由原方程得f(0)=0且f'(x)=2xf(2x/2)·2 +2x即f'(x)=4xf(x)+2xd[f(x)]/[2f(x)+1] = 2xdxd[2f(x)+1]/[2f(x)+1] = 4xdxln|2f(x)+1| = 2x²+C2f(x)+1=C e^(2x²)2f(0)+1=C，又f(0)=0，故C=1故2f(x)+1=e^(2x²)f(x)=[e^(2x²) -1]/2
