Question

求解高数 微分方程题

Answer 1

φ(x)= e^x + ∫(0->x) tφ(t) dt -x∫(0->x) φ(t) dt
x=0, =>φ(0)= 1
φ(x)= e^x + ∫(0->x) tφ(t) dt -x∫(0->x) φ(t) dt
两边取导
φ'(x)= e^x + xφ(x) -xφ(x) -∫(0->x) φ(t) dt
= e^x -∫(0->x) φ(t) dt
x=0 =>φ'(0)= 1
两边取导

φ''(x)= e^x -φ(x)
φ''(x)+φ(x) = e^x
φp = Ce^x
φ''p +φp = e^x
2Ce^x = e^x
C=1/2
let
φ(x) = Acosx +Bsinx + (1/2)e^x
φ(0)=1
A+ 1/2 =1
A=1/2
φ'(x) = -(1/2)sinx +Bcosx + (1/2)e^x
φ'(0) = 1
B + 1/2=1
B=1/2
ie
φ(x) = (1/2)cosx +(1/2)sinx + (1/2)e^x