
Arturo O. answered 11/07/17
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I suggest integration.
y’ = 3xy3
y'/y3 = 3x
y-3dy = 3xdx
y-2/2 = 3x2/2 + c
y(0) = 1 ⇒
(1)-2/2 = 3(0)2/2 + c
1/2 = c
y-2/2 = 3x2/2 + 1/2 = (1/2)(3x2 + 1)
y = √[4/(3x2 + 1)]
Now plug in x = 1 to get y(1).