# [SOLVED]b.1.3.18 >values of r for de

#### karush

##### Well-known member
$y'''-3y''+2y'=0$
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?

#### HallsofIvy

##### Well-known member
MHB Math Helper
$y'''-3y''+2y'=0$
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?
Does your text really say that, using the word "rewrite"? You 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 $e^{rx}$ is never 0, we must have $r^3- 3r^2+ 2r= r(r-2)(r-1)= 0$.

#### karush

##### Well-known member
Does your text really say that, using the word "rewrite"? You 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 $e^{rx}$ is never 0, we must have $r^3- 3r^2+ 2r= r(r-2)(r-1)= 0$.

#### karush

##### Well-known member
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

Last edited:

#### topsquark

##### Well-known member
MHB Math Helper
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
What??

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

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.