// Numbas version: exam_results_page_options {"name": "Numbastest uke 43: Differensiallikninger", "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": "Hvilke av likninger er differensiallikningene?", "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": "All rights reserved"}, "statement": "

Hvilke av likningene er differensiallikninger?

", "advice": "

En differensiallikning (førsteordens) er en likning som inkluderer en ukjent funksjon (f.eks. $y(x), y(t), y, x(y), x(t)$ osv.), den deriverte av denne funksjonen og variabel.

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$y'+y+y^2=x$ er en differensiallikning med $y(x)$ som en ukjente funksjon

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$y'=y$ er en differensiallikning med $y(x)$ som en ukjente funksjon

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$y'+5e^{xy}=\\sin(x)$ er en differensiallikning med $y(x)$ som en ukjente funksjon

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$x'+x=y$ er en differensiallikning med $x(y)$ som en ukjente funksjon

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$x^2+y^2=y'$ er en differensiallikning med $y(x)$ som en ukjente funksjon

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$\\cos(y'(x))=\\sin(x)$ er en differensiallikning med $y(x)$ som en ukjente funksjon

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$y'(t)=cos(t)$ er en differensiallikning med $y(t)$ som en ukjente funksjon

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$y(t)=cos(t)$ er ikke en differensiallikning fordi den ikke inkluderer noen deriverte 

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$x^2+x=y$ er ikke en differensiallikning fordi den ikke inkluderer noen deriverte

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$z=x^2+y^2$ er ikke en differensiallikning fordi den ikke inkluderer noen deriverte

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$x^2+2x+1=0$ er ikke en differensiallikning fordi den ikke inkluderer noen deriverte

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$y'(x)+y(x)$ er ikke en likning

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Avgjør ved regning hvilke av følgende funksjoner er løsninger til differensiallikningene i listen til venstre

", "advice": "

1) $\\var{out}(\\var{inp})=\\simplify{{a}e^({b}{inp})+{c}}$ er løsningen til  $\\var{out}'=\\simplify{{a*b-a*d}e^({b}*{inp})-{c*d}+{d}{out}}$ fordi

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VS: $\\var{out}'=\\simplify{{a}*{b}e^({b}{inp})}$

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HS: $\\simplify{{a*b-a*d}e^({b}*{inp})-{c*d}+{d}{out}}=\\simplify{{a*b-a*d}e^({b}*{inp})-{c*d}}+(\\var{d})\\cdot(\\simplify{{a}e^({b}{inp})+{c}})=\\simplify{{a}*{b}e^({b}{inp})}$

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2) $\\var{out}(\\var{inp})=\\simplify{{a}e^({b}{inp})+{c}}$ er løsningen til  $\\var{out}''=\\simplify{{a*(b^2)}e^({b}{inp})}$ fordi

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$\\var{out}''=(\\simplify{{a}*{b}e^({b}{inp})})'=\\simplify{{a*(b^2)}e^({b}{inp})}$

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3) $\\var{out}(\\var{inp})=\\simplify{{a}sin({b}{inp})}$ er løsningen til  $\\var{out}''=\\simplify{{-a*(b^2)}}\\sin(\\var{b}\\var{inp})$ fordi

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$\\var{out}'=\\simplify{{a*b}cos({b}{inp})}$

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$\\var{out}''=(\\simplify{{a*b}cos({b}{inp})})'=\\simplify{-{a*b^2}sin({b}{inp})}$

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4) $\\var{out}(\\var{inp})=\\simplify{{a}sin({b}{inp})}$ er løsningen til $\\var{out}''=\\simplify{-{b^2}{out}}$ fordi

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VS: $\\var{out}''=\\simplify{{-a*(b^2)}}\\sin(\\var{b}\\var{inp})$

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HS: $\\simplify{-{b^2}{out}}=\\simplify{-{b^2}}\\cdot\\simplify{{a}sin({b}{inp})}=\\simplify{{-b^2*a}sin({b}{inp})}$ 

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5) $\\var{out}(\\var{inp})=\\simplify{{a}{inp}*e^({b}{inp})}$ er løsningen til $\\dfrac{\\var{out}'+\\var{out}}{\\simplify{{a}e^({b}{inp})}}=\\simplify{{b+1}{inp}+1}$ fordi

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$\\var{out}'=\\simplify{{a}*e^({b}{inp})+{a*b}{inp}*e^({b}{inp})}=\\simplify{{a}*e^({b}{inp})*(1+{b}{inp})}$

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$\\var{out}'+\\var{out}=\\simplify{{a}*e^({b}{inp})*(1+{b}{inp})}+\\simplify{{a}{inp}*e^({b}{inp})}=\\simplify{{a}*e^({b}{inp})*(1+{b+1}{inp})}$

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$\\dfrac{\\var{out}'+\\var{out}}{\\simplify{{a}e^({b}{inp})}}=\\simplify{{b+1}{inp}+1}$

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6)$\\var{out}(\\var{inp})=\\simplify{{a}{inp}*e^({b}{inp})}$ er løsningen til $\\var{out}''=\\dfrac{\\simplify{{b^2}{inp}+{2b}}}{\\var{inp}}\\var{out}$ fordi

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$\\var{out}''=(\\simplify{{a}*e^({b}{inp})*(1+{b}{inp})})'=\\simplify{{a*b}*e^({b}{inp})*(2+{b}{inp})}=\\simplify{{2*a*b}*e^({b}{inp})+{a*b^2}{inp}*e^({b}{inp})}$

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$\\dfrac{\\simplify{{b^2}{inp}+{2b}}}{\\var{inp}}\\var{out}=\\dfrac{\\simplify{{b^2}{inp}+{2b}}}{\\var{inp}}\\cdot\\simplify{{a}{inp}*e^({b}{inp})}=\\simplify{{2*a*b}*e^({b}{inp})+{a*b^2}{inp}*e^({b}{inp})}$

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

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

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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 $y(\\frac{1}{\\var{a}})=e$

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

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