anonymous
  • anonymous
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 ).
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
i'm not sure
freckles
  • freckles
I don't think it helped
anonymous
  • anonymous
can you please rewrite it i wanna make sure

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freckles
  • freckles
I just solved for y... I don't know how that would help with p
anonymous
  • anonymous
yeah i need to solve it for y
freckles
  • freckles
you will have to solve y in terms of the integral of p
anonymous
  • anonymous
i didn't understand how did you get (y' v)' =0
freckles
  • freckles
\[y''v+v'y'=(vy')'\]
freckles
  • freckles
yep mistake I imagined y as y' for some reason
anonymous
  • anonymous
yeah
freckles
  • freckles
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
freckles
  • freckles
and that would mean it would have to be p(t)=0 since as x goes to infinity p goes to 0
freckles
  • freckles
\[p(x)=0 \\ \text{ so the equation really is } y''=0\]
anonymous
  • anonymous
y needs to be bonded by p
freckles
  • freckles
did I write the given parts above correctly?
anonymous
  • anonymous
yes
freckles
  • freckles
if you didn't understand my reasoning for p(x) being 0 above \[\int\limits |p'(t)| dt =g(t) <\infty \\ \text{ means } g(t) \text{ is a constant number so } g(t)=C \in \mathbb{R}\\ \int\limits |p'(t)| dt=C \\ |p'(t)|=0 \\ p'(t)=0 \implies p(t)=K \in \mathbb{R} \\ \text{ and we are also given } \\ \lim_{x \rightarrow \infty} p(x)=0 \\ \text{ so we have } \\ \lim_{x \rightarrow \infty} K=0 \implies K=0 \\ \text{ so } p(x)=0\]
anonymous
  • anonymous
i got it
freckles
  • freckles
maybe I'm missing something...
anonymous
  • anonymous
so y< 0?
freckles
  • freckles
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
anonymous
  • anonymous
thanks a lot for your help

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