lecture_39 (dragged) 3

lecture_39 (dragged) 3 - u t = ² t Ψ τ − 1 g τ dτ =...

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4 MA 36600 LECTURE NOTES: WEDNESDAY, APRIL 29 We substitute this into the original equation: d dt x ( p ) = P ( t ) x ( p ) + g ( t ) Ψ · d dt u + d dt Ψ · u = P ( t ) Ψu + g ( t ) ° Ψ ( t ) · d dt u g ( t ) ± = ° d dt Ψ P ( t ) Ψ ± · u = 0 = Ψ ( t ) · d dt u = g ( t ) . Upon inverting and integrating, we fnd that
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Unformatted text preview: u ( t ) = ² t Ψ ( τ ) − 1 g ( τ ) dτ = ⇒ x ( p ) = Ψ ( t ) ² t Ψ ( τ ) − 1 g ( τ ) dτ. We have seen this same Formula many times beFore....
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