[理工] [工數]-ODE
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
我是這樣算的
(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
卡關了 c1 c2怎麼求~
有請高手幫忙 謝謝!!
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