nikita2
  • nikita2
Could you help me with this equation y'' = 1/y^2
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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JamesJ
  • JamesJ
This is a tricky equation. I'll come back in a minute and help you with it.
anonymous
  • anonymous
are u asking for y?
Shayaan_Mustafa
  • Shayaan_Mustafa
it is a differential equation.

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Shayaan_Mustafa
  • Shayaan_Mustafa
ask me what are you trying to ask?
nikita2
  • nikita2
yes it is differential eq.
Shayaan_Mustafa
  • Shayaan_Mustafa
so what are asking for? finding general solution or anything else?
nikita2
  • nikita2
general
Shayaan_Mustafa
  • Shayaan_Mustafa
ok you need a general solution. There are two solution. first is complementary solution and other is particular solution. do you know this?
Shayaan_Mustafa
  • Shayaan_Mustafa
you just get relax nikita2. I am here for help. take time and post a correct and complete question.
nikita2
  • nikita2
ok, I will )
Shayaan_Mustafa
  • Shayaan_Mustafa
ok
JamesJ
  • JamesJ
@shayaan, do you know how to find the particular solution? This is quite non-standard.
Shayaan_Mustafa
  • Shayaan_Mustafa
yes i know. there are several methods. but good one is variation of parameters .
JamesJ
  • JamesJ
No, that's not correct. This is non-linear equation.
nikita2
  • nikita2
\[ (D _{2}^{t}u)^{2} (D _{2}^{x}D _{2}^{t}u) = 1\] And i should to solve Cauchy problem
Shayaan_Mustafa
  • Shayaan_Mustafa
is it non-linear?
nikita2
  • nikita2
yes(
Shayaan_Mustafa
  • Shayaan_Mustafa
hmmm... then fine.
JamesJ
  • JamesJ
y = y(t), y'' - 1/y^2 = 0. Not linear, unfortunately.
Shayaan_Mustafa
  • Shayaan_Mustafa
do you know how to start with cauchy euler method?? you get start then we are here to volunteerily
nikita2
  • nikita2
yes i'll try it now
JamesJ
  • JamesJ
ok ... multiply first both sides by y' dt \[ y'' y' dt = \frac{y'}{y^2} dt \] Writing \( d(y') = y' dt \) and \( dy = y' dt \) we have \[ (y')' d(y') = \frac{1}{y^2} dy \] Now integrate both sides and we have ....
Shayaan_Mustafa
  • Shayaan_Mustafa
@JamesJ give him reason why we did so? So that he could get clear concepts. Thanks.
nikita2
  • nikita2
Now integrate both sides and we have .... \[(y')^2/2 = -1/3*1/y^3\]
nikita2
  • nikita2
+ Const
JamesJ
  • JamesJ
\[ \frac{1}{2} (y')^2 = -\frac{1}{y} + C \]
nikita2
  • nikita2
Oh! i've done mistake here!
JamesJ
  • JamesJ
Now.... \[ \frac{dy}{dt} = \pm \sqrt{ \frac{ 2(Cy-1)}{y}} \] This is separable and it isn't too hard to find an equation for t, t = t(y).
nikita2
  • nikita2
i see...
JamesJ
  • JamesJ
Interesting, eh? I only figured this out recently myself, working it out in this problem: http://openstudy.com/study#/updates/4f0de661e4b084a815fcff56
nikita2
  • nikita2
Oh! I've not notise that it is forse of atraktion!
nikita2
  • nikita2
But anyway if i'll try to do next. Solve \[D _{2}^{t}u(t,x) = y(t,x)\] where y we found
JamesJ
  • JamesJ
Tricky, but probably not impossible analytically. What's the original motivation for this problem?
nikita2
  • nikita2
I solving the Cauchy problem \[(D _{2}^{t}u(t,x))^2*(D _{2}^{x}D _{2}^{t}u(t,x)) =\] the initial condition is on {t=0}
nikita2
  • nikita2
=1
JamesJ
  • JamesJ
right. btw, you might find this site helpful: math.stackexchange.com
JamesJ
  • JamesJ
but in the meantime, it would be great if you wanted to help out here answering some elementary questions now and again!
nikita2
  • nikita2
I did it and I will )
nikita2
  • nikita2
Thank you very much!
JamesJ
  • JamesJ
Happy to help. This was a fun problem.

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