x^2*y'' + xy' + 9y = 0
請問此題該如何使用Cauchy–Euler equation求出general solution？
已知答案為y=c1cos(3lnx)+c2sin(3lnx)
利用柯西-歐拉方程求ODE的general solution
2014-11-18 9:52 am
回答 (2)
2014-11-18 12:46 pm
✔ 最佳答案
x^2*y" + xy' + 9y = 0Set x=e^u => du/dx=e^(-u) & u=ln(x)y'=(dy/du)*du/dx=Y'*e^(-u)y"=d[Y'e^(-u)]/du*(du/dx)=[Y"e^(-u)-Y'e^(-u)]*e^(-u)=(Y"-Y')e^(-2u)代入上式裡面:0=Y"-Y'+Y'+9=Y"+9 輔助方程式為:0=D^2+9 => D=+-3j; j=√(-1)=> y(x)=Y(u)=c1*cos(u)+c2*sin(u)=c1*cos[ln(x)]+c2*sin[ln(x)]=版主答案2014-11-18 04:47:27 補充：
Y'=dy/du
2014-11-18 4:55 pm
x^2*y'' + xy' + 9y = 0
令y=x^m
y' =mx^(m-1)
y" = (m-1)mx^(m-2)
=> (m-1)mx^m + mx^m + 9x^m = 0
(m-1)m + m +9 =0
m^2 +9 =0
m= +3i, -3i
x^3i = e^( i 3lnx) = cos( 3lnx) + i sin(3lnx)
x^-3i = e^{ i( -3lnx) } = cos( -3lnx) + i sin(-3lnx) =cos( 3lnx) - i sin(3lnx)
通解為 cos( 3lnx) 與sin(3lnx)之線性組合
=>y = C1cos (3 lnx) + C2 sin (3 lnx)
令y=x^m
y' =mx^(m-1)
y" = (m-1)mx^(m-2)
=> (m-1)mx^m + mx^m + 9x^m = 0
(m-1)m + m +9 =0
m^2 +9 =0
m= +3i, -3i
x^3i = e^( i 3lnx) = cos( 3lnx) + i sin(3lnx)
x^-3i = e^{ i( -3lnx) } = cos( -3lnx) + i sin(-3lnx) =cos( 3lnx) - i sin(3lnx)
通解為 cos( 3lnx) 與sin(3lnx)之線性組合
=>y = C1cos (3 lnx) + C2 sin (3 lnx)
收錄日期: 2021-04-30 19:16:37
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20141118000015KK00280