solve the differential equation?

Question

y'= (cube root x)(2x+8/x) if y(1)=3

Answer 1

Use superposition

y1' = 2x^(4/3)

y2' = 8x^(-2/3)

You can intergate these directly

y1 = 6/7*x^(7/3) + K

y2 = 24 x^(1/3)

Now add y1 and y2

y = y1 + y2 = 6/7*x^(7/3) +24 x^(1/3) +K

Apply y(1) = 3

3 = 6/7 +24 + K --->-21 - 6/7 = K =-153/7

y = 6/7*x^(7/3) +24 x^(1/3) -153/7

Answer 2

dy = x^(1/3)*(2x + 8/x)dx = [2x^(4/3) + 8x^(-2/3)]dx, so
y = [2x^(7/3)]/(7/3) + [8x^(1/3)]/[1/3) + C. Using y = 3 when x = 1, we find C = -93/35.