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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?

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2 Answers 2

up vote 5 down vote accepted

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.

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I am not sure what is meant by the weird 0 with a tail and the upside down As. I am not sure on how to find a solution but with trial and error I found r = 1/2 –  user138246 Jun 13 '12 at 23:31
    
@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. $(x+b/2)^2 = b^2/4 - c$. –  Ben Millwood Jun 14 '12 at 1:02

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.

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I do not understand why x goes away. –  user138246 Jun 13 '12 at 23:35
    
@Jordan: we divide $(2r^2+r-1)e^{rx}=0$ by $e^{rx}$, which is possible because $e^{rx}\ne0$, to obtain $2r^2+r-1=0$. –  anon Jun 13 '12 at 23:54
    
Oh okay I originally did that but I was not sure if it was legal because of the confusion with x. –  user138246 Jun 13 '12 at 23:56

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