Could you help me with this equation y'' = 1/y^2

- nikita2

Could you help me with this equation y'' = 1/y^2

- jamiebookeater

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- JamesJ

This is a tricky equation. I'll come back in a minute and help you with it.

- anonymous

are u asking for y?

- Shayaan_Mustafa

it is a differential equation.

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## More answers

- Shayaan_Mustafa

ask me what are you trying to ask?

- nikita2

yes it is differential eq.

- Shayaan_Mustafa

so what are asking for? finding general solution or anything else?

- nikita2

general

- 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

you just get relax nikita2. I am here for help. take time and post a correct and complete question.

- nikita2

ok, I will )

- Shayaan_Mustafa

ok

- JamesJ

@shayaan, do you know how to find the particular solution? This is quite non-standard.

- Shayaan_Mustafa

yes i know. there are several methods. but good one is variation of parameters .

- JamesJ

No, that's not correct. This is non-linear equation.

- nikita2

\[ (D _{2}^{t}u)^{2} (D _{2}^{x}D _{2}^{t}u) = 1\]
And i should to solve Cauchy problem

- Shayaan_Mustafa

is it non-linear?

- nikita2

yes(

- Shayaan_Mustafa

hmmm... then fine.

- JamesJ

y = y(t), y'' - 1/y^2 = 0. Not linear, unfortunately.

- Shayaan_Mustafa

do you know how to start with cauchy euler method??
you get start then we are here to volunteerily

- nikita2

yes i'll try it now

- 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

@JamesJ
give him reason why we did so?
So that he could get clear concepts.
Thanks.

- nikita2

Now integrate both sides and we have ....
\[(y')^2/2 = -1/3*1/y^3\]

- nikita2

+ Const

- JamesJ

\[ \frac{1}{2} (y')^2 = -\frac{1}{y} + C \]

- nikita2

Oh! i've done mistake here!

- 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

i see...

- JamesJ

Interesting, eh?
I only figured this out recently myself, working it out in this problem:
http://openstudy.com/study#/updates/4f0de661e4b084a815fcff56

- nikita2

Oh! I've not notise that it is forse of atraktion!

- 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

Tricky, but probably not impossible analytically.
What's the original motivation for this problem?

- 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

=1

- JamesJ

right.
btw, you might find this site helpful:
math.stackexchange.com

- JamesJ

but in the meantime, it would be great if you wanted to help out here answering some elementary questions now and again!

- nikita2

I did it and I will )

- nikita2

Thank you very much!

- JamesJ

Happy to help. This was a fun problem.

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