# Toronto Math Forum

## MAT244--2018F => MAT244--Tests => Quiz-3 => Topic started by: Victor Ivrii on October 12, 2018, 06:05:54 PM

Title: Q3 TUT 0601
Post by: Victor Ivrii on October 12, 2018, 06:05:54 PM
Find a differential equation whose general solution is
$$y=c_1e^{2t}+c_2e^{-3t}.$$
Title: Re: Q3 TUT 0601
Post by: Nick Callow on October 12, 2018, 06:06:11 PM
Find a differential equation with general solution $c_1e^{2t} + c_2e^{-3t}$.

Given that $r = -3, 2$ we know the characteristic equation of this general solution will be of the form $(r-2)(r+3)$. Expanding this, we get that the differential equation has the form $r^2 + r - 6$. Therefore, a differential equation with the above general solution will be $y''(t) + y'(t) - 6y(t) = 0$.

We can check this by working in reverse. Suppose we have a differential equation $y''(t) + y'(t) - 6y(t) = 0$. Then we can try a solution of the form $y(t) = e^{rt}$. Consequently, $y'(t) = re^{rt}$ and $y''(t) = r^2e^{rt}$. Subbing this into the differential equation we have, $r^2e^{rt} + re^{rt} - 6e^{rt} = 0$. We can factor out an $e^{rt}$ and the characteristic equation becomes $r^2 + r - 6$. Factoring this, we have $(r-2)(r+3)$. Since our $r = -3,2$ we know two particular solutions are $y_1(t) = e^{2t}$ and $y_2(t) = e^{-3t}$. The general form of this will be $c_1e^{2t} + c_2e^{-3t}$. This matches what the question was asking, so we are finished.
Title: Re: Q3 TUT 0601
Post by: Keyue Xie on October 12, 2018, 06:53:18 PM
$$y=c_1e^{2t} + c_2e^{-3t}$$
$$r_1 = 2, r_2 = -3$$
$$(r-2)(r+3) = 0$$
$$r^2+ r - 6 = 0$$
$$y''(t) + y'(t) - 6y(t) = 0$$
Title: Re: Q3 TUT 0601
Post by: Shengying Yang on October 12, 2018, 07:04:07 PM
I agree with Nick's answer. Here is just a version without word explanation which is easier to see.
\begin{align*} ∵ r=2 , r=-3 \end{align*}
\begin{align*} ∴(r-2)(r+3)=0 \end{align*}
\begin{align*} ∴ r^2-2r+3r-6=0 \end{align*}
\begin{align*} ∴r^2+r-6=0 \end{align*}

\begin{align*} ∴ y''+y'-6y=0 \end{align*}
Title: Re: Q3 TUT 0601
Post by: Victor Ivrii on October 12, 2018, 07:46:06 PM
Keyue , Shengying

Do not post after perfect solution was posted