MAT244--2018F > Final Exam

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Victor Ivrii:

Find the general solution of
\begin{equation*}
y'''-2y'' -y '+2y = \frac{12e^{2t}}{e^t+1}.
\end{equation*}
Hint: All roots are integers (or complex integers).

Qi Cui:
$$Homo: r^3 -2r^2-r+2=0$$
$$(r+1)(r-1)(r-2)=0$$
$$r_1=1, r_2=-1, r_3=2$$
$$\therefore y_c(t)=c_1e^t+c_2e^{-t}+c_3e^{2t}$$
$$W = \left| \begin {array}{ccc} {e^t}&e^{-t}&e^{2t}\\ e^t& -e^{-t}&2e^{2t}\\ e^t&e^{-t}& 4e^{2t} \end {array} \right| = -6e^{2t}$$
$$W_1 = \left| \begin {array}{ccc} {0}&e^{-t}&e^{2t}\\ 0& -e^{-t}&2e^{2t}\\ 1&e^{-t}& 4e^{2t} \end {array} \right| = 3e^{t}$$
$$W_2 = \left| \begin {array}{ccc} {e^t}&0&e^{2t}\\ e^t&0&2e^{2t}\\ e^t&1& 4e^{2t} \end {array} \right| = e^{3t}$$
$$W_3 = \left| \begin {array}{ccc} {e^t}&e^{-t}&0\\ e^t& -e^{-t}&0\\ e^t&e^{-t}& 1 \end {array} \right| = -2$$
substitute above into the formula:
$$y_p(t)=y_1{\int}{{w_1(s)}{g(s)}\over{W(s)}} ds+y_2{\int}{{w_2(s)}{g(s)}\over{W(s)}}ds+y_3{\int}{{w_2(s)}{g(s)}\over{W(s)}}ds$$
$$=-e^t{\int}{{6e^s}\over{e^s+1}} ds-e^{-t} {\int}{{2e^{3s}}\over{e^s+1}}ds+e^{2t}{\int}{{4}\over{e^s+1}}ds$$
$$=-6e^tln|e^t+1|-e^{-t}ln|e^t+1|-e^{-t}(e^t-1)^2-e^{2t}4ln|e^{-x}+1|$$
$$y(t)=y_c(t)+y_p(t)$$

Doris Zhuomin Jia:
homo: r3 - 2r2 - r+2 = 0
(r-1)(r+1)(r-2) =0   r = 1, -1, 2
y1= et   y2=e-t   y3=e2t

W(t,y) = et       e-t     e2t
et      -e-t    2e2t
et       e-t    4e2t
= et(-e-t·4e2t-e-t·2e2t) - e-t(et· 4e2t- et· 2e2t）+ e2t(et·e-t+et·e-t)
= -6e2t - 2e2t + 2e2t =-6e2t
W1 = e-t·2e2t+e-t·e2t= et
W2 = et· 2e2t - et·e2t  = e3t
W3 = et· -e-t -  et·e-t = -2

u1 = ∫ [(2e2t/et +1)·et]/-6e2t = -2ln|et+1| +C1
u2 = ∫[(2e2t/et +1)·e3t]/-6e2t = 2et-e2t+3 -2ln|et+1| +C2
u3 = ∫[(2e2t/et +1)·(-2)]/-6e2t =4(-ln|et+1|+t)+C3

y= u1y1+u2y2+u3y3
y= et(-2ln|et+1| +C1)+e-t(2et-e2t+3 -2ln|et+1| +C2)+e2t(4(-ln|et+1|+t)+C3)

Jiexuan Wei:
forget to see 'readme before posting', sorry

Chengyin Ye:
I think that instead of writing e-t(2et-e2t+3 -2ln|et+1| +C2), we should write e-t(2et -2ln|et+1| +C2) because e-t(-e2t) and 3et  are in the homogeneous solution.

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