// Numbas version: exam_results_page_options {"name": "Numbastest uke 44: Differensiallikninger 2", "metadata": {"description": "", "licence": "None specified"}, "duration": 0, "percentPass": 0, "showQuestionGroupNames": false, "shuffleQuestionGroups": false, "showstudentname": true, "question_groups": [{"name": "Group", "pickingStrategy": "all-ordered", "pickQuestions": 1, "questionNames": ["", "", "", ""], "variable_overrides": [[], [], [], []], "questions": [{"name": "Line\u00e6re f\u00f8rsteordens differensiallikninger", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Elena Malyutina", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/7213/"}], "tags": [], "metadata": {"description": "", "licence": "None specified"}, "statement": "

Hint: For å skrive flere eksponenter etter hverandre, sett en parentes rundt eksponentene:  ^(...)

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Gitt differensiallikningen $y'(\\var{t})=\\var{a}y(\\var{t})+\\var{b}$

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Finn den integrerende faktoren

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$\\rho(\\var{t})=$[[0]]

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Ganger vi begge sider av likningen med $\\rho(\\var{t})$ og omformer litt får vi

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$\\rho(\\var{t})y'(\\var{t})-\\var{a}\\rho(\\var{t})y(\\var{t})=\\var{b}\\rho(\\var{t})$.

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Deriver $\\rho(\\var{t})y(\\var{t})$ på papir, sett først inn svaret du fikk a) for $\\rho(\\var{t})$. (Numbas vil ikke gi deg svaret her ;)

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Resultatet fra b) viser at vi kan bytte ut høyre side av likningen fra b), $\\rho(\\var{t})y'(\\var{t})-\\var{a}\\rho(\\var{t})y(\\var{t})$, med $\\frac{d}{d\\var{t}}[\\rho(\\var{t})y(\\var{t})]$.

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Vi vil så integrere begge sider av likningen

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$\\frac{d}{d\\var{t}}[\\rho(\\var{t})y(\\var{t})]=\\var{b}\\rho(\\var{t})$

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Integrer $\\frac{d}{d\\var{t}}[\\rho(\\var{t})y(\\var{t})]$ (sett først inn svaret du fikk a) for $\\rho(\\var{t})$) 

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$\\int\\frac{d}{d\\var{t}}[\\rho(\\var{t})y(\\var{t})]d\\var{t}=$[[1]]

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\n

Integrer $\\var{b}e^{\\var{a}\\var{t}}$

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$\\int\\var{b}e^{\\var{a}\\var{t}}d\\var{t}=$[[0]]+C

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Finn den generelle løsningen.

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$y(\\var{t})=$[[0]]

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Finn løsningen som oppfyller initialbetingelsen $y(0)=0$

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$y(\\var{t})=$[[0]]

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Hint: For å skrive flere eksponenter etter hverandre, sett en parentes rundt eksponentene:  ^(...)

\n

Gitt differensiallikningen $y'+\\frac{1}{\\var{t}}y=\\var{b}\\var{t}+\\var{a}$

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Finn den integrerende faktoren

\n

$\\rho(\\var{t})=$[[0]]

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Integrer $\\var{t}(\\var{b}\\var{t}+\\var{a})$

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$\\int\\var{t}(\\var{b}\\var{t}+\\var{a})d\\var{t}=$[[0]]+C

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Finn den generelle løsningen.

\n

$y(\\var{t})=$[[0]]

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Finn løsningen som oppfyller initialbetingelsen $y(1)=\\simplify{{b}/3}$

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$y(\\var{t})=$[[0]]

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Hint: For å skrive flere eksponenter etter hverandre, sett en parentes rundt eksponentene:  ^(...)

\n

Gitt differensiallikningen $y'=\\var{t}^{\\var{a}}y^2$

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Finn den generelle løsningen.

\n

$y(\\var{t})=$[[0]]

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Finn løsningen som oppfyller initialbetingelsen $y(0)=1

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$y(\\var{t})=$[[0]]

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Hint: For å skrive flere eksponenter etter hverandre, sett en parentes rundt eksponentene:  ^(...)

\n

Gitt differensiallikningen $\\frac{dy}{d\\var{t}}=\\simplify{y^2+({b}+{c})y+{b}{c}}$

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Faktoriser høyre side av likningen

\n

$\\simplify{y^2+({b}+{c})y+{b}{c}}$=[[0]]

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Finn den generelle løsningen.

\n

$y(\\var{t})=$[[0]]

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Konstatløsningene til differensiallikningen er $y\\equiv$ [[0]] og $y\\equiv$ [[1]]

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