// Numbas version: exam_results_page_options {"name": "Aoife's copy of Separable variables 1 violeta", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"tags": [], "rulesets": {"std": ["all", "!collectNumbers", "!noLeadingMinus"]}, "advice": "
The differential equation is separable, so we can write
\n\\[\\int{\\!\\frac{1}{\\var{b1}+y}\\,\\mathrm{d}y} = \\int{\\!\\frac{1}{\\var{a1}+x}\\,\\mathrm{d}x},\\]
\nthen
\n\\[\\ln\\lvert\\var{b1}+y\\rvert=\\ln\\lvert\\var{a1}+x\\rvert+c,\\]
\nso
\n\\[y=\\simplify{A({a1}+x)-{b1}},\\]
\nwhich is the general solution of the equation.
\nNow,
\n\\[\\var{d1}=y(\\var{c1})=\\simplify[std]{A({a1}+{c1})-{b1}},\\]
\nso
\n\\[A=\\simplify[std]{({d1}+{b1})/({a1}+{c1})}=\\simplify{{d1+b1}/{a1+c1}},\\]
\nand then the full solution is
\n\\[y=\\simplify[std]{{d1+b1}/{a1+c1}({a1}+x)-{b1}}=\\simplify{{(d1*a1-b1*c1)}/{a1+c1}+{(d1+b1)}*x/{a1+c1}}.\\]
", "statement": "Equations which can be written in the form
\n\\[\\dfrac{\\mathrm{d}y}{\\mathrm{d}x} = f(x), \\dfrac{\\mathrm{d}y}{\\mathrm{d}x} = f(y), \\dfrac{\\mathrm{d}y}{\\mathrm{d}x} = f(x)f(y)\\]
\ncan all be solved by integration.
\nIn each case it is possible to separate the $x$'s to one side of the equation and the $y$'s to the other
\nSolving such equations is therefore known as solution by separation of variables
\n\nQuestion
\nFind the solution of the differential equation
\n\\[(\\var{a1}+x)\\dfrac{\\mathrm{d}y}{\\mathrm{d}x}=\\var{b1}+y\\]
\n\nsatisfying the condition that $y = \\var{d1}$ when $x = \\var{c1}$
\n", "preamble": {"css": "", "js": ""}, "ungrouped_variables": ["a1", "c1", "b1", "d1"], "metadata": {"licence": "Creative Commons Attribution 4.0 International", "description": "Equations which can be written in the form
\n\\[\\dfrac{\\mathrm{d}y}{\\mathrm{d}x} = f(x), \\dfrac{\\mathrm{d}y}{\\mathrm{d}x} = f(y), \\dfrac{\\mathrm{d}y}{\\mathrm{d}x} = f(x)f(y)\\]
\ncan all be solved by integration.
\nIn each case it is possible to separate the $x$'s to one side of the equation and the $y$'s to the other
\nSolving such equations is therefore known as solution by separation of variables
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$y=$ [[0]] (Do not enter decimals in your answer.)
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", "partialCredit": 0}, "answer": "{(d1*a1-b1*c1)}/{a1+c1}+{(d1+b1)}*x/{a1+c1}", "showPreview": true, "variableReplacements": [], "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "vsetRangePoints": 5, "checkVariableNames": false, "vsetRange": [0, 1], "showFeedbackIcon": true, "answerSimplification": "all", "marks": "10", "type": "jme", "unitTests": [], "checkingType": "absdiff"}], "showFeedbackIcon": true, "marks": 0, "type": "gapfill", "unitTests": []}], "name": "Aoife's copy of Separable variables 1 violeta", "variablesTest": {"condition": "", "maxRuns": 100}, "type": "question", "contributors": [{"name": "Stephen Bowlzer", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/206/"}, {"name": "Violeta CIT", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/1030/"}, {"name": "Aoife O'Brien", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/1693/"}]}]}], "contributors": [{"name": "Stephen Bowlzer", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/206/"}, {"name": "Violeta CIT", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/1030/"}, {"name": "Aoife O'Brien", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/1693/"}]}