### Author Topic: Q3 TUT 0601  (Read 1823 times)

#### Victor Ivrii

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##### Q3 TUT 0601
« on: October 12, 2018, 06:05:54 PM »
Find a differential equation whose general solution is
$$y=c_1e^{2t}+c_2e^{-3t}.$$

#### Nick Callow

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##### Re: Q3 TUT 0601
« Reply #1 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.

#### Keyue Xie

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##### Re: Q3 TUT 0601
« Reply #2 on: October 12, 2018, 06:53:18 PM »
$$y=c_1e^{2t} + c_2e^{-3t}$$
$$r_1 = 2, r_2 = -3$$
$$(r-2)(r+3) = 0$$
$$r^2+ r - 6 = 0$$
$$y''(t) + y'(t) - 6y(t) = 0$$

#### Shengying Yang

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##### Re: Q3 TUT 0601
« Reply #3 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*}