Find general solution y" + 2y' = 3 + 4sin(2t)

Question

Answer: c1 + c2e^-2t + 3/2t - 1/2sin(2t) - 1/2cos(2t)

Answer 1

For y'' + 2y' equated to 0, write u² + 2u or u(u + 2) = 0, which gives u = 0 or u = -2.

Then build the homogeneous or complementary solution as y_c = C₁e^0t + C₂e^-2t or C₁ + C₂e^-2t.

For y'' + 2y' = 3 alone, one attempts y_p = A₀ (an undetermined constant), 

but A₀ is part of y_c (for C₁ = A₀). A₀ must then be modified.

Equate y_p to A₀t (which is not part of y_c) to gain the correct form for a particular solution.

For [y_p, y_p'_, y_p''] = [A₀t, A_0, 0], build 0 + 2A₀ = 3 or A₀ = 3/2. Then y_p is 1.5t.

For y'' + 2y' = 4sin 2t alone, assume a particular solution of the form

y_p = Asin 2t + Bcos 2t with

y_p' = 2Acos 2t − 2Bsin 2t and

y_p'' = -4Asin 2t − 4Bcos 2t.

From y_p'' + 2y_p' = 4sin 2t, come to

(-4A − 4B)sin 2t + (4A − 4B)cos 2t = 4sin 2t.

{-4A − 4B = 4; 4A − 4B = 0} has a simultaneous solution of A = B = -0.5.  

Then adjust y_p = Asin 2t + Bcos 2t to y_p = -0.5sin 2t + -0.5cos 2t.

The solution sought is then "tacked together" as y = C₁ + C₂e^-2t + 1.5t − 0.5sin 2t − 0.5cos 2t.