Author Topic: TUT0402 Quiz5  (Read 301 times)

Jingjing Cui

  • Full Member
  • ***
  • Posts: 16
  • Karma: 5
    • View Profile
TUT0402 Quiz5
« on: November 01, 2019, 01:57:17 PM »
$$
(1-t)y''+ty'-y=2(t-1)^2e^{-t} ,\;\; 0<t<1\\
y_1(t)=e^t\\
y_2(t)=t\\
Verify \;\;y_1(t) \;\;and\;\; y_2(t)\;\; satisfy\;\; the\;\; corresponding\;\; homogeneous\;\; equation:\\
y_1'(t)=y_1''(t)=e^t\\
(1-t)e^t+te^t-e^t=0\\
y_2'(t)=t \;\; y_2''(t)=1\\
(1-t)t+t^2-t=0\\
y''+\frac{t}{1-t}y'-\frac{1}{1-t}y=-2(t-1)e^{-t}\\
g(t)=-2(t-1)e^{-t}\\
W=det\begin{vmatrix}
e^t&t\\
e^t&1\\
\end{vmatrix}
=e^t-te^t=e^t(1-t)\\
W_1=det\begin{vmatrix}
0&t\\
1&1\\
\end{vmatrix}
=-t\\
W_2=det\begin{vmatrix}
e^t&0\\
e^t&1\\
\end{vmatrix}
=e^t\\
Y(t)=y_1(t)\int\frac{W_1g(t)}{W}dt+y_2(t)\int\frac{W_2g(t)}{W}dt\\
=e^t\int\frac{2t(t-1)e^{-t}}{e^t(1-t)}dt-t\int\frac{2e^t(t-1)e^{-t}}{e^t(1-t)}dt\\
=-e^t\int(2te^{-2t})dt+2t\int(e^{-t})dt\\
=(t+\frac{1}{2})e^{-t}-2te^{-t}\\
=\frac{1}{2}e^{-t}-te^{-t}\\
y(t)=c_1e^t+c_2t+\frac{1}{2}e^{-t}-te^{-t}\\
$$