Question

帮忙解答一下这道题，步骤要详细？

Answer 1

f(x) = sinx - ∫(x-t)f(t)dt
= sinx - ∫x * f(t)dt + ∫t * f(t)dt
= sinx - x * ∫f(t)dt + ∫t * f(t)dt
两边对 x 取微分，得到：
f'(x) = cosx - [∫f(t)dt + x * f(x)] + [x * f(x)]
= cosx - ∫f(t)dt
再进行一次求微分，得到：
f"(x) = -sinx - f(x)
所以，f(x) 满足的微分方程为：
f"(x) + f(x) + sinx = 0

Answer 2

f(x)
= sinx - ∫(0->x) (x-t) f(t) dt
= sinx - x∫(0->x) f(t) dt + ∫(0->x)tf(t) dt
f'(x)= cosx - ∫(0->x) f(t) dt - xf(x) + xf(x)
=cosx - ∫(0->x) f(t) dt
f''(x) = -sinx - f(x)
f''(x) +f(x) = -sinx
The aux. equation
r^2 +1 =0
r=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)+2Dcosx -2Csinx
yp'' +yp = -sinx
x(-Ccosx- Dsinx)+2Dcosx -2Csinx +x(Ccosx+ Dsinx) = -sinx
2Dcosx -2Csinx = -sinx
=> D=0 , C=1/2
ie
yp=(1/2)xsinx
y= yg+ yp = Acosx+Bsinx +(1/2)xsinx
y(0) =0 =>A=0
y= Bsinx +(1/2)xsinx
y'=Bcosx +(1/2)(xcosx + sinx)
y'(0)=1 => B=1
ie
y= sinx +(1/2)xsinx

Answer 3

f(x) = sinx - ∫(x-t)f(t)dt
= sinx - ∫x * f(t)dt + ∫t * f(t)dt
= sinx - x * ∫f(t)dt + ∫t * f(t)dt
两边对 x 取微分，得到：
f'(x) = cosx - [∫f(t)dt + x * f(x)] + [x * f(x)]
= cosx - ∫f(t)dt
再进行一次求微分，得到：
f"(x) = -sinx - f(x)
所以，f(x) 满足的微分方程为：
f"(x) + f(x) + sinx = 0