Anonymous
Anonymous asked in 科學及數學數學 · 1 decade ago

2nd ODE again....

Find the general solution to

y ′′ − 2y ′ − 3y = 64xe−x.

Update:

Find the general solution to

y ′′ − 2y ′ − 3y = 64xe−x.

Update 2:

y ′′ − 2y ′ − 3y = 64xe^−x.

Update 3:

y '' − 2y ′ − 3y = 64xe^−x.

1 Answer

Rating
  • 1 decade ago
    Favorite Answer

    The complementary solution is, with characteristic equation:

    λ2 - 2λ - 3 = 0

    λ = 3 or -1

    So yc = Ae3x + Be-x

    where A and B are some constants.

    For the particular solutions:

    圖片參考:http://i388.photobucket.com/albums/oo325/loyitak19...

    So the complete solution is:

    y = Ae3x + Be-x - e-x(x2/2 + 4x)

    since B is an arbitrary constant.

    Source(s): Myself
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