# Thread: solving D.E by undetermined coefficients.

1. ## 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: 8
A1 + 2A1 = x+2 , A1= (x+2)/10
but where does the constant(16) come from?

2. 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: 8
A1 + 2A1 = x+2 , A1= (x+2)/10
but where does the constant(16) come from?
You $\displaystyle y_h$ is correct, but the form of your particular solution is incorrect.

$\displaystyle y_p=Ae^{4x}+Bxe^{4x}+C$

but since $\displaystyle e^{4x}$ is part of the homogeneous solution we need to multiply exponential part of the particular solution by $\displaystyle x$ this gives

$\displaystyle y_p=Axe^{4x}+Bx^2e^{4x}+C$

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# y" 2y'-24y=16-(x 2)e^4x

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