求f（x）的表达式
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f(x)+2∫(0->x) f(t) dt =x^2
x=0, =>f(0) = 0
f(x)+2∫(0->x) f(t) dt =x^2
两边求导
f'(x)+2f(x) =2x
let
yg= A.e^(-2x)
yp= Bx+C
yp' = B
yp'+2yp = 2x
B+2(Bx+C) =2x
2Bx + (B+2C) = 2x
=>
2B=2 and B+2C =0
B=1 and C=-1/2
ie
yp=x - 1/2
f(x) = yg+ yp = A.e^(-2x) +x - 1/2
f(0) =0
A - 1/2 =0
A=1/2
ie
f(x) =(1/2)e^(-2x) +x - 1/2
x=0, =>f(0) = 0
f(x)+2∫(0->x) f(t) dt =x^2
两边求导
f'(x)+2f(x) =2x
let
yg= A.e^(-2x)
yp= Bx+C
yp' = B
yp'+2yp = 2x
B+2(Bx+C) =2x
2Bx + (B+2C) = 2x
=>
2B=2 and B+2C =0
B=1 and C=-1/2
ie
yp=x - 1/2
f(x) = yg+ yp = A.e^(-2x) +x - 1/2
f(0) =0
A - 1/2 =0
A=1/2
ie
f(x) =(1/2)e^(-2x) +x - 1/2
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