MAT244--2018F > Final Exam

FE-P3

<< < (2/2)

Qinger Zhang:

--- Quote from: Qi Cui on December 14, 2018, 09:23:35 AM ---
$$W_2 = \left|
\begin {array}{ccc}
  {e^t}&0&e^{2t}\\
  e^t&0&2e^{2t}\\
  e^t&1& 4e^{2t}
\end {array}
\right| = e^{3t}$$

--- End quote ---
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.

Navigation

[0] Message Index

[*] Previous page