How to solve a complicated equation?

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!!!

[Star Strider]

There must be some sort of substitution or other assumption involved.

The Symboilic Math Toolbox arraives at an entirely different solution —

syms alpha beta f Omega tau
G=-(-Omega^3*tau + (alpha + beta)*Omega)^2 - (Omega^2 - alpha*f)^2

G = 

Gs = solve(G == 0, [alpha beta]);
alpha = Gs.alpha

alpha = 

alpha = simplify(alpha, 500)

alpha = 

beta = Gs.beta

beta = 

.

[Cola]

Thank you very much for the help!!!

[Cola]

@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 ] )

[Star Strider]

My pleasure!

Answer 1

[David Goodmanson]

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.

[Cola]

I really thank you and sincerely wish you all the best.