How to solve a complicated equation?

5 次查看（过去 30 天）
Cola
Cola 2021-7-7
编辑： David Goodmanson 2021-7-16
There is a Equation G. How to obtain the values of α and β when G=0？
G=-(-Omega^3*tau + (alpha + beta)*Omega)^2 - (Omega^2 - alpha*f)^2.
The answer:
alpha = Omega^2*cos(Omega*tau)/f;
beta = Omega*(f*sin(Omega*tau) - Omega*cos(Omega*tau))/f.
CAN anyone help me with this issue??? Thanks!!!
  4 个评论
显示 2更早的评论
Cola
Cola 2021-7-16
@Star Strider I obtain the values, and thank you very much for the help!!!
syms Omega tau alpha beta f
[ A, B ] = solve( [ -Omega^3*tau + (alpha + beta)*Omega == 0, Omega^2 - alpha*f == 0 ], [ alpha, beta ] )

请先登录，再进行评论。

请先登录，再回答此问题。

采纳的回答

David Goodmanson
David Goodmanson 2021-7-16
编辑：David Goodmanson 2021-7-16
Hi Cola,
Since there is one equation and two unknowns, it must be possible to define, say, beta in terms of alpha, where alpha can be anything. For G = 0 we have
(-Om^3*t + (a+b)*Om)^2 = -(Om^2 - a*f)^2
so
-Om^3*t + (a+b)*Om = +-*i*(Om^2 - a*f)
where there are obvious notational substitutions for Omega, tau, alpha, beta, and the +- choice gives two different solutions. Solving for b,
b = (1/Om)*( Om^3*t -a*Om +-i*(Om^2 - a*f) )
where 'a' can be anything. Solving instead for a (this does not give a different family of solutions, rather the same ones expressed differently) gives
a = (Om^3*t +-i*Om^2 -b*Om)/(Om +-i*f)
Here the sign in the denominator (+ or -) has to match the sign in the denominator, and b can be anything. The choice b = 0 gives the solutions from Star Strider.

更多回答（0 个）

类别

在 Help Center 和 File Exchange 中查找有关 Calculus 的更多信息

标签

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by