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# 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

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- 六呎將軍Lv 71 decade agoFavorite 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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