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14 September, 12:14

Y''+Y'+Y=1

solve ordinary differential equation

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  1. 14 September, 12:42
    0
    y (t) = c₁ e^ (-1/2 t) cos (√3/2 t) + c₂ e^ (-1/2 t) sin (√3/2 t) + 1

    Step-by-step explanation:

    y" + y' + y = 1

    This is a second order nonhomogenous differential equation with constant coefficients.

    First, find the roots of the complementary solution.

    y" + y' + y = 0

    r² + r + 1 = 0

    r = [ - 1 ± √ (1² - 4 (1) (1)) ] / 2 (1)

    r = [ - 1 ± √ (1 - 4) ] / 2

    r = - 1/2 ± i√3/2

    These roots are complex, so the complementary solution is:

    y = c₁ e^ (-1/2 t) cos (√3/2 t) + c₂ e^ (-1/2 t) sin (√3/2 t)

    Next, assume the particular solution has the form of the right hand side of the differential equation. In this case, a constant.

    y = c

    Plug this into the differential equation and use undetermined coefficients to solve:

    y" + y' + y = 1

    0 + 0 + c = 1

    c = 1

    So the total solution is:

    y (t) = c₁ e^ (-1/2 t) cos (√3/2 t) + c₂ e^ (-1/2 t) sin (√3/2 t) + 1

    To solve for c₁ and c₂, you need to be given initial conditions.
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