f(x)=sinx+∫(0->x) (t-x).f(t)dt

x=0, => f(0) =0

f(x)

=sinx+∫(0->x) (t-x).f(t)dt

=sinx+∫(0->x) tf(t)dt -x∫(0->x) f(t)dt

f"(x)

=cosx + xf(x) -xf(x) -∫(0->x) f(t)dt

=cosx -∫(0->x) f(t)dt

x=0, => f"(0) = 1

f""(x)

=-sinx -f(x)

f""(x)+f(x) = -sinx

The aux.equation

p^2 +1=0

p=i or -i

let

yg= Acosx +Bsinx

yp=x(Ccosx +Dsinx)

yp" = x(-Csinx +Dcosx) +(Ccosx +Dsinx)

yp""

= x(-Ccosx -Dsinx) +(-Csinx +Dcosx)+ (-Csinx +Dcosx)

= x(-Ccosx -Dsinx) +2(-Csinx +Dcosx)

yp""+yp = -sinx

x(-Ccosx -Dsinx) +2(-Csinx +Dcosx) +x(Ccosx +Dsinx) =-sinx

2(-Csinx +Dcosx) =-sinx

-2C =-1 and 2D=0

C=1/2 and D=0

通解

f(x) = yg+yp=Acosx +Bsinx + (1/2)xsinx

f(0) =0, => A=0

f"(x) =Bcosx + (1/2)sinx +(1/2)x.cosx

f"(0) = 1, =>B=1

ie

f(x) =sinx + (1/2)xsinx