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

[wakko101]

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. =)

 

[jhicks]

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.

 

[Dick]

Even worse 1/x is not defined at x=0. You are forgetting to put in an integration constant.

 

[wakko101]

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.

 

[Dick]

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.

 

[wakko101]

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.

 

[Dick]

Yep, that's it.

 

[wakko101]

Thanks...much appreciated. =)