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IX. Opgaver – 3. Januar 2004 580 Egenværdierne er rødderne −2 og 6 ([LA] Sætning 14). 2) De partielle afledede er I (0,0) er gradienten fx = 2x + 4y + 2xy, fy = 2y + 4x + x 2 . ∇f(0,0) = (fx(0,0),fy(0,0)) = (0,0) , så den nødvendige betingelse for et lokalt ekstremum er opfyldt. De dobbelte partielle afledede er I (0,0) er Hessematricen fxx fxy fxx = 2 + 2y, fxy = fyx = 4 + 2x, fyy = 2 . fyx fyy = 2 4 4 2 netop matricen fra 1). Da egenværdierne er = 0 og ikke har samme fortegn, så er (0,0) ikke et lokalt ekstremum ([LA] 13 side 89). Alternativt giver andenordenstesten ([S] 11.7 Theorem 3) en størrelse D = fxxfyy − f 2 xy = −12. Da D < 0, er (0,0) ikke et lokalt ekstremum. Opgave 7. Betragt differentialligningen y ′ = −2xy + x . 1) Angiv den fuldstændige løsning. 2) Angiv den løsning y(x), der opfylder y(0) = 1 2 .
IX. Opgaver – 3. Januar 2004 581 y 1 0 1 Retningsdiagram og to skitserede grafer Løsning. 1) Graferne for den fuldstædige løsning kan skitseres ud fra retningsdiagrammet af små linjestykker igennem (x,y) med hældning y ′ (x) Med notationen ([DL] 1.5, 1.6) er stamfunktionerne a(x) = −2x, b(x) = x A(x) = a(x)dx = −2x dx = −x 2 , B(x) = e −A(x) b(x)dx = e x2 x dx 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 - 6. Maksimum/mi
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I. Differentiation - 7. Lagrangemet
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II. Integration - 1. Dobbelt integr
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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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1. Vektorer og matricer V Matricer
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V. Matricer - 4. Determinanter = (
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