Initial Value Prob: y' = (3x^2)/[(3y^2)-4] where y(0)=1

mammothrob

Junior Member
Joined
Nov 12, 2005
Messages
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Im having a little issue figuring out this intial value problem.


Solve the Initial Value Problem
y' = (3x^2)/[(3y^2)-4] where y(0)=1

Looks like I can solve it as a seperable DE.

dy/dx = (3x^2)/[(3y^2)-4]

[(3y^2)-4] dy = (3x^2) dx

Integrating both sides.....

(y^3) - 4y = (x^3) + c

I don't see how to get this in terms of y = (explicitly)
to find my c...

Am I just missing some easy algebra or did I use the worng method for this one?


Thanks,
Rob
 
Re: Initial Value Problem

mammothrob said:
Im having a little issue figuring out this intial value problem.


Solve the Initial Value Problem
y' = (3x^2)/[(3y^2)-4] where y(0)=1

Looks like I can solve it as a seperable DE.

dy/dx = (3x^2)/[(3y^2)-4]

[(3y^2)-4] dy = (3x^2) dx

Integrating both sides.....

(y^3) - 4y = (x^3) + c

Now apply y = 1 when x = 0

(1^3) - 4*1 = 0 + C

C = -3

Easy algebra


I don't see how to get this in terms of y = (explicitly)
to find my c...

Am I just missing some easy algebra or did I use the worng method for this one?


Thanks,
Rob
 
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