m2 微分與積分

"Suppose that y'(0)=0"
y'*y"=y, then 3(y')^2*y"=3y*y'
Integrating both sides with respect to x, then (y')^3= (3/2)y^2+C
x=0, y=0, y'(0)=0, then (y')^3=(3/2)y^2
y'=k y^(2/3), k^3=3/2
y^(-2/3) y'= k, then 3y^(1/3)= kx+C
x=0, y=0, then 3y^(1/3)= kx
so, y=(kx/3)^3= x^3/18

Note: Suppose that y'(0)=a, then
(y')^3=(3/2)y^2+a^3= k^3 (y^2+b^2) [where, b^2=(2/3)a^3, k^3=3/2]
y'= k(y^2+b^2)^(1/3)
∫dy/(y^2+b^2)^(1/3) = kx+C
(No closed form for this integral)

唔係好肯定......
y=(e^x)-1