Toronto Math Forum

MAT244--2018F => MAT244--Tests => Term Test 1 => Topic started by: Victor Ivrii on October 16, 2018, 05:32:34 AM

Title: TT1 Problem 3 (noon)
Post by: Victor Ivrii on October 16, 2018, 05:32:34 AM
(a) Find the general solution for equation
\begin{equation*}
y''+8y'+7y=-8e^{t} + 24 e^{-t}.
\end{equation*}
(b)  Find solution, satisfying $y(0)=0$, $y'(0)=0$.
Title: Re: TT1 Problem 3 (noon)
Post by: Jialu Lin on October 16, 2018, 07:42:57 AM
Here is my solution.
Title: Re: TT1 Problem 3 (noon)
Post by: Shengying Yang on October 16, 2018, 08:05:30 AM
There is a mistake in your answer. Plugging in $y(0)=0$ , you should get $0=C_1+C_2-\frac{1}{2}$ . Therefore, $C_1=\frac{1}{2}, C_2=0$
$$∴y(t)=\frac{1}{2}e^{-7t}-\frac{1}{2}e^t+4te^{-t}$$
Title: Re: TT1 Problem 3 (noon)
Post by: Victor Ivrii on October 18, 2018, 04:08:44 AM
Jialu did everything right (almost, there is an error in the calculation of the constants, but the answer is correct).

Shengying, the error is in Jialu's solution, the answer is correct.