# [SOLVED]de1.3.19 Determine the values of r for DE

#### karush

##### Well-known member
$\textsf{Determine the values of$r$for which the given differential equation has solutions of the form$ y = t^r$for$t > 0 $}$
$t^2y''+4ty'+2y = 0$
$\color{red}{r=-1,-2}$
$t^2y''-4ty'+4y=0$
$\color{red}{r=1,4}$
ok the book answers are in red, and obviously this is from a factored quadradic but I couldn't an example of the steps.
probably just a couple!

#### MarkFL

Staff member
$\textsf{Determine the values of$r$for which the given differential equation has solutions of the form$ y = t^r$for$t > 0 $}$
$t^2y''+4ty'+2y = 0$
$\color{red}{r=-1,-2}$
$t^2y''-4ty'+4y=0$
$\color{red}{r=1,4}$
ok the book answers are in red, and obviously this is from a factored quadradic but I couldn't an example of the steps.
probably just a couple!
Okay, if we are to consider a solution of the form:

$$\displaystyle y=t^r$$

Then:

$$\displaystyle y'=rt^{r-1}$$

$$\displaystyle y''=r(r-1)t^{r-2}$$

And thus the ODE becomes:

$$\displaystyle t^2r(r-1)t^{r-2}+4trt^{r-1}+2t^r=0$$

Or:

$$\displaystyle r(r-1)t^{r}+4rt^{r}+2t^r=0$$

$$\displaystyle t^r\left(r(r-1)+4r+2\right)=0$$

Now, if we exclude the trivial solution $$y\equiv0$$ this leaves:

$$\displaystyle r(r-1)+4r+2=0$$

$$\displaystyle r^2+3r+2=0$$

$$\displaystyle (r+1)(r+2)=0$$

Thus:

$$\displaystyle r\in\{-2,-1\}$$

Can you do the second one?

#### karush

##### Well-known member
given
$t^2y''-4ty'+4y=0$
let
$\displaystyle y=t^r, \quad y''=r(r-1)t^{r-2}, \quad y''=r(r-1)t^{r-2}$
then
$t^2(r(r-1)t^{r-2}-4rt^{r-1}+4t^r=0$
simplify
$(r(r-1)t^r-4rt^{r-1}+4t^r=r(r-1)-4r+4=r^2-5r+4=0$
factor
(r-1)(r-4)=0
thus
$r\in{1,4}$

typos maybe

#### MarkFL

Staff member
There are some typos/issues with your work. See if you can spot them... #### karush

##### Well-known member
$\displaystyle y=t^r, \quad y'=rt^{r-1}, \quad y''=r(r-1)t^{r-2}$
then
$t^2(r(r-1)t^{r-2})-4t(rt^{r-1})+4(t^r)=0$
then simplify
$(r - 4) (r - 1) t^r=0$
so $r\in\{1,4\}$