Diff. Equation Help: (1+y^3)y' = x^2

mammothrob

Junior Member
Joined
Nov 12, 2005
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91
Determine the region in the xy-plane for which

(1+y^3)y' = x^2

This has a unique solution.

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Not really understanding what this is asking of me?

I solved this as a seperable diff. Equation.

(1+ y^3)dy = (x^2) dx

y + (1/4)y^4 = (1/3) x^3 + c

3y + (3/4)y^4 = x^3


not sure what this one wants from me...

the general solution to which all the solution curves can be reached from?


any help is appreciated.

Rob
 
I do not know how to solve this in closed form, but I offer a solution.
Perhaps someone will supply a answer in closed form


[1+y^3]dy/dx=x^2
[1+y^3] dy = x^2 dx
y+y^4/4 =x^3/3 multiply both sides by 12
12y +3y^4 =4x^3
x^3= 1/4 [ 12y +3y^4]
x^3=[3y +3/4 y^4]
x=[3y+3/4 y^4]^1/3

of the 3 roots only one is real ,and two are complex. We are only interested in the real root
y=0 x=0
y=1 x=1.55
y=2 x=2.62
y=3 x=4.12

this curve is the upper limit

the lower limit is:
y=-1 x=imaginary
y=-1.59 x=0
y=-2 x=2.29
y=-3 x=3.73

the area in the xy plane is the area bounded by the upper curve and the lower curve

Arthur
 
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