Question

consider y''+ (1+ p(x) ) y= 0, where p is continuous in [x0, infinity ), lim p(x)= 0, integral of |p'(t)| dt < infinity.

show that all solution of this DE are bounded in [x0, infinity ).

Answer 1

i'm not sure

Answer 2

I don't think it helped

Answer 3

can you please rewrite it i wanna make sure

Answer 4

I just solved for y...
I don't know how that would help with p

Answer 5

yeah i need to solve it for y

Answer 6

you will have to solve y in terms of the integral of p

Answer 7

i didn't understand how did you get (y' v)' =0

Answer 8

\[y''v+v'y'=(vy')'\]

Answer 9

yep mistake I imagined y as y' for some reason

Answer 10

yeah

Answer 11

so we want to show...
\[x_0 \le y < \infty \\ \text{ given } \\ p \text{ is continuous on } [x_0,\infty) \\ \lim_{x \rightarrow \infty}p(x)=0 \\ \int\limits |p'(t)| dt <\infty \]
doesn't the last condition mean p(t) is a constant function

Answer 12

and that would mean it would have to be p(t)=0 since as x goes to infinity p goes to 0

Answer 13

\[p(x)=0 \\ \text{ so the equation really is } y''=0\]

Answer 14

y needs to be bonded by p

Answer 15

did I write the given parts above correctly?

Answer 16

yes

Answer 17

if you didn't understand my reasoning for p(x) being 0 above
\[\int\limits |p'(t)| dt =g(t) <\infty \\ \text{ means } g(t) \text{ is a constant number so } g(t)=C \in \mathbb{R}\\ \int\limits |p'(t)| dt=C \\ |p'(t)|=0 \\ p'(t)=0 \implies p(t)=K \in \mathbb{R} \\ \text{ and we are also given } \\ \lim_{x \rightarrow \infty} p(x)=0 \\ \text{ so we have } \\ \lim_{x \rightarrow \infty} K=0 \implies K=0 \\ \text{ so } p(x)=0\]

Answer 18

i got it

Answer 19

maybe I'm missing something...

Answer 20

so y< 0?

Answer 21

I don't know how you got that...
I'm gonna mention some people that I think might be better at this than me:
@jim_thompson5910 @sithsandgiggles @dan815 @ganeshie8 @IrishBoy123 @Kainui
someone correct me on sith's name pretty please

Answer 22

thanks a lot for your help