Toronto Math Forum
MAT2442018F => MAT244Tests => Quiz3 => Topic started by: Victor Ivrii on October 12, 2018, 06:05:54 PM

Find a differential equation whose general solution is
$$y=c_1e^{2t}+c_2e^{3t}.$$

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 $(r2)(r+3)$. 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 $(r2)(r+3)$. 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.

$$
y=c_1e^{2t} + c_2e^{3t}
$$
$$
r_1 = 2, r_2 = 3
$$
$$
(r2)(r+3) = 0
$$
$$
r^2+ r  6 = 0
$$
$$
y''(t) + y'(t)  6y(t) = 0
$$

I agree with Nick's answer. Here is just a version without word explanation which is easier to see.
$$
\begin{align*}
∵ r=2 , r=3
\end{align*}
$$
$$
\begin{align*}
∴(r2)(r+3)=0
\end{align*}
$$
$$
\begin{align*}
∴ r^22r+3r6=0
\end{align*}
$$
$$
\begin{align*}
∴r^2+r6=0
\end{align*}
$$
$$
\begin{align*}
∴ y''+y'6y=0
\end{align*}
$$

Keyue , Shengying
Do not post after perfect solution was posted