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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.
\n$y'+y+y^2=x$ er en differensiallikning med $y(x)$ som en ukjente funksjon
\n$y'=y$ er en differensiallikning med $y(x)$ som en ukjente funksjon
\n$y'+5e^{xy}=\\sin(x)$ er en differensiallikning med $y(x)$ som en ukjente funksjon
\n$x'+x=y$ er en differensiallikning med $x(y)$ som en ukjente funksjon
\n$x^2+y^2=y'$ er en differensiallikning med $y(x)$ som en ukjente funksjon
\n$\\cos(y'(x))=\\sin(x)$ er en differensiallikning med $y(x)$ som en ukjente funksjon
\n$y'(t)=cos(t)$ er en differensiallikning med $y(t)$ som en ukjente funksjon
\n$y(t)=cos(t)$ er ikke en differensiallikning fordi den ikke inkluderer noen deriverte
\n$x^2+x=y$ er ikke en differensiallikning fordi den ikke inkluderer noen deriverte
\n$z=x^2+y^2$ er ikke en differensiallikning fordi den ikke inkluderer noen deriverte
\n$x^2+2x+1=0$ er ikke en differensiallikning fordi den ikke inkluderer noen deriverte
\n$y'(x)+y(x)$ er ikke en likning
\n", "rulesets": {}, "builtin_constants": {"e": true, "pi,\u03c0": true, "i": true}, "constants": [], "variables": {}, "variablesTest": {"condition": "", "maxRuns": 100}, "ungrouped_variables": [], "variable_groups": [], "functions": {}, "preamble": {"js": "", "css": ""}, "parts": [{"type": "m_n_2", "useCustomName": false, "customName": "", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "minMarks": 0, "maxMarks": 0, "shuffleChoices": true, "displayType": "checkbox", "displayColumns": 0, "minAnswers": 0, "maxAnswers": 0, "warningType": "none", "showCellAnswerState": true, "markingMethod": "sum ticked cells", "choices": ["$y'+y+y^2=x$", "$y'=y$", "$y'+5e^{xy}=\\sin(x)$", "$y(t)=cos(t)$", "$y'(t)=cos(t)$", "$x'+x=y$", "$x^2+x=y$", "$x^2+y^2=y'$", "$z=x^2+y^2$", "$x^2+2x+1=0$", "$y'(x)+y(x)$", "$\\cos(y'(x))=\\sin(x)$"], "matrix": ["0.5", "0.5", "0.5", "0", "0.5", "0.5", 0, "0.5", 0, 0, 0, "0.5"], "distractors": ["", "", "", "", "", "", "", "", "", "", "", ""]}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "Avgj\u00f8r ved regning hvilke av f\u00f8lgende funksjoner er l\u00f8sninger til difflikninger", "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": "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
\nVS: $\\var{out}'=\\simplify{{a}*{b}e^({b}{inp})}$
\nHS: $\\simplify{{a*b-a*d}e^({b}*{inp})-{c*d}+{d}{out}}=\\simplify{{a*b-a*d}e^({b}*{inp})-{c*d}}+(\\var{d})\\cdot(\\simplify{{a}e^({b}{inp})+{c}})=\\simplify{{a}*{b}e^({b}{inp})}$
\n2) $\\var{out}(\\var{inp})=\\simplify{{a}e^({b}{inp})+{c}}$ er løsningen til $\\var{out}''=\\simplify{{a*(b^2)}e^({b}{inp})}$ fordi
\n$\\var{out}''=(\\simplify{{a}*{b}e^({b}{inp})})'=\\simplify{{a*(b^2)}e^({b}{inp})}$
\n3) $\\var{out}(\\var{inp})=\\simplify{{a}sin({b}{inp})}$ er løsningen til $\\var{out}''=\\simplify{{-a*(b^2)}}\\sin(\\var{b}\\var{inp})$ fordi
\n$\\var{out}'=\\simplify{{a*b}cos({b}{inp})}$
\n$\\var{out}''=(\\simplify{{a*b}cos({b}{inp})})'=\\simplify{-{a*b^2}sin({b}{inp})}$
\n4) $\\var{out}(\\var{inp})=\\simplify{{a}sin({b}{inp})}$ er løsningen til $\\var{out}''=\\simplify{-{b^2}{out}}$ fordi
\nVS: $\\var{out}''=\\simplify{{-a*(b^2)}}\\sin(\\var{b}\\var{inp})$
\nHS: $\\simplify{-{b^2}{out}}=\\simplify{-{b^2}}\\cdot\\simplify{{a}sin({b}{inp})}=\\simplify{{-b^2*a}sin({b}{inp})}$
\n5) $\\var{out}(\\var{inp})=\\simplify{{a}{inp}*e^({b}{inp})}$ er løsningen til $\\dfrac{\\var{out}'+\\var{out}}{\\simplify{{a}e^({b}{inp})}}=\\simplify{{b+1}{inp}+1}$ fordi
\n$\\var{out}'=\\simplify{{a}*e^({b}{inp})+{a*b}{inp}*e^({b}{inp})}=\\simplify{{a}*e^({b}{inp})*(1+{b}{inp})}$
\n$\\var{out}'+\\var{out}=\\simplify{{a}*e^({b}{inp})*(1+{b}{inp})}+\\simplify{{a}{inp}*e^({b}{inp})}=\\simplify{{a}*e^({b}{inp})*(1+{b+1}{inp})}$
\n$\\dfrac{\\var{out}'+\\var{out}}{\\simplify{{a}e^({b}{inp})}}=\\simplify{{b+1}{inp}+1}$
\n6)$\\var{out}(\\var{inp})=\\simplify{{a}{inp}*e^({b}{inp})}$ er løsningen til $\\var{out}''=\\dfrac{\\simplify{{b^2}{inp}+{2b}}}{\\var{inp}}\\var{out}$ fordi
\n$\\var{out}''=(\\simplify{{a}*e^({b}{inp})*(1+{b}{inp})})'=\\simplify{{a*b}*e^({b}{inp})*(2+{b}{inp})}=\\simplify{{2*a*b}*e^({b}{inp})+{a*b^2}{inp}*e^({b}{inp})}$
\n$\\dfrac{\\simplify{{b^2}{inp}+{2b}}}{\\var{inp}}\\var{out}=\\dfrac{\\simplify{{b^2}{inp}+{2b}}}{\\var{inp}}\\cdot\\simplify{{a}{inp}*e^({b}{inp})}=\\simplify{{2*a*b}*e^({b}{inp})+{a*b^2}{inp}*e^({b}{inp})}$
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\nGitt differensiallikningen $y'(\\var{t})=\\var{a}y(\\var{t})$
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\n$y(\\var{t})=$[[0]]
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\n$y(\\var{t})=$[[0]]
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