2
$\begingroup$

I am trying to find what values of r in $y = e^{rx}$ satsify $2y'' + y' - y = 0$

I thought I was being clever and knew how to do this so this is how I proceeded.

$y' = re^{rx}$ $y'' = r^2 e^{rx}$

$2(r^2 e^{rx}) +re^{rx} -e^{rx} = 0 $

I am not sure how to proceed from here, the biggest thing I am confused on is that I am working with a variable x, with no input conditions at all, and a variable r (the constant) so how do I do this?

2 Answers 2

5

The final equation you have is $2r^2 \exp(rx) + r \exp(rx) - \exp(rx) = 0$ $(2r^2 + r - 1)\exp(rx) = 0$ Now $\exp(rx) \neq 0$, for all $r$ and $x$. Hence, you get that $2r^2 + r - 1 =0$ Can you proceed from here by solving the quadratic equation for $r$?

Move your cursor over the gray area below for complete solution.

Note that we can write $2r^2 + r - 1$ as shown below. $2r^2 + r - 1 = 2r^2 + 2r -r -1 = 2r(r+1)-1 ( r+1) = (2r-1)(r+1)$ Hence, $2r^2 + r - 1 = 0 \implies (2r-1)(r+1) = 0 \implies r = \dfrac12 \text{ or } r= - 1$. Hence, $y$ is either $\exp(x/2)$ or $\exp(-x)$. In general, we find that $y = c_1 \exp(x/2) + c_2 \exp(-x)$ where $c_1,c_2$ are constants, satisfies the differential equation.

  • 0
    @Jordan: you should probably memorise the formula for solutions of a quadratic, it's very often useful. If you forget it, you can rederive it by writing $x^2 + bx + c = 0$ as $(x+b/2)^2 + c - b^2/4 = 0$, i.e. $(x+b/2)^2 = b^2/4 - c$.2012-06-14
2

Here's the best part: $e^{rx}$ is never zero. Thus, if we factor that out, it is simply a quadratic in $r$ that remains.

  • 0
    Oh okay I originally did that but I was not sure if it was legal because of the confusion with x.2012-06-13