Question

设连续函数f(x)满足f(x)=sinax-（tf(x-t)0到x的积分）（a>0）,求f(x)

Answer 1

∫(0,x)tf(x-t)dt 令u=x-t
=-∫(x,0)(x-u)f(u)du
=∫(0,x)(x-u)f(u)du
∴f(x)=sin(ax)+∫(0,x)(u-x)f(u)du
=sin(ax)+∫(0,x)uf(u)du-x∫(0,x)f(u)du
则f'(x)=acos(ax)-∫(0,x)f(u)du
f"(x)=-a^2*sin(ax)-f(x)
即f"(x)+f(x)=-a^2*sin(ax)
当a=1,特解f0(x)=x*cosx/2
当a≠1,特解f0(x)=a^2*sin(ax)/(a^2-1)
f"(x)+f(x)=0 特征根方程为λ^2+1=0,得λ=±i
则f(x)=f0(x)+p*sinx+q*cosx
又f(0)=0,f'(0)=a
解得：
a=1,f(x)=(sinx+x*cosx)/2
a≠1,f(x)=[a^2*sin(ax)-a*sinx]/(a^2-1)