y”+y’-2y=2t Please find general solution?

How did you get "-t-1/2" part? Thankx though!

2 Answers

  • General solution= Ke^(-2t)+Ke^t-t-1/2

  • Note that the above answer is wrong as there should be two arbitrary constants.

    Find the complementary function by solving the auxiliary equation:

    y'' - y' - 2 = 0

    m² + m - 2 = 0

    (m + 2)(m - 1) = 0

    m + 2 = 0 OR m - 1 = 0

    m = -2 OR m = 1

    yᶜ = C₁℮ᵗ + C₂℮^(-2t)

    yᶜ = C₁℮ᵗ + C₂ / ℮^(2t)

    Find the particular integral by comparing coefficients:

    yᵖ = At + B

    yᵖ' = A

    yᵖ'' = 0

    yᵖ'' - yᵖ' - 2yᵖ = 2t

    A - 2(At + B) = 2t

    A - 2At - 2B = 2t

    -2At + (A - 2B) = 2t

    -2A = 2

    A = -1

    A - 2B = 0

    2B = A

    B = A / 2

    B = -½

    yᵖ = -t - ½

    Find the general solution by combining these two parts:

    y = yᶜ + yᵖ

    y = C₁℮ᵗ + C₂ / ℮^(2t) - t - ½

Leave a Reply

Your email address will not be published. Required fields are marked *

Related Posts