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These are all separable first order differential equations.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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Separate the variables:

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

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

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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.

rebelmaths

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