Asked by Amy
                Find the equation of y if dy/dx = 4x (x^2 + 8)^(1/3) and y(0)=0.
            
            
        Answers
                    Answered by
            Reiny
            
    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
    
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
                    Answered by
            oobleck
            
    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
                    Answered by
            Reiny
            
    Thanks oobleck
you are right
    
you are right
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