Question

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

Answer 1

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

Answer 2

are u asking for y?

Answer 3

it is a differential equation.

Answer 4

ask me what are you trying to ask?

Answer 5

yes it is differential eq.

Answer 6

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

Answer 7

general

Answer 8

ok you need a general solution.
There are two solution.
first is complementary solution and other is particular solution. do you know this?

Answer 9

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

Answer 10

ok, I will )

Answer 11

ok

Answer 12

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

Answer 13

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

Answer 14

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

Answer 15

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

Answer 16

is it non-linear?

Answer 17

yes(

Answer 18

hmmm... then fine.

Answer 19

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

Answer 20

do you know how to start with cauchy euler method??

you get start then we are here to volunteerily

Answer 21

yes i'll try it now

Answer 22

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 ....

Answer 23

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

Answer 24

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

Answer 25

+ Const

Answer 26

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

Answer 27

Oh! i've done mistake here!

Answer 28

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).

Answer 29

i see...

Answer 30

Interesting, eh?

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

Answer 31

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

Answer 32

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

Answer 33

Tricky, but probably not impossible analytically.

What's the original motivation for this problem?

Answer 34

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}

Answer 35

=1

Answer 36

right.

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

Answer 37

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

Answer 38

I did it and I will )

Answer 39

Thank you very much!

Answer 40

Happy to help. This was a fun problem.