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#### karush

##### Well-known member

- Jan 31, 2012

- 2,886

rewrite as

$y^3-3y^2+2y=0$

then factor

$y(y-2)(y-1)=0$

so then

$r=0,1,2$

OK this one is probably obvious but I was wondering why y isn't referred to r in the first place?

- Thread starter karush
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- Thread starter
- #1

- Jan 31, 2012

- 2,886

rewrite as

$y^3-3y^2+2y=0$

then factor

$y(y-2)(y-1)=0$

so then

$r=0,1,2$

OK this one is probably obvious but I was wondering why y isn't referred to r in the first place?

- Jan 29, 2012

- 1,151

Does your text really say that, using the word "rewrite"? You

rewrite as

$y^3-3y^2+2y=0$

then factor

$y(y-2)(y-1)=0$

so then

$r=0,1,2$

OK this one is probably obvious but I was wondering why y isn't referred to r in the first place?

- Thread starter
- #3

- Jan 31, 2012

- 2,886

cannot"rewrite" the differential equation $y''- 3y''+ 2y'= 0$ as "$y^3- y^2+ 2y= 0$" because the derivatives are NOT powers. What your text book should (and probably does) say before this point is that if we look for solutions of the form $y= e^{rx}$ then $y'= re^{rx}$, $y''= r^2e^{rx}$, and $y'''= r^3e^{rx}$ so the equation becomes $r^3e^{rx}- 3r^2e^{rx}+ 2re^{rx}= (r^3- 3r^2+ 2r)e^{rx}$. And because $e^{rx}$ isnever0, we must have $r^3- 3r^2+ 2r= r(r-2)(r-1)= 0$.

- Thread starter
- #4

- Jan 31, 2012

- 2,886

ok well let me try this one..

Determine the values of**r** for which the given differential equation has solutions of the form $y = e^{rt}$

$y''-y=0$

then

$\exp\int 1 \, dx =e^{1t}=e^{rt}$

so

$r=1$

ok I wasn't sure about - sign

Determine the values of

$y''-y=0$

then

$\exp\int 1 \, dx =e^{1t}=e^{rt}$

so

$r=1$

ok I wasn't sure about - sign

Last edited:

- Aug 30, 2012

- 1,180

What??ok well let me try this one..

Determine the values ofrfor which the given differential equation has solutions of the form $y = e^{rt}$

$y''-y=0$

then

$\exp\int 1 \, dx =e^{1t}=e^{rt}$

so

$r=1$

ok I wasn't sure about - sign

How about this. Given the differential equation

\(\displaystyle a_n y^{(n)}(x) + a_{n-1} y^{n-1}(x) + \text{ ... } + a_1 y'(x) + a_0 y(x) = 0\)

We have a corresponding "characteristic equation"

\(\displaystyle a_n r^n + a_{n-1} r^{n-1} + \text{ ... } + a_1 r^1 + a_0 = 0\)

where the solutions for r give n solutions to the differential equation of the form

\(\displaystyle y(x) = be^{rx}\)

So in your example we have

\(\displaystyle y''(x) - y(x) = 0\)

This has a characteristic equation

\(\displaystyle r^2 - 1 = 0 \implies r = \{-1, ~ 1 \}\)

So we have solutions to the differential equation of the form \(\displaystyle y_1(x) = b_1 e^{-x}\) and \(\displaystyle y_2(x) = b_2 e^{x}\)

-Dan

Addendum: What HallsofIvy did proves what I did up here. He showed you where the characteristic equation comes from and why it works.