## nikita2 4 years ago Could you help me with this equation y'' = 1/y^2

1. JamesJ

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

2. anonymous

3. Shayaan_Mustafa

it is a differential equation.

4. Shayaan_Mustafa

5. nikita2

yes it is differential eq.

6. Shayaan_Mustafa

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

7. nikita2

general

8. 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?

9. Shayaan_Mustafa

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

10. nikita2

ok, I will )

11. Shayaan_Mustafa

ok

12. JamesJ

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

13. Shayaan_Mustafa

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

14. JamesJ

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

15. nikita2

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

16. Shayaan_Mustafa

is it non-linear?

17. nikita2

yes(

18. Shayaan_Mustafa

hmmm... then fine.

19. JamesJ

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

20. Shayaan_Mustafa

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

21. nikita2

yes i'll try it now

22. 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 ....

23. Shayaan_Mustafa

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

24. nikita2

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

25. nikita2

+ Const

26. JamesJ

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

27. nikita2

Oh! i've done mistake here!

28. 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).

29. nikita2

i see...

30. JamesJ

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

31. nikita2

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

32. nikita2

But anyway if i'll try to do next. Solve $D _{2}^{t}u(t,x) = y(t,x)$ where y we found

33. JamesJ

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

34. 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}

35. nikita2

=1

36. JamesJ

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

37. JamesJ

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

38. nikita2

I did it and I will )

39. nikita2

Thank you very much!

40. JamesJ

Happy to help. This was a fun problem.