Question

微分方程 设二阶可微函数f(x)满足方程∫[0到x](t+1)f'(x-t)dt=x^2+e^x-f(x) 求f(x)

Answer 1

∫[0,x](t+1)f'(x-t)dt=x^2+e^x-f(x),设F'(t) = f(t)
x=0时,左边=0,右边=1-f(0),故f(0) = 1
左边 = -∫[0,x](t+1)d[f(x-t)]
= ∫[0,x]f(x-t)dt - (t+1)f(x-t)|(0,x)
= ∫[0,x]f(x-t)dt + f(x) - (x+1)
即∫[0,x]f(x-t)dt + f(x) - (x+1) = x^2 + e^x - f(x)
∫[0,x]f(x-t)dt = x^2 + e^x - 2f(x) + (x+1)
左边对x求导得
[∫[0,x+Δx]f(x+Δx-t)dt - ∫[0,x]f(x-t)dt]/Δx
= ∫[0,x][f(x+Δx-t)-f(x-t)]dt/Δx + ∫[x,x+Δx]f(x+Δx-t)dt/Δx
= ∫[0,x]f'(x-t)dt + [F(t)/Δx]|(x,x+Δx)
= -f(x-t)|(0,x) + f(x)
= 2f(x) - 1
右边对x求导得
2x + e^x - 2f'(x) + 1
2f(x) - 1 = 2x + e^x - 2f'(x) + 1
整理得
f'(x) + f(x) = x + (e^x)/2 + 1
解这个微分方程得
f(x) = x + (e^x)/4 + Ce^(-x)