推 wil0829ly:感謝y大 我懂了~ 11/21 11:31
※ 引述《wil0829ly (汪汪)》之銘言：
: 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!!
u(x)=sinx:
|
y(0) = sinx| = 0
|x=0
|
y'(0) = cosx| = 1
|x=0
求得IC拿來用
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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