# Toronto Math Forum

## MAT244--2019F => MAT244--Test & Quizzes => Quiz-3 => Topic started by: AllanLi on October 11, 2019, 02:02:30 PM

Title: quiz 3 tut 0401
Post by: AllanLi on October 11, 2019, 02:02:30 PM

y'' + 3y' = 0 , y(0) = -2, y'(0) = 3
we get

r^2 + 3r = 0
Solve for r, we get the solution for r.

r1= -3, r2 = -3
For the repeated roots, the solution for y is

y(t) = C1e^{rt} + C2te^{rt}
So we have

y(t) = C1e^{-3t}+tC2e^{-3t}
Since

y(0) = -2, y'(0) = 3
We will have two equations about C1 and C2

C1 = -2, -3C1 + C2(1+0) = 3
So we have

C1 = -2, C2 = -3
Then the solution is

y(t) = -2e^{-3t}-3te^{-3t}

y(t) = - (2+3t)e^{-3t}

Title: Re: quiz 3 tut 0401
Post by: Coollight 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*}