MAT244--2018F > Quiz-3

Q3 TUT 0601

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**Victor Ivrii**:

Find a differential equation whose general solution is

$$y=c_1e^{2t}+c_2e^{-3t}.$$

**Nick Callow**:

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**:

$$

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**:

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*}

$$

**Victor Ivrii**:

Keyue , Shengying

Do not post after perfect solution was posted

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