Thread: solving D.E by undetermined coefficients.

question: y'' + 2y' - 24y = 16 - (x+2)e^4x

what i've done:

yh : λ^2 + 2λ - 24 = 0
(λ+6)(λ-4) = 0
yh = c1e^(-6x) + c2e^4x

Try: yp = (A1x)e^4x <--- this is ok because it doesn't appear in yh

y'p= (4A1x)e^4x + A1e^4x

y''p= (16A1x)e^4x + 4A1e^4x + 4A1e^4x = (16A1x)e^4x + 8A1e^4x

subbing into the D.E gives...

(16A1x)e^4x + 8A1e^4x + 2[(4A1x)e^4x + A1e^4x] - 24[(A1x)e^4x] = 16 - (x+2)e^4x

so grouping like terms gives: xe^4x: 16A1 + 8A1 - 24A1= 0 which is ok
e^4x: 8A1 + 2A1 = x+2 , A1= (x+2)/10
but where does the constant(16) come from?

Originally Posted by linalg123

question: y'' + 2y' - 24y = 16 - (x+2)e^4x

what i've done:

yh : λ^2 + 2λ - 24 = 0
(λ+6)(λ-4) = 0
yh = c1e^(-6x) + c2e^4x

Try: yp = (A1x)e^4x <--- this is ok because it doesn't appear in yh

y'p= (4A1x)e^4x + A1e^4x

y''p= (16A1x)e^4x + 4A1e^4x + 4A1e^4x = (16A1x)e^4x + 8A1e^4x

subbing into the D.E gives...

(16A1x)e^4x + 8A1e^4x + 2[(4A1x)e^4x + A1e^4x] - 24[(A1x)e^4x] = 16 - (x+2)e^4x

so grouping like terms gives: xe^4x: 16A1 + 8A1 - 24A1= 0 which is ok
e^4x: 8A1 + 2A1 = x+2 , A1= (x+2)/10
but where does the constant(16) come from?

You is correct, but the form of your particular solution is incorrect.



but since is part of the homogeneous solution we need to multiply exponential part of the particular solution by this gives