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IV. Differentialligninger – 2. 1. ordens ligninger 340 Opgave 1 - fortsat y(x) = Ce 3x 1 2 som opfylder (y1(0),y2(0)) = C(1,2) = (1,2) fås for C = 1. Den ønskede løsning skrevet ud y1(x) = e 3x y2(x) = 2e 3x 2.32. Opgave ☞ [LA] 15 Lineært system Opgave 2 Angiv den fuldstændige løsning til differentialligningssystemet Løsning Koefficientmatricen er y ′ 1 = y1 + 2y2 − 8 y ′ 2 = 2y1 + y2 − 7 A = 1 2 2 1 2.33. Opgave ☞ [LA] 15 Lineært system
IV. Differentialligninger – 2. 1. ordens ligninger 341 Opgave 2 - fortsat Egenværdierne findes som rødder i det karakteristiske polynomium |A − λI2| = 1 − λ 2 2 1 − λ Egenværdier = λ 2 − 2λ − 3 λ1 = −1, λ2 = 3 2.34. Opgave ☞ [LA] 15 Lineært system Opgave 2 - fortsat Egenvektorer hørende til egenværdien −1: A + I = 2 2 ∼ 2 2 giver egenvektorer x1 −x2 = x2 x2 = x2 1 1 0 0 −1 1 2.35. Opgave ☞ [LA] 15 Lineært system Opgave 2 - fortsat
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CALCULUS "SLIDES" TIL CALCULUS 1 +
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3. Generelle områder 201 4. Koordi
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Forord "Slides" til forelæsningern
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I. Differentiation - 1. Kontinuitet
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I. Differentiation - 4. Kædereglen
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I. Differentiation - 7. Lagrangemet
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II. Integration - 1. Dobbelt integr
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III. Potensrækker - 3. Potensrækk
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V. Matricer - 2. Lineære afbildnin
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V. Matricer - 3. Lineære ligninger
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V. Matricer - 4. Determinanter 419
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V. Matricer - 4. Determinanter = (
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1. Egenvektorer VI Egenvektorer og
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VI. Egenvektorer og diagonalisering
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VIII. Appendiks - 1. Polære koordi
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Litteratur • Arnold, Ordinary dif
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0-række/søjle, 423 n-te afsnitssu
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Identitetsmatricen, 383 Identitetsm
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