What Values of r Satisfy Differential Equations with Exponential Solutions?

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SUMMARY

This discussion focuses on determining the values of \( r \) for which specific differential equations have solutions of the form \( y = e^{rt} \). The equations analyzed include \( y' + 2y = 0 \), \( y'' - y = 0 \), \( y'' + y' - 6y = 0 \), and \( y''' - 3y'' + 2y' = 0 \). The solutions yield \( r = 2 \) for the first equation, \( r = 0 \) for the second, and \( r = 4 \) or \( r = 5 \) for the third. The fourth equation requires further exploration to determine its roots.

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karush
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$ \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. $\quad y'''-3y''+2y'=0$}$
 
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karush said:
$ \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.
 
doesn't $(y^{rt})^n=ry^{rt}$
 
$\dfrac{d^{n}}{dt^{n}}e^{rt}=r^{n}e^{rt}$
 
karush said:
$ \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
 
Gryfen said:
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?
 

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