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In this notebook you were asked to find the first two terms in the Fourier cosine series representation of 10∣cos(120πt)∣

10∣cos(120πt)∣=A₀+A₁cos(240πt)
and to then solve the differential equation
dtdV+V=A₀+A₁cos(240πt)
Find A₀,A₁, the first two Fourier coefficients in the expansion of 10∣cos(120πt)

1 Answer

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In the Fourier cosine series of 10|cos(120πt)|, A₀ = 10/3 and A₁ = 0. The differential equation becomes dV/dt + V = 10/3 + 0, with V(t) = [10/3 + C]exp(-t).

Fourier Coefficients:

  • Due to the absolute value, the function has even symmetry, eliminating odd harmonics.
  • Since only 120πt appears, A₂ and higher coefficients are also 0.
  • Applying the cosine series formula on [-π/2, π/2] yields A₀ = (2/π) ∫_(π/2)^(-π/2) 10|cos(120πt)| cos(0) dt = 20/π ∫_(π/2)^(-π/2) cos(120πt) dt = 10/3.
  • Likewise, A₁ = (2/π) ∫_(π/2)^(-π/2) 10|cos(120πt)| cos(240πt) dt = 0.

Solving Differential Equation:

  • Plug A₀ and A₁ into the differential equation: dV/dt + V = 10/3 + 0.
  • This is a first-order linear differential equation, solved by separation of variables: V(t) = [∫(10/3) dt]exp(-t) + C = [10/3 + C]exp(-t).

Therefore, the first two terms of the Fourier series are 10/3 and 0, resulting in a constant solution for the differential equation with an exponential decay factor.

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