设f(x)=sinx-∫x 0(x-t)f(t)dt其中f为连续函数,求f(x) 要详细说明,尤其是∫x 0(x-t)f(t)dt怎么求的导
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![设f(x)=sinx-∫x 0(x-t)f(t)dt其中f为连续函数,求f(x) 要详细说明,尤其是∫x 0(x-t)f(t)dt怎么求的导](/uploads/image/z/6798405-21-5.jpg?t=%E8%AE%BEf%28x%29%3Dsinx-%E2%88%ABx+0%28x-t%29f%28t%29dt%E5%85%B6%E4%B8%ADf%E4%B8%BA%E8%BF%9E%E7%BB%AD%E5%87%BD%E6%95%B0%2C%E6%B1%82f%28x%29+%E8%A6%81%E8%AF%A6%E7%BB%86%E8%AF%B4%E6%98%8E%2C%E5%B0%A4%E5%85%B6%E6%98%AF%E2%88%ABx+0%28x-t%29f%28t%29dt%E6%80%8E%E4%B9%88%E6%B1%82%E7%9A%84%E5%AF%BC)
设f(x)=sinx-∫x 0(x-t)f(t)dt其中f为连续函数,求f(x) 要详细说明,尤其是∫x 0(x-t)f(t)dt怎么求的导
设f(x)=sinx-∫x 0(x-t)f(t)dt其中f为连续函数,求f(x) 要详细说明,尤其是∫x 0(x-t)f(t)dt怎么求的导
设f(x)=sinx-∫x 0(x-t)f(t)dt其中f为连续函数,求f(x) 要详细说明,尤其是∫x 0(x-t)f(t)dt怎么求的导
f(x) = sinx - ∫(0~x) (x - t) f(t) dt
= sinx - x∫(0~x) f(t) dt + ∫(0~x) tf(t) dt,之后两边对x求导
f'(x) = cosx - [x' · ∫(0~x) f(t) dt + x · f(x)] + xf(x)
f'(x) = cosx - ∫(0~x) f(t) dt,两边再对x求导
f''(x) = - sinx - f(x)
==> y'' + y = - sinx,解微分方程
特征方程:r² + 1 = 0 => r = ±i
y = Acosx + Bsinx
p = x · (Acosx + Bsinx) = Axcosx + Bxsinx
p'' = - Axcosx - 2Asinx + 2Bcosx - Bxsinx,代入微分方程中
p'' + p = - sinx
(- Axcosx - 2Asinx + 2Bcosx - Bxsinx) + (Axcosx + Bxsinx) = - sinx
- 2Asinx + 2Bcosx = - sinx
解得A = 1/2,B = 0
p = (1/2)xcosx
通解为y = (1/2)xcosx + Acosx + Bsinx
所以f(x) = (1/2)xcosx + Acosx + Bsinx,其中A和B都是任意常数