f(x)=∫(0到X)f(x-t)sint dt+x
f(0)=0
在积分∫(0到X)f(x-t)sint dt中
令x-t=u,t=x-u,t=0得u=x,t=x,得u=0
∫(0到X)f(x-t)sint dt=∫(x到0)f(u)sin(x-u)( -du)
∫(0到x)f(u)(sinxcosu-cosxsinu)du
=sinx∫(0到x)f(u)cosudu-cosx∫(0到x)f(u)sinudu
所以
f(x)=sinx∫(0到x)f(u)cosudu-cosx∫(0到x)f(u)sinudu+x
两边同时对x求导,得
f'(x)=cosx∫(0到x)f(u)cosudu+sinx*f(x)cosx+sinx∫(0到x)f(u)sinudu-cosx*f(x)sinx+1
f'(x)=cosx∫(0到x)f(u)cosudu+sinx∫(0到x)f(u)sinudu+1
f'(0)=1
再求导,得
f''(x)=-sinx∫(0到x)f(u)cosudu+cos^2xf(x)+cosx∫(0到x)f(u)sinudu+sin^2x*f(x)
=f(x)-sinx∫(0到x)f(u)cosudu+cosx∫(0到x)f(u)sinudu
=f(x)+(x-f(x))
=x
所以
f'(x)=x^2/2+c1
由f'(0)=1,得
c1=1
f'(x)=x^2/2+1
继续积分得
f(x)=x^3/6+x+c2
f(0)=0
得c2=0
所以
f(x)=x^3/6+x.
太难了,希望满意哦!