Second order homogeneous differential equation with constant coefficients?

mx''+kx=0

x''+x=0

Solve the corresponding auxiliary equation r²+1=0, r=±i, so the homogeneous solution is

x(t)=Acost+Bsint

Plugging in the initial values:

1=Acos(0)+Bsin(0)

1=A

x'(t)=-Asint+Bcost

0=-sin(0)+Bcos(0)

0=B

The solution is then x(t)=cos(t)

The above answerer has confused the particular solution with the particular integral. Just to clarify their misunderstanding, the particular integral is one part of the solution that depends on the right hand side of the equation, the particular solution is the final solution after all the constants have been solved for. Find the complementary function by solving the auxiliary equation: y'' + 2y' + 5 = 0 m² + 2m + 5 = 0 (m + 1)² + 4 = 0 (m + 1)² = -4 m + 1 = ±2i m = -1 ± 2i yᶜ = ℮ˉ˟[Asin(2x) + Bcos(2x)] yᶜ = [Asin(2x) + Bcos(2x)] / ℮˟ Find the particular integral by comparing coefficients: yᵖ = Cx² + Ex + F yᵖ' = 2Cx + E yᵖ'' = 2C yᵖ'' + 2yᵖ' + 5yᵖ = 15x² 2C + 2(2Cx + E) + 5(Cx² + Ex + F) = 15x² 2C + 4Cx + 2E + 5Cx² + 5Ex + 5F = 15x² 5Cx² + (4C + 5E)x + (2C + 2E + 5F) = 15x² 5C = 15 C = 3 4C + 5E = 0 5E = -4C E = -4C / 5 E = -12 / 5 2C + 2E + 5F = 0 5F = -2C - 2E F = (-2C - 2E) / 5 F = (-6 + 24 / 5) / 5 F = (6 / 5) / 5 F = -6 / 25 yᵖ = 3x² - 12x / 5 - 6 / 25 Find the general solution by combining these two parts: y = yᶜ + yᵖ y = [Asin(2x) + Bcos(2x)] / ℮˟ + 3x² - 12x / 5 - 6 / 25 y = 3x² - 12x / 5 - 6 / 25 + [Asin(2x) + Bcos(2x)] / ℮˟