※ 引述《wil0829ly (汪汪)》之铭言： : 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!! u(x)=sinx: | y(0) = sinx| = 0 |x=0 | y'(0) = cosx| = 1 |x=0 求得IC拿来用 u(x)=cosx: | y(0) = c1 cos(√2x) + c2 sin(√2x) + cosx| = c1 +1 = 0 |x=0 | y'(0) = -√2 c1 sin(√2x) + √2 c2 cos(√2x) - sinx| =c2 √2 = 1 |x=0 得 c1 = -1 , c2 = 1/√2 故 y(x) = -cos(√2x) + 1/√2 sin(√2x) + cosx --

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