// Numbas version: exam_results_page_options {"name": "MA1002 First order differential equations 1", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"preventleave": false, "showfrontpage": false, "allowregen": true}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"preamble": {"js": "", "css": ""}, "name": "MA1002 First order differential equations 1", "variablesTest": {"condition": "", "maxRuns": 100}, "extensions": [], "ungrouped_variables": ["a1", "a3", "a2", "a4"], "contributors": [{"name": "Kieran Mulchrone", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/1243/"}], "advice": "\n

These are all separable first order differential equations.

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a)

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$\\displaystyle{\\frac{dy}{dx}=\\frac{y}{\\var{a1}x} \\Rightarrow \\int \\frac{1}{y}\\;dy = \\frac{1}{\\var{a1}}\\int\\frac{1}{x}\\;dx \\Rightarrow \\ln(y)=\\frac{1}{\\var{a1}}\\ln(x)+C}$

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Exponentiation of both sides then gives $y=Ax^{1/\\var{a1}}$ where we have renamed the constant of integration.

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To find the particular solution satisfying $y=1$ at $x=2$, we have $\\displaystyle{1=A \\times 2^{1/\\var{a1}} \\Rightarrow A = \\frac{1}{2^{1/\\var{a1}}}}$

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Hence the solution is $\\displaystyle{y=\\left(\\frac{x}{2}\\right)^{1/\\var{a1}}}$

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b)

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$\\displaystyle{\\frac{dy}{dx}=-\\var{a2}\\frac{y}{x} \\Rightarrow \\int \\frac{1}{y}\\;dy = -\\var{a2}\\int\\frac{1}{x}\\;dx \\Rightarrow \\ln(y)=-\\var{a2}\\ln(x)+C}$

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Exponentiation of both sides then gives $y=Ax^{-\\var{a2}}$ where we have renamed the constant of integration.

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The particular solution satisfying $y=1$ at $x=2$, gives $A = 2^{\\var{a2}}$

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Hence the solution is $\\displaystyle{y=\\left(\\frac{2}{x}\\right)^{\\var{a2}}}$

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c)

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$\\displaystyle{\\frac{dy}{dx}=\\var{a3}\\frac{x}{y} \\Rightarrow \\int y\\;dy = \\var{a3}\\int x\\;dx \\Rightarrow \\frac{y^2}{2}=\\var{a3}\\frac{x^2}{2}+C\\Rightarrow y^2=\\var{a3}x^2+A}$

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The particular solution satisfying $y=1$ at $x=2$, gives $A = \\var{1-4*a3}$.

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Hence the solution is $\\displaystyle{y^2=\\simplify[std]{{a3}x^2+{1-4*a3}}}$.

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d)

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$\\displaystyle{\\frac{dy}{dx}=-\\var{a4}\\frac{x}{y} \\Rightarrow \\int y\\;dy = -\\var{a4}\\int x\\;dx \\Rightarrow y^2=-\\var{a4}x^2+A}$

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The particular solution satisfying $y=1$ at $x=2$, gives $A = \\var{1+4*a4}$.

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Hence the solution is $\\displaystyle{y^2=\\simplify[std]{{-a4}x^2+{1+4*a4}}}$.

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Solve 4 first order differential equations of two types:$\\displaystyle \\frac{dy}{dx}=\\frac{ax}{y},\\;\\;\\frac{dy}{dx}=\\frac{by}{x},\\;y(2)=1$ for all 4.

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rebelmaths

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Input numbers as fractions or integers, not as decimals

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$\\displaystyle{\\frac{dy}{dx}=\\frac{y}{\\var{a1}x}}$

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$y=\\;\\;$[[0]]

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Do not enter decimals in your answer; use only fractions or integers.

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Input numbers as fractions or integers, not as decimals

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$\\displaystyle{\\frac{dy}{dx}=-\\var{a2}\\frac{y}{x}}$

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$y=\\;\\;$[[0]]

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Do not enter decimals in your answer; use only fractions or integers.

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Input numbers as fractions or integers, not as decimals

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$\\displaystyle{\\frac{dy}{dx}=\\var{a3}\\frac{x}{y}}$

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The solution can be written in the form $y^2=f(x)$. Enter $f(x)$ in the box below

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$y^2=\\;\\;$[[0]]

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Do not enter decimals in your answer; use only fractions or integers.

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Input numbers as fractions or integers, not as decimals

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$\\displaystyle{\\frac{dy}{dx}=-\\var{a4}\\frac{x}{y}}$

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The solution can be written in the form $y^2=g(x)$. Enter $g(x)$ in the box below

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$y^2=\\;\\;$[[0]]

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Do not enter decimals in your answer; use only fractions or integers.

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Separate the variables:

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Find the solutions of the following ordinary differential equations satisfying the condition $y=1$ at $x=2$.

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You may find it instructive to sketch your various solutions (but this is not required for this CBA).

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