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

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JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.2This is a tricky equation. I'll come back in a minute and help you with it.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0it is a differential equation.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ask me what are you trying to ask?

nikita2
 4 years ago
Best ResponseYou've already chosen the best response.0yes it is differential eq.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so what are asking for? finding general solution or anything else?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok you need a general solution. There are two solution. first is complementary solution and other is particular solution. do you know this?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0you just get relax nikita2. I am here for help. take time and post a correct and complete question.

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.2@shayaan, do you know how to find the particular solution? This is quite nonstandard.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yes i know. there are several methods. but good one is variation of parameters .

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.2No, that's not correct. This is nonlinear equation.

nikita2
 4 years ago
Best ResponseYou've already chosen the best response.0\[ (D _{2}^{t}u)^{2} (D _{2}^{x}D _{2}^{t}u) = 1\] And i should to solve Cauchy problem

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.2y = y(t), y''  1/y^2 = 0. Not linear, unfortunately.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0do you know how to start with cauchy euler method?? you get start then we are here to volunteerily

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.2ok ... 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 ....

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@JamesJ give him reason why we did so? So that he could get clear concepts. Thanks.

nikita2
 4 years ago
Best ResponseYou've already chosen the best response.0Now integrate both sides and we have .... \[(y')^2/2 = 1/3*1/y^3\]

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.2\[ \frac{1}{2} (y')^2 = \frac{1}{y} + C \]

nikita2
 4 years ago
Best ResponseYou've already chosen the best response.0Oh! i've done mistake here!

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.2Now.... \[ \frac{dy}{dt} = \pm \sqrt{ \frac{ 2(Cy1)}{y}} \] This is separable and it isn't too hard to find an equation for t, t = t(y).

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.2Interesting, eh? I only figured this out recently myself, working it out in this problem: http://openstudy.com/study#/updates/4f0de661e4b084a815fcff56

nikita2
 4 years ago
Best ResponseYou've already chosen the best response.0Oh! I've not notise that it is forse of atraktion!

nikita2
 4 years ago
Best ResponseYou've already chosen the best response.0But anyway if i'll try to do next. Solve \[D _{2}^{t}u(t,x) = y(t,x)\] where y we found

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.2Tricky, but probably not impossible analytically. What's the original motivation for this problem?

nikita2
 4 years ago
Best ResponseYou've already chosen the best response.0I 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}

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.2right. btw, you might find this site helpful: math.stackexchange.com

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.2but in the meantime, it would be great if you wanted to help out here answering some elementary questions now and again!

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.2Happy to help. This was a fun problem.
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