MHB How Do I Find Constants A, B, C, D, and ω in This Coupled DE System?

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The discussion revolves around solving a coupled system of differential equations involving W(t) and V(t), expressed in terms of constants A, B, C, D, and ω. The user has differentiated the functions and substituted them back into the original equations, leading to two simultaneous equations that must equal zero. To find the constants, the coefficients of cos(ωt) and sin(ωt) must also equal zero, resulting in a system of four homogeneous linear equations. The solution typically yields trivial results, but specific values of ω may allow for nontrivial solutions. This approach highlights the complexity of finding the constants in relation to the parameters α, β, γ, and δ.
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We've been asked to solve this system of equations

dW/dt = = αW−βV
dV dt = −γV+δW

by proposing that

W (t) = A sin(ωt) + B cos(ωt)
V (t) = C sin(ωt) + D cos(ωt).

so I've differentiated both W(t) and V(t) to give W'(t) = ωA cos(ωt) - ωB sin(wt) and V'(t) = ωC cos(ωt) - ωD sin(wt) and replaced with the dW/dt and dV/dt in the system to give the expressions

ωA cos(ωt) - ωB sin(wt) = αW−βV

ωC cos(ωt) - ωD sin(wt) = −γV+δWand now I've replaced the W and V's on the RHS with W (t) = A sin(ωt) + B cos(ωt)
V (t) = C sin(ωt) + D cos(ωt) (originally given in the question) and I've factored out the constant and the parameters of cos(ωt) and sin (ωt) to give two simultaneous equations equal to zero which are

(ωA-αB+βD) cos(ωt) + (βC-αA-ωB) sin(ωt) = 0

(ωC+γD-δB) cos(ωt) + (γC-ωD-δA) sin(ωt) = 0

So how do I find the value of the constants A,B,C and D in terms of the parameters α,β,γ,δ ω?

Please can somebody help me :(

Thanks,

Charlotte
 
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Re: Difficult System help please :(

charlottewill said:
We've been asked to solve this system of equations

dW/dt = = αW−βV
dV dt = −γV+δW

by proposing that

W (t) = A sin(ωt) + B cos(ωt)
V (t) = C sin(ωt) + D cos(ωt).

so I've differentiated both W(t) and V(t) to give W'(t) = ωA cos(ωt) - ωB sin(wt) and V'(t) = ωC cos(ωt) - ωD sin(wt) and replaced with the dW/dt and dV/dt in the system to give the expressions

ωA cos(ωt) - ωB sin(wt) = αW−βV

ωC cos(ωt) - ωD sin(wt) = −γV+δWand now I've replaced the W and V's on the RHS with W (t) = A sin(ωt) + B cos(ωt)
V (t) = C sin(ωt) + D cos(ωt) (originally given in the question) and I've factored out the constant and the parameters of cos(ωt) and sin (ωt) to give two simultaneous equations equal to zero which are

(ωA-αB+βD) cos(ωt) + (βC-αA-ωB) sin(ωt) = 0

(ωC+γD-δB) cos(ωt) + (γC-ωD-δA) sin(ωt) = 0

So how do I find the value of the constants A,B,C and D in terms of the parameters α,β,γ,δ ω?

That last line should read: So how do I find the value of the constants A,B,C, D and ω in terms of the parameters α,β,γ,δ? In other words, ω is not one of the fixed parameters, it is up for grabs along with A,B,C and D.

Next, you want those two equations to hold for all values of t. The only way for that to happen is if all those coefficients of cos(ωt) and sin(ωt) are zero. Thus you are looking at the system of equations $$\begin{aligned}\omega A - \alpha B \phantom{{}+\gamma C} +\beta D &=0 \\ -\alpha A -\omega B +\beta C \phantom{{}+\gamma C} &=0 \\ -\delta B + \omega C + \gamma D &=0 \\ -\delta A \phantom{{}+\gamma B} +\gamma C -\omega D &=0. \end{aligned}$$

The is a set of four homogeneous linear equations in A,B,C and D, so normally the only solution will be when they are all 0. But for some values of the (variable) parameter $\omega$ there will be a nontrivial solution.

The details of solving that set of equations look a bit gruesome, but it looks as though that is what you are expected to do.
 

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