令x＝y＝0得f(0)＝0
f(x+h)＝e^x×f(h)＋e^h×f(x)
f(x+h)－f(x)＝(e^h－1)×f(x)＋e^x×f(h)＝(e^h－1)×f(x)＋e^x×(f(h)－f(0))
f'(x)
＝lim(h→0) [f(x+h)－f(x)]/h
＝lim(h→0) (e^h－1)/h×f(x)＋lim(h→0) e^x×(f(h)－f(0))/h
＝f(x)＋e^x×f'(0)
＝f(x)＋2e^x
此微分方程是一阶线性方程,套用通解公式得f(x)＝Ce^x＋2(x+1)e^x,C是任意常数. 由f(0)＝0得C＝－2
所以f(x)＝2xe^x