设f(x)满足f”(x)+2f’(x)+f(x)=0,且 f(0)=1,f’(0)=0,an=∫n+∞f(x)dx. 求f(x)及an;

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问题 设f(x)满足f”(x)+2f’(x)+f(x)=0,且
    f(0)=1,f’(0)=0,an=∫n+∞f(x)dx.
求f(x)及an

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答案已知方程的特征方程为r2+2r+1=0,得特征根为r1=r2=-1,故通解为 f(x)=C1e-x+C2xe-x. 由f(0)=1,f’(0)=0,得C1=C2=1.所以 f(x)=e-x+xe-x=(1+x)e-x. an=∫n+∞f(x)dx=∫n+∞(1+x)e-xdx=-∫n+∞(1+x)de-x=-[(1+x)e-xn+∞-∫n+∞e-xdx] =(1+n)e-n+e-n=ne-n+2e-n

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