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Solve the equation, and enter the value of $\\alpha$ and the expression for $f(x)$ in the boxes. Do not enter decimals in your answers.
\n$\\alpha=$ [[0]].
\n$f(x)=$ [[1]]. (Expand $f(x)$ fully, so that no parentheses appear in the expression.)
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", "strings": ["."]}, "answersimplification": "all", "vsetrangepoints": 5, "marks": 1, "checkingtype": "absdiff", "checkingaccuracy": 0.001, "answer": "{-a1}", "vsetrange": [0, 1], "showCorrectAnswer": true}, {"variableReplacementStrategy": "originalfirst", "checkvariablenames": false, "showpreview": true, "scripts": {}, "expectedvariablenames": [], "type": "jme", "variableReplacements": [], "notallowed": {"partialCredit": 0, "showStrings": true, "message": "Do not enter decimals in your answer, and expand $f(x)$ fully, so that no parentheses appear in the expression.
", "strings": [".", "(", ")"]}, "vsetrangepoints": 5, "marks": 1, "checkingtype": "absdiff", "checkingaccuracy": 0.001, "answer": "x^2+{2*b1}*x+{(a1+d1)^2-c1^2-2*b1*c1}", "vsetrange": [0, 1], "showCorrectAnswer": true}], "showCorrectAnswer": true}], "tags": ["checked2015", "MAS1603", "MAS2105"], "variable_groups": [], "metadata": {"licence": "Creative Commons Attribution 4.0 International", "description": "Find the solution of a first order separable differential equation of the form $(a+y)y'=b+x$.
", "notes": "Jan 2016 (WHF)
\nThe bddy condition determines the solution, so not correct to have $\\pm$ in the solution.
"}, "statement": "You are given the differential equation
\n\\[(\\var{a1}+y)y'=\\var{b1}+x,\\]
\nsatisfying $y(\\var{c1})=\\var{d1}$.
\nThe solution can be written in the form $y=\\alpha\\pm\\sqrt{f(x)}$, where $\\alpha$ is a constant, and $f(x)$ is some function of $x$.
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\n\\[\\simplify{{a1}*y+(1/2)*y^2}=\\simplify{{b1}*x+(1/2)*x^2+c},\\]
\nor
\n\\[\\simplify{(1/2)*(y+{a1})^2-{a1^2}/2}=\\simplify{{b1}*x+(1/2)*x^2+c},\\]
\nthen the general solution of the equation is
\n\\[y=\\var{-a1}\\pm\\simplify{sqrt(x^2+{2*b1}*x+2c+{a1^2})}\\]
\nor, upon redefining the constant $c$,
\n\\[y=\\var{-a1}\\pm\\simplify{sqrt(x^2+{2*b1}*x+c)}.\\]
\nThen we have
\n\\[\\var{d1}=y(\\var{c1})=\\var{-a1}\\pm\\simplify[std]{sqrt({c1}^2+{2*b1}*{c1}+c)}=\\var{-a1}\\pm\\simplify{sqrt({c1^2+2*b1*c1}+c)},\\]
\nso
\n\\[c=\\simplify[std]{({a1}+{d1})^2-{c1^2+2*b1*c1}}=\\simplify{{(a1+d1)^2-c1^2-2*b1*c1}}.\\]
\nThen the full solution is
\n\\[y=\\var{-a1}\\pm\\simplify{sqrt(x^2+{2*b1}*x+{(a1+d1)^2-c1^2-2*b1*c1})}.\\]
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