-1.3.15 find r in e^{rt}

  • MHB
  • Thread starter karush
  • Start date
In summary: No values of r satisfy that. Equation 17 is y''+ y'-6y= 0. Taking y= e^{rx} then y'= re^{rx} and y''= r^2e^{rx} so y''+ y'-6y= r^2e^{rx}+ 6e^{rx}= (r^2+ 6)e^{rx}= 6. e^{rx} is not 6 so we can divide both sides by it to get r^2+ 6= 6. What values of r satisfy that? Values of r satisfy that. Equation 18 is y'''- 3y''+ 2y'= 0. Taking y= e^{rx} then
  • #1
karush
Gold Member
MHB
3,269
5
$ \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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  • #2
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.
 
  • #3
doesn't $(y^{rt})^n=ry^{rt}$
 
  • #4
$\dfrac{d^{n}}{dt^{n}}e^{rt}=r^{n}e^{rt}$
 
  • #5
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
 
  • #6
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 [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?
 

1. What is the meaning of "-1.3.15" in the equation "e^{rt}"?

In this equation, "-1.3.15" is most likely a value for the variable "t". It represents a specific point in time for which we are trying to find the value of "r".

2. What does the "e" stand for in the equation "e^{rt}"?

In this equation, "e" represents the mathematical constant known as Euler's number, which is approximately equal to 2.71828. It is commonly used in many mathematical and scientific equations.

3. What does the "r" represent in the equation "e^{rt}"?

In this equation, "r" is a variable that represents the rate of change. It is often used in exponential growth or decay equations to represent the growth or decay rate.

4. How is the value of "r" determined in the equation "e^{rt}"?

The value of "r" can be determined by rearranging the equation and solving for it. Alternatively, it can also be found by using a graphing calculator or computer program to plot the data points and find the best-fit line for the equation.

5. Can the equation "e^{rt}" be used to model real-life situations?

Yes, the equation "e^{rt}" can be used to model various real-life situations, such as population growth, radioactive decay, and compound interest. It is a commonly used equation in many fields of science and mathematics.

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