c15 - 18.03 Class 15, March 13, 2004 Operators: Exponential...

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18.03 Class 15, March 13, 2004 Operators: Exponential shift law Undetermined coefficients [1] Operators. The ERF is based on the following calculation: D e^{rt} = r e^{rt} = rI e^{rt} so D^n e^{rt} = r^n I e^{rt} and (a_n D^n + . .. + a_0 I) e^{rt} = (a_n r^n + . .. + a_0) e^{rt} or p(D) e^{rt} = p(r) e^{rt} So to solve p(D) x = A e^{rt} , try x_p = B e^{rt} ; p(D) (B e^{rt}) = B p(D) e^{rt} = B p(r) e^{rt} so we should take B = A/p(r) : x_p = e^{rt}/p(r) . What if p(r) = 0? eg x" - x = e^{-t} . (*) The key to solving this problem is the behavior of D on products: (d/dt) (xy) = x' y + x y' In terms of operators: D(vu) = v Du + u Dv Especially: D(e^{rt} u) = e^{rt} Du + u r e^{rt} = e^{rt} ( Du + ru ) = e^{rt} ( D + rI ) u Apply D again: D^2 (e^{rt} u) = D( e^{rt} (D+rI)u ) = e^{rt} (D+rI)(D+rI) u = e^{rt} (D+rI)^2 u Use: let's try a variation of parameters approach to solving (*): Try for x = e^{-t} u Then D^2 x = e^{-t} (D-I)^2 u -1] x = e^{-t} I u ------------------------- e^{-t} = e^{-t} ( (D-I)^2 - I ) u so want ( (D-I)^2 - I ) u = 1
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This note was uploaded on 01/31/2011 for the course MAT 17A taught by Professor Staff during the Winter '08 term at UC Davis.

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c15 - 18.03 Class 15, March 13, 2004 Operators: Exponential...

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