點樣solve 呢條 linear ordinary differential equation?
y'' + 6y' + 8y = 0
Thank you
點樣solve 呢條 linear ordinary differential equation?
2007-02-07 10:57 pm
回答 (2)
2007-02-07 11:15 pm
let y = exp(ax)
Then the eqn becomes
a^2 exp(ax) + 6a exp(ax) + 8 exp(ax) = 0
In another words, a^2+6a+8 = 0. So (a+4)(a+2) = 0, so a = -2 or -4
So y1(x) = exp(-2x), y2 = exp(-4x). So the general solution is
y = A*y1 + B*y2, where A and B can be any constants.
That is, y(x) = A*exp(-2x) + B*exp(-4x)
Then the eqn becomes
a^2 exp(ax) + 6a exp(ax) + 8 exp(ax) = 0
In another words, a^2+6a+8 = 0. So (a+4)(a+2) = 0, so a = -2 or -4
So y1(x) = exp(-2x), y2 = exp(-4x). So the general solution is
y = A*y1 + B*y2, where A and B can be any constants.
That is, y(x) = A*exp(-2x) + B*exp(-4x)
2007-02-07 11:08 pm
y'' + 6y' + 8y = 0?
如果你無打錯
6y^1=6y
y^2+6y+8y=0
y^2+14y=0
y=0 or -14
如果你打錯左, 我估係
y^3+6y^2+8y=0
y(y^2+6y+8)=0
y=0 or -2 or -4
如果你無打錯
6y^1=6y
y^2+6y+8y=0
y^2+14y=0
y=0 or -14
如果你打錯左, 我估係
y^3+6y^2+8y=0
y(y^2+6y+8)=0
y=0 or -2 or -4
收錄日期: 2021-04-12 23:18:45
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20070207000051KK01530