MHB What Values of r Satisfy Differential Equations with Exponential Solutions?

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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?
 
I have the equation ##F^x=m\frac {d}{dt}(\gamma v^x)##, where ##\gamma## is the Lorentz factor, and ##x## is a superscript, not an exponent. In my textbook the solution is given as ##\frac {F^x}{m}t=\frac {v^x}{\sqrt {1-v^{x^2}/c^2}}##. What bothers me is, when I separate the variables I get ##\frac {F^x}{m}dt=d(\gamma v^x)##. Can I simply consider ##d(\gamma v^x)## the variable of integration without any further considerations? Can I simply make the substitution ##\gamma v^x = u## and then...

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