Solving ODE Problem: y'=(y^3)/2, y(0)=1

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Homework Help Overview

The discussion revolves around solving the ordinary differential equation (ODE) y'=(y^3)/2 with the initial condition y(0)=1. Participants explore methods of separation of variables and integration while addressing the implications of complex numbers in their solutions.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to separate variables but encounters issues with negative values under a square root. Some participants suggest the inclusion of integration constants and question the handling of initial conditions. There is discussion about the necessity of complex numbers in the solution.

Discussion Status

Participants have provided various insights regarding the integration process and the use of initial conditions. There is an ongoing exploration of the implications of complex solutions, and while some guidance has been offered, multiple interpretations of the problem are still being discussed.

Contextual Notes

There is a noted concern about the undefined nature of 1/x at x=0 and the original poster's hesitation to use complex numbers due to the course context. The discussion reflects uncertainty about the correct application of initial conditions and the role of integration constants.

wakko101
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I'm having trouble finding the solution to the following ODE:

y'=(y^3)/2 with initial value y(0) = 1

I try to separate it but end up with

y^2=-1/x

which makes no sense, since you can't take the square root of a negative number.

Any help?

Cheers,
Lauren. =)
 
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You *can* take the square root of a negative number, but the answer will be complex. Don't forget your constants of integration also, which should help you solve this problem.
 
Last edited:
Even worse 1/x is not defined at x=0. You are forgetting to put in an integration constant.
 
Good point with the constants. So...

1/(y^3) dy = 1/2 dx Integrate both sides

-1/(2y^2) = x/2 + c

y^2 = 1/(-x - 2c)

this solves the problem with it being undefined at 0, but at this point, I guess I need to add my complex number i?

y = i/(x+2c), which would make c=i/2 and so y = 2i/(2x + ci)...I think.

Is this right?

we haven't really used complex numbers so far in this particular course, which was why I was hesitant to use it in the answer.

Cheers,
L.
 
Nooo. Just put y=0 and x=1 and solve for C. Those are your initial conditions. I don't think you have to solve for y.
 
okay...so, then, plugging in my initial values at the Y^2 point, I get:

1^2=-1/(0+2c) ==> c=-1/2 ==> y^2=-1/(x-1) ==>

y^2 = 1/(1-x) ==> y = sqrt(1/(1-x))

Is that it?

Cheers,
L.
 
Yep, that's it.
 
Thanks...much appreciated. =)
 

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