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

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This discussion focuses on solving a coupled system of differential equations defined by dW/dt = αW − βV and dV/dt = −γV + δW. The proposed solutions are W(t) = A sin(ωt) + B cos(ωt) and V(t) = C sin(ωt) + D cos(ωt). The user seeks to determine the constants A, B, C, D, and ω by setting up a system of equations derived from differentiating W(t) and V(t) and equating coefficients of sin(ωt) and cos(ωt) to zero. The resulting homogeneous linear equations indicate that nontrivial solutions exist for specific values of ω.

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charlottewill
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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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