-b.1.3.17 y''+y'-6y=0 Find values of r of the form y = e^{rt}

[karush]

Spoiler: problem source

#17

Find values of $r$ of the form $y = e^{rt}$
$y''+y'-6y=0$
$r^2+r-6=(r-2)(r+3)\quad \therefore r=2 \quad r=-3$

well so far

it that all there is to do?
However I didn't see clearly what the purpose of this was or how it is eventually used.

 

[topsquark]

You are done.

The point is for y'' + y' - 6y = 0, subbing in \(\displaystyle y = e^{rt}\) gives you both possible, thus all, solutions in this case. So a second order linear homogeneous differential equation with constant coefficients has a solution \(\displaystyle y = Ae^{r_1t} + Be^{r_2t}\). This can be extended in general to all orders.

-Dan

 

[HOI]

The problem specifically asked for a solution of the form $y= e^{rx}$.

Presumably you know that the derivative of $e^{rx}$ is $re^{rx}$ and that its second derivative is r times the derivative of $e^{rx}$ so is $r(re^{rx})= r^2e^{rx}$.

So taking y in $y''+ y'- 6y= 0$ to be $e^{rx}$ we get $r^2e^{rx}+ re^{rx}- 6e^{rx}= 0$. There is a "$e^{rx}$" in each term and $e^{rx}$ is never 0 so we can divide by it, eliminating x entirely, and leaving $r^2+ r- 6= (r+ 3)(r- 2)= 0$.

That is, $e^{rx}$ satisfies the differential equation if and only if r is either 2 or -3. Since this is a linear homogeneous equation, any linear combination, $Ae^{2x}+ Be^{-3x}$, where A and B can be any constants, is also a solution. Since, further, this is a second order linear homotenous equation, the set of all solutions forms a two dimensional vector space so every solution is of that form- $e^{2x}$ and $e^{-3x}$ form a basis for that vector space.

 

[karush]

https://dl.orangedox.com/geAQogxCM0ZYQLaUju

a little bit updated