FE-P3

[Victor Ivrii]

Typed solutions only. No uploads

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= e^t   y₂=e^-t   y₃=e^2t

W(t,y) = e^t       e^-t     e^2t 
             e^t      -e^-t    2e^2t
             e^t       e^-t    4e^2t
= e^t(-e^-t·4e^2t-e^-t·2e^2t) - e^-t(e^t· 4e^2t- e^t· 2e^2t）+ e^2t(e^t·e^-t+e^t·e^-t)
= -6e^2t - 2e^2t + 2e^2t =-6e^2t
W₁ = e^-t·2e^2t+e^-t·e^2t= e^t
W₂ = e^t· 2e^2t - e^t·e^2t  = e^3t
W₃ = e^t· -e^-t -  e^t·e^-t = -2

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

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

[Jiexuan Wei]

forget to see 'readme before posting', sorry

[Chengyin Ye]

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

[Qinger Zhang]

$$W_2 = \left|
\begin {array}{ccc}
  {e^t}&0&e^{2t}\\
  e^t&0&2e^{2t}\\
  e^t&1& 4e^{2t}
\end {array}
\right| = e^{3t}$$

I think W2 is negative.
$$W_2 = \left|
\begin {array}{ccc}
  {e^t}&0&e^{2t}\\
  e^t&0&2e^{2t}\\
  e^t&1& 4e^{2t}
\end {array}
\right| = -e^{3t}$$
$$y_p(t)=y_1{\int}{{w_1(s)}{g(s)}\over{W(s)}} ds+y_2{\int}{{w_2(s)}{g(s)}\over{W(s)}}ds+y_3{\int}{{w_2(s)}{g(s)}\over{W(s)}}ds$$
$$=-e^t{\int}{{6e^s}\over{e^s+1}} ds+e^{-t}
{\int}{{2e^{3s}}\over{e^s+1}}ds+e^{2t}{\int}{{4}\over{e^s+1}}ds$$
$$=-6e^tln|e^t+1|+e^{-t}ln|e^t+1|-e^{-t}(e^t-1)^2-e^{2t}4ln|e^{-x}+1| $$
$$y(t)=c_1e^t+c_2e^{-t}+c_3e^{2t}-6e^tln|e^t+1|+e^{-t}ln|e^t+1|-e^{-t}(e^t-1)^2-e^{2t}4ln|e^{-x}+1|$$

[Victor Ivrii]

Writing characteristic equation: $L(k):= k^3-2k^2-k+2=0$, with $L(k)=(k-2)k^2 -(k-2)=(k-2)(k^2-1)$; then $k_1=1, k_2=-1, k_3=2$. Then
\begin{equation}
y^*= C_1e^{t} + C_2 e^{-t}  + C_3 e^{2t}
\label{eq-3-1}
\end{equation}
is a general solution to the homogeneous equation.

We are looking for solution to the inhomogeneous equation as (\ref{eq-3-1}) with unknown functions $C_1,C_2,C_3$ s.t.
\begin{align*}
&\left\{\begin{aligned}
&C_1' e^{t} + C_2' e^{-t}+ C_3' e^{2t}=0,\\
&C_1' e^{t} - C_2' e^{-t} + 2C_3' e^{2t}=0,\\
&C_1' e^{t} + C_2' e^{-t}+4 C_3' e^{2t}=\frac{12e^{2t}}{e^t+1};
\end{aligned}\right.\implies\\[4pt]
&\left\{\begin{aligned}
2&C_1' e^{t} + 3C_3' e^{2t}=0,\\
&C_1' e^{t} - C_2' e^{-t} + 2C_3' e^{2t}=0,\\
2&C_1' e^{t} +  6 C_3' e^{2t}=\frac{12e^{2t}}{e^t+1};
\end{aligned}\right.\implies\\[4pt]
&C_3'e^{2t}=\frac{4e^{2t}}{e^t+1}, \qquad C_1'e^{t}= -\frac{6e^{2t}}{e^t+1},\qquad C_2'e^{-t}=\frac{2e^{2t}}{e^t+1}\implies\\[4pt]
&C_1=-\int \frac{6e^{t}\,dt}{e^t+1}= -6 \ln (e^t+1) +c_1,\\[4pt]
&C_2= \int \frac{2e^{3t}\,dt}{e^t+1}= \int \Bigl[2e^{2t}- 2 e^{t}\Bigr] \,dt +\int  \frac{2e^{t}\,dt}{e^t+1}
=e^{2t} -2e^{t} +2\ln (e^t+1)+c_2,\\[4pt]
&C_3=\int \frac{4\,dt}{e^t+1}=\int \frac{4e^{-t}\,dt}{1+e^{-t}}= -4\ln (1+e^{-t})+c_3= -4\ln (e^t+1)+4t +c_3.
\end{align*}
Then
\begin{align*}
y= &\bigl[-6 \ln (e^t+1) +c_1\bigr]e^{t}+\bigl[e^{2t} -2e^{t} +2\ln (e^t+1)+c_2\bigr]e^{-t}+
\bigl[-4\ln (e^t+1)+4t +c_3\bigr]e^{2t}=\\
&-6  e^{t}\ln (e^t+1)+ 2e^{-t}\ln (e^t+1)
-4e^{2t}\ln (e^t+1)+4te^{2t}-2 +
c_1e^{t} + c_2 e^{-t}  + c_3 e^{2t}
\end{align*}
with $c_1:= c_1+1$ in the last transition.