-b.1.3.19 Determine the values of r for t^2y''+4ty'+2y = 0

[karush]

1000
$\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]

$\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]

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]

There are some typos/issues with your work. See if you can spot them...:)

 

[karush]

$\displaystyle y=t^r, \quad y'=rt^{r-1}, \quad y''=r(r-1)t^{r-2}$
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 $r\in\{1,4\}$
a few minor adjustments