# Toronto Math Forum

## MAT244--2019F => MAT244--Test & Quizzes => Quiz-5 => Topic started by: Jingjing Cui on November 01, 2019, 01:57:17 PM

Title: TUT0402 Quiz5
Post by: Jingjing Cui on November 01, 2019, 01:57:17 PM
$$(1-t)y''+ty'-y=2(t-1)^2e^{-t} ,\;\; 0<t<1\\ y_1(t)=e^t\\ y_2(t)=t\\ Verify \;\;y_1(t) \;\;and\;\; y_2(t)\;\; satisfy\;\; the\;\; corresponding\;\; homogeneous\;\; equation:\\ y_1'(t)=y_1''(t)=e^t\\ (1-t)e^t+te^t-e^t=0\\ y_2'(t)=t \;\; y_2''(t)=1\\ (1-t)t+t^2-t=0\\ y''+\frac{t}{1-t}y'-\frac{1}{1-t}y=-2(t-1)e^{-t}\\ g(t)=-2(t-1)e^{-t}\\ W=det\begin{vmatrix} e^t&t\\ e^t&1\\ \end{vmatrix} =e^t-te^t=e^t(1-t)\\ W_1=det\begin{vmatrix} 0&t\\ 1&1\\ \end{vmatrix} =-t\\ W_2=det\begin{vmatrix} e^t&0\\ e^t&1\\ \end{vmatrix} =e^t\\ Y(t)=y_1(t)\int\frac{W_1g(t)}{W}dt+y_2(t)\int\frac{W_2g(t)}{W}dt\\ =e^t\int\frac{2t(t-1)e^{-t}}{e^t(1-t)}dt-t\int\frac{2e^t(t-1)e^{-t}}{e^t(1-t)}dt\\ =-e^t\int(2te^{-2t})dt+2t\int(e^{-t})dt\\ =(t+\frac{1}{2})e^{-t}-2te^{-t}\\ =\frac{1}{2}e^{-t}-te^{-t}\\ y(t)=c_1e^t+c_2t+\frac{1}{2}e^{-t}-te^{-t}\\$$