求解积分方程{∫0to1f(xt)dt}=nf(x),答案是f（x）=C*（nx）^(1/n-1)

Question

求解积分方程{∫【0 to 1】f(xt)dt}=nf(x),
答案是f（x）=C*（nx）^(1/n-1)

Answer 1

令u=xt,则t=u/x
dt=du/x
∫【0 to 1】f(xt)dt
=∫(0,x) f(u)du/x
=(1/x)∫(0,x) f(u)du
=nf(x)
所以
∫(0,x) f(u)du=nxf(x)
两边对x求导
f(x)=nxf'(x)+nf(x)
nxdf(x)/dx=(1-n)f(x)
df(x)/f(x)=(1-n)/n * dx/x
两边积分
lnf(x)=(1-n)/n * lnx + C1
f(x)=C*x^[(1-n)/n]