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[SOLVED] 1.3.15 find r in e^{rt}

karush

Well-known member
Jan 31, 2012
2,928
$ \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
Jan 27, 2012
267
$ \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
Jan 31, 2012
2,928
doesn't $(y^{rt})^n=ry^{rt}$
 

tkhunny

Well-known member
MHB Math Helper
Jan 27, 2012
267
$\dfrac{d^{n}}{dt^{n}}e^{rt}=r^{n}e^{rt}$
 

Gryfen

New member
Aug 29, 2018
1
$ \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
Jan 30, 2018
756
for 16 I got r=0, and for 17 I got r=4 or 5
No. Equation 16 is y''- y= 0. Taking [tex]y= e^{rx}[/tex] then [tex]y'= re^{rx}[/tex] and [tex]y''= r^2e^{rx}[/tex] so [tex]y''- y= r^2e^{rx}- e^{rx}= (r^2- 1)e^{rx}= 0[/tex]. [tex]e^{rx}[/tex] is not 0 we can divide both sides by it to get [tex]r^2- 1= 0[/tex]. What values of r satisfy that?