Undetermined Coefficients

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March 6th 2011, 10:49 AM
JJ007

Undetermined Coefficients

Set up the correct linear combination of functions with undetermined literal coefficients to use in finding a particular integral by the method of UC. (Do not actually find the particular integrals.)

$y''-6y'+8y = x^3+x+e^{-2x}$

$m^2-6m+8=0$

$y_c=C_1e^{2x}+C_2e^{4x}$

$S_1=(x^3,x^2,x,1)$
$S_2=(x,1)$
$S_3=(e^{-2x})$

S_2 is completely contained in S_1 but none are included in the complimentary function?

Thanks.
March 6th 2011, 11:39 AM
TheEmptySet

Quote:

Originally Posted by JJ007

Set up the correct linear combination of functions with undetermined literal coefficients to use in finding a particular integral by the method of UC. (Do not actually find the particular integrals.)

$y''+6y'+8y = x^3+x+e^{-2x}$

$m^2-6m+8=0$

$y_c=C_1e^{2x}+C_2e^{4x}$

$S_1=(x^3,x^2,x,1)$
$S_2=(x,1)$
$S_3=(e^{-2x})$

S_2 is completely contained in S_1 but none are included in the complimentary function?

Thanks.

You have a typo in your characteristic equation. It should read

$m^2+6m+8=(m+2)(m+4)$
March 6th 2011, 11:58 AM
JJ007

Quote:

Originally Posted by TheEmptySet

You have a typo in your characteristic equation. It should read

$m^2+6m+8=(m+2)(m+4)$

Sorry. The typo was in the original equation. So it's still (m-2)(m-4).
March 6th 2011, 12:03 PM
TheEmptySet

Since the complimentary solution does not appear in the right hand side you can just "guess" the most simple form

$y=Ax^3+Bx^2+Cx+D+Me^{-2x}$

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