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IX. Opgaver – 2. Januar 2003 564 A(x) = a(x)dx = −x −1 dx = −lnx B(x) = e −A(x) b(x)dx = e ln x 2x −1 dx = 2x 2.26. Løs differentialligning ☞ Matematik Alfa 1, Januar 2003 Opgave 5 - løsning Som giver fuldstændig løsning y(x) = Ce A(x) + B(x)e A(x) = Ce − ln x − ln x + 2xe = C 1 + 2 x 2.27. Løs differentialligning ☞ Matematik Alfa 1, Januar 2003
IX. Opgaver – 2. Januar 2003 565 Opgave 5 - retningsfelt y 0 1 I punktet (x,y) tegnes et kort linjestykke med hældning y ′ (x) = − y 2 x + x . 1 2.28. Løs differentialligning ☞ Matematik Alfa 1, Januar 2003 Opgave 5 - løsning I den partikulære løsning bestemmes C ved y(2) = 5. I alt er løsningen x y(2) = C 1 + 2 = 5 2 y(x) = 6 + 2 x
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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 - 2. Partielle a
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I. Differentiation - 3. Tangentplan
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I. Differentiation - 4. Kædereglen
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I. Differentiation - 5. Gradient 88
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I. Differentiation - 6. Maksimum/mi
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I. Differentiation - 7. Lagrangemet
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II. Integration - 1. Dobbelt integr
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II. Integration - 2. Itereret integ
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II. Integration - 3. Generelle omr
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II. Integration - 4. Koordinatskift
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III. Potensrækker - 1. l’Hospita
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III. Potensrækker - 2. Talfølger
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III. Potensrækker - 3. Potensrækk
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III. Potensrækker - 4. Taylorpolyn
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1. Grafiske/numeriske metoder IV Di
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IV. Differentialligninger - 1. Graf
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IV. Differentialligninger - 2. 1. o
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IV. Differentialligninger - 3. Gene
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1. Vektorer og matricer V Matricer
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V. Matricer - 1. Vektorer og matric
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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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VII. Skalarprodukt og projektion -
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VIII. Appendiks - 1. Polære koordi
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Page 514 and 515: VIII. Appendiks - 1. Polære koordi
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Page 607 and 608: Identitetsmatricen, 383 Identitetsm
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