final Exam / Re: FE-P3
« on: December 17, 2018, 08:12:41 PM »I think W2 is negative.
$$W_2 = \left|
\begin {array}{ccc}
{e^t}&0&e^{2t}\\
e^t&0&2e^{2t}\\
e^t&1& 4e^{2t}
\end {array}
\right| = e^{3t}$$
$$W_2 = \left|
\begin {array}{ccc}
{e^t}&0&e^{2t}\\
e^t&0&2e^{2t}\\
e^t&1& 4e^{2t}
\end {array}
\right| = -e^{3t}$$
$$y_p(t)=y_1{\int}{{w_1(s)}{g(s)}\over{W(s)}} ds+y_2{\int}{{w_2(s)}{g(s)}\over{W(s)}}ds+y_3{\int}{{w_2(s)}{g(s)}\over{W(s)}}ds$$
$$=-e^t{\int}{{6e^s}\over{e^s+1}} ds+e^{-t}
{\int}{{2e^{3s}}\over{e^s+1}}ds+e^{2t}{\int}{{4}\over{e^s+1}}ds$$
$$=-6e^tln|e^t+1|+e^{-t}ln|e^t+1|-e^{-t}(e^t-1)^2-e^{2t}4ln|e^{-x}+1| $$
$$y(t)=c_1e^t+c_2e^{-t}+c_3e^{2t}-6e^tln|e^t+1|+e^{-t}ln|e^t+1|-e^{-t}(e^t-1)^2-e^{2t}4ln|e^{-x}+1|$$
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Final Exam / Re: FE-P1
« on: December 17, 2018, 08:05:41 PM »
$\psi(x,y)=x^2 \sin^4(y)+x \sin^3(y)+2\sin^5 (y)$
there should be a 'c', so we get $\psi(x,y)=x^2 \sin^4(y)+x \sin^3(y)+2\sin^5 (y)$+c
Does not matter where we put it--to the right or the left. V.I.
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Final Exam / Re: FE-P2
« on: December 17, 2018, 07:36:56 PM »$$Homo: r^3 -3r^2+4r-2=0$$$$ y_{p2}(t)=-e^{-t} $$ because W2 is negative
$$r_1=1r_2=1+ir_3=1-i$$
$$\therefore y_c(t)=c_1e^t+c_2e^tcost+c_3e^tsint$$
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:
$$y^{'''}-3y{''}+4y{'}-2y=10e^{-t}$$
I think A should be -1 here.
$$\therefore y_{p2}(t)=-e^{-t} $$
$$y^{'''}-3y{''}+4y{'}-2y=20cost$$
$$y_{p3}(t)=Acost+Bsint $$
$$y^{'}= -Asint+Bcost$$
$$y^{''}= -Acost-Bsint $$
$$y^{'''}= Asint-Bcost$$
substitute above into the function:
we have A=2, B=6
$$\therefore y_{p3}(t)=2cost+6sint $$
$$\therefore y(t)= c_1e^t+c_2e^tcost+c_3e^tsint-10e^t- e^{-t} + 2cost+6sint $$
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