Originally Posted by

**HallsofIvy** Notice that you are [b]not[/b ]asked to solve this equation!

You are, instead, given two functions and asked to show that they are solutions. That is much easier. (Think about the difference between solving the equation [/tex]xe^{x-1}+ x^2- 2= 0[/tex] and showing that x= 1 **is** a solution!)

If y= 0, then, of course, y'= 0 so the equation, y'= (3/2)y^{1/3} becomes "0= 0".

If $\displaystyle y= x^{3/2}$, then $\displaystyle y'= (3/2)x^{3/2- 1}= (3/2)x^{1/2}= (2/3)y^{1/3}$ because $\displaystyle y= x^{3/2}$ gives immediately that $\displaystyle y^{1/3}= x^{1/2}$