Consider the differential equation with constant coefficients y"(x)+ay'(x)+by(x)=u(x) with some unknown but fixed initial condition. It is known that y(x)=sinx when u(x)=sinx. Please calculate y(x)= ? when u(x)=cosx. 1 <解>　　y(x)=-cos(√2x)+──sin(√2x)+cos(x) √2 我是这样算的 (sinx)"+a(sinx)'+bsinx=sinx -sinx+acosx+bsinx=sinx acosx+bsinx=2sinx a=0 , b=2 原式=> y"+2y=u(x)=cosx yh=c1cos(√2x)+c2sin(√2x) yp=cosx y(x)=c1cos(√2x)+c2sin(√2x)+cosx 卡关了 c1 c2怎麽求~ 有请高手帮忙 谢谢!! --

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