Author Topic: quiz 3 tut 0401  (Read 256 times)

AllanLi

  • Jr. Member
  • **
  • Posts: 9
  • Karma: 0
    • View Profile
quiz 3 tut 0401
« on: October 11, 2019, 02:02:30 PM »
\begin{equation}
y'' + 3y' = 0 , y(0) = -2, y'(0) = 3
\end{equation}we get
\begin{equation}
r^2 + 3r = 0
\end{equation} Solve for r, we get the solution for r.
\begin{equation}
r1= -3, r2 = -3
\end{equation} For the repeated roots, the solution for y is
\begin{equation}
y(t) = C1e^{rt} + C2te^{rt}
\end{equation}So we have
\begin{equation}
y(t) = C1e^{-3t}+tC2e^{-3t}
\end{equation} Since
\begin{equation}
y(0) = -2, y'(0) = 3
\end{equation} We will have two equations about C1 and C2
\begin{equation}
C1 = -2, -3C1 + C2(1+0) = 3
\end{equation}So we have
\begin{equation}
C1 = -2, C2 = -3
\end{equation} Then the solution is
\begin{equation}
y(t) = -2e^{-3t}-3te^{-3t}
\end{equation}
\begin{equation}
y(t) = - (2+3t)e^{-3t}
\end{equation}

Coollight

  • Jr. Member
  • **
  • Posts: 7
  • Karma: 0
    • View Profile
Re: quiz 3 tut 0401
« Reply #1 on: October 11, 2019, 02:37:00 PM »
if I am wrong, please correct me.

\begin{align*}
\ r^{2} + 3r = 0 \ should \ be \\
\ r(r+3) &= 0 \\
\ r_{1} = 0 ,& \ r_{2} = -3 \\
\ so \ the \ general \ solution \ should \ be \\
\ y &= c_{1} + c_{2} e^{-3t} \\
\ since \ y(0) = -2 \\
\ c_{1}+ c_{2} e^{0} &= -2 \\
\ c_{1}+ c_{2}  &= -2 \\
\ since \ y'(0) = 3 \\
\ y' &= -3c_{2} e^{-3t} \\
\ y' &= -3c_{2} e^{0} \\
\ -3c_{2} &= 3 \\
\ so \ c_{1} =-1 ,& \ c_{2} = -1 \\
\ so \ the \  solution \ will \ be \\
\ y &= -1 - e^{-3t}
\end{align*}