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.