# [SOLVED]1.3.15 find r in e^{rt}

#### karush

##### Well-known member
$\textsf{Determine the values of$r$}$
$\textsf{for which the given differential equation has solutions }$
$\textsf{of the form$y = e^{rt}$}$
$\textit{15.$\quad y'+2y=0$}$
\begin{align*}
2\exp\int \, dx &=2e^{t}+c\\
&=e^{2t}+c\\
\therefore r&=2
\end{align*}
no sure about the rest due to mutiple differentiation
$\textit{16$\quad y''-y=0$}$
$\textit{17.$\quad y''+y'-6y=0$}$
$\textit{18.$\quad y'''-3y''+2y'=0$}$

#### tkhunny

##### Well-known member
MHB Math Helper
$\textsf{Determine the values of$r$}$
$\textsf{for which the given differential equation has solutions }$
$\textsf{of the form$y = e^{rt}$}$
$\textit{15.$\quad y'+2y=0$}$
\begin{align*}
2\exp\int \, dx &=2e^{t}+c\\
&=e^{2t}+c\\
\therefore r&=2
\end{align*}
no sure about the rest due to mutiple differentiation
$\textit{16$\quad y''-y=0$}$
$\textit{17.$\quad y''+y'-6y=0$}$
$\textit{18.$\quad y'''-3y''+2y'=0$}$
You don't seem to have the spirit of the problem statement. You are not instructed to "solve the DE". You are just finding values of r for that form. Take all the derivatives of $e^{rt}$ that you need, in each case.

#### karush

##### Well-known member
doesn't $(y^{rt})^n=ry^{rt}$

#### tkhunny

##### Well-known member
MHB Math Helper
$\dfrac{d^{n}}{dt^{n}}e^{rt}=r^{n}e^{rt}$

#### Gryfen

##### New member
$\textsf{Determine the values of$r$}$
$\textsf{for which the given differential equation has solutions }$
$\textsf{of the form$y = e^{rt}$}$
$\textit{15.$\quad y'+2y=0$}$
\begin{align*}
2\exp\int \, dx &=2e^{t}+c\\
&=e^{2t}+c\\
\therefore r&=2
\end{align*}
no sure about the rest due to mutiple differentiation
$\textit{16$\quad y''-y=0$}$
$\textit{17.$\quad y''+y'-6y=0$}$
$\textit{18.$\quad y'''-3y''+2y'=0$}$

for 16 I got r=0, and for 17 I got r=4 or 5

#### Country Boy

##### Well-known member
MHB Math Helper
for 16 I got r=0, and for 17 I got r=4 or 5
No. Equation 16 is y''- y= 0. Taking $$y= e^{rx}$$ then $$y'= re^{rx}$$ and $$y''= r^2e^{rx}$$ so $$y''- y= r^2e^{rx}- e^{rx}= (r^2- 1)e^{rx}= 0$$. $$e^{rx}$$ is not 0 we can divide both sides by it to get $$r^2- 1= 0$$. What values of r satisfy that?