Consider the initial value problem dy/dx = y^(1/3), y(0)=0. Verify that for each constant b ≥ 0,?
(a) Verify that for each constant b ≥ 0, the function
y(x) = 0 if x ≤ b
y(x) = ((2/3)(x-b))^(3/2)
is a valid solution of the initial value problem, even at x = b.
Update:y(x) = ((2/3)(x-b))^(3/2) if x>b
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Verified answer
dy/dx = y^(1/3)
dy / y^(1/3) = dx
∫ y^(-1/3) dy = ∫ dx
y^(-1/3+1)/(-1/3+1) = x + b
(3/2) y^(2/3) = x + b
multiply both sides by 2
3 y^(2/3) = 2x+2b
divide both sides by 3
y^(2/3) = (2/3) x + (2/3) b
y^(2/3) = (2/3)(x-b)
Raise both sides to power (3/2)
y = ((2/3) (x-b))^(3/2)
y(0)=0
0 = (2/3) (0-b))
y(x) = ((2/3) (x-b))^(3/2) , if x > b