Linear differential equation

Find the solution for the following differential equation which satisfies the given conditions:

2(d²x/dt²) + 5(dx/dt) + 2x = 5sin(t)

x = 2
dx/dt = -3
t = 0

You're supposed to change it to 2λ² + 5λ + 2 = 0 so then λ = -0.5 or λ = -2

So x = Ae^-0.5t + Be^-2t + a

Not sure what constitutes 'a' here.

But I'm not sure what to do next. x = asin(t)?

Thank you.

The general solution comprises the homogeneous complimentary(xc) solution and non-homogeneous particular(xp) solution.

x(t) = xc + xp
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2(d²x/dt²) + 5(dx/dt) + 2x = 5Sin(t)

Complimentary
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characteristic equation
2λ² + 5λ + 2 = 0
2λ² + 4λ + λ + 2 = 0
2λ(λ + 2) + 1(λ + 2) = 0
λ = -0.5, -2

therefore
xc = Ae^(-0.5t) + Be^(-2t)

Particular Solution
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since the non-homogeneous part is 5Sin(t) xp as to be in the form

xp = C₁Cos(t) + C₂Sin(t)
xp' = -C₁Sin(t) + C₂Cos(t)
xp'' = -C₁Cos(t) - C₂Sin(t)

Replace these in the original d.e 2(d²x/dt²) + 5(dx/dt) + 2x = 5Sin(t)

2(d²x/dt²) + 5(dx/dt) + 2x = 5sin(t)
2(-C₁Cos(t) - C₂Sin(t) ) + 5( -C₁Sin(t) + C₂Cos(t) ) + 2( C₁Cos(t) + C₂Sin(t)) = 5Sin(t)
5C₂Cos(t) - 5C₁Sin(t) = 5Sin(t)

equate like-terms
5C₂Cos(t) = 0
- 5C₁Sin(t) = 5Sin(t)

C₂ = 0
-5C₁ = 5
C₁ = -1

therefore
xp = C₁Cos(t) + C₂Sin(t) becomes
xp = -Cos(t)

general solution
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x(t) = xc + xp
x(t) = Ae^(-0.5t) + Be^(-2t) - Cos(t)
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Using initial Condisions
x = 2
dx/dt = -3
t = 0


x = Ae^(-0.5t) + Be^(-2t) - Cos(t)
x' = -0.5Ae^(-0.5t) - 2Be^(-2t) + Sin(t)

x = Ae^(-0.5t) + Be^(-2t) - Cos(t)
2 = Ae^(0) + Be^(0) - Cos(0)
2 = A + B - 1
A + B = 3

x' = -0.5Ae^(-0.5t) - 2Be^(-2t) + Sin(t)