if dy/dx = 4x (x^2 + 8)^(1/3)
y = (4x/2x)(x^2 + 8)^(1/3+1) + c
= 2(x^2 + 8)^(4/3) + c
given: when x = 0, y = 0
0 = 2(0+8)^(4/3) + c
0 = 16 + c
y = 2(x^2 + 8)^(4/3) - 16
find dy/dx to verify
Find the equation of y if dy/dx = 4x (x^2 + 8)^(1/3) and y(0)=0.
3 answers
I get (using u = x^2+8)
dy = 2 u^(1/3) du
y = 2 * 3/4 u^(4/3) + c
y = 3/2 (x^2 + 8)^(4/3) + c
dy = 2 u^(1/3) du
y = 2 * 3/4 u^(4/3) + c
y = 3/2 (x^2 + 8)^(4/3) + c
Thanks oobleck
you are right
you are right