Calculus Examples

$x d y = y (\ln (x) - \ln (y)) d x$

Step 1

Rewrite the differential equation to fit the Exact differential equation technique.

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Step 1.1

Subtract $y (\ln (x) - \ln (y)) d x$ from both sides of the equation.

$x d y - y (\ln (x) - \ln (y)) d x = 0$

Step 1.2

Rewrite.

$- y (\ln (x) - \ln (y)) d x + x d y = 0$

Step 2

Find $\frac{\partial M}{\partial y}$ where $M (x, y) = - y (\ln (x) - \ln (y))$ .

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Step 2.1

Differentiate $M$ with respect to $y$ .

$\frac{\partial M}{\partial y} = \frac{d}{d y} [- y (\ln (x) - \ln (y))]$

Step 2.2

Use the quotient property of logarithms, $\log_{b} (x) - \log_{b} (y) = \log_{b} (\frac{x}{y})$ .

$\frac{\partial M}{\partial y} = \frac{d}{d y} [- y \ln (\frac{x}{y})]$

Step 2.3

Since $- 1$ is constant with respect to $y$ , the derivative of $- y \ln (\frac{x}{y})$ with respect to $y$ is $- \frac{d}{d y} [y \ln (\frac{x}{y})]$ .

$\frac{\partial M}{\partial y} = - \frac{d}{d y} [y \ln (\frac{x}{y})]$

Step 2.4

Differentiate using the Product Rule which states that $\frac{d}{d y} [f (y) g (y)]$ is $f (y) \frac{d}{d y} [g (y)] + g (y) \frac{d}{d y} [f (y)]$ where $f (y) = y$ and $g (y) = \ln (\frac{x}{y})$ .

$\frac{\partial M}{\partial y} = - (y \frac{d}{d y} [\ln (\frac{x}{y})] + \ln (\frac{x}{y}) \frac{d}{d y} [y])$

Step 2.5

Differentiate using the chain rule, which states that $\frac{d}{d y} [f (g (y))]$ is $f' (g (y)) g' (y)$ where $f (y) = \ln (y)$ and $g (y) = \frac{x}{y}$ .

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Step 2.5.1

To apply the Chain Rule, set $u$ as $\frac{x}{y}$ .

$\frac{\partial M}{\partial y} = - (y (\frac{d}{d u} [\ln (u)] \frac{d}{d y} [\frac{x}{y}]) + \ln (\frac{x}{y}) \frac{d}{d y} [y])$

Step 2.5.2

The derivative of $\ln (u)$ with respect to $u$ is $\frac{1}{u}$ .

$\frac{\partial M}{\partial y} = - (y (\frac{1}{u} \frac{d}{d y} [\frac{x}{y}]) + \ln (\frac{x}{y}) \frac{d}{d y} [y])$

Step 2.5.3

Replace all occurrences of $u$ with $\frac{x}{y}$ .

$\frac{\partial M}{\partial y} = - (y (\frac{1}{\frac{x}{y}} \frac{d}{d y} [\frac{x}{y}]) + \ln (\frac{x}{y}) \frac{d}{d y} [y])$

Step 2.6

Multiply by the reciprocal of the fraction to divide by $\frac{x}{y}$ .

$\frac{\partial M}{\partial y} = - (y (1 \frac{y}{x} \frac{d}{d y} [\frac{x}{y}]) + \ln (\frac{x}{y}) \frac{d}{d y} [y])$

Step 2.7

Multiply $\frac{y}{x}$ by $1$ .

$\frac{\partial M}{\partial y} = - (y (\frac{y}{x} \frac{d}{d y} [\frac{x}{y}]) + \ln (\frac{x}{y}) \frac{d}{d y} [y])$

Step 2.8

Combine $\frac{y}{x}$ and $y$ .

$\frac{\partial M}{\partial y} = - (\frac{y \cdot y}{x} \frac{d}{d y} [\frac{x}{y}] + \ln (\frac{x}{y}) \frac{d}{d y} [y])$

Step 2.9

Raise $y$ to the power of $1$ .

$\frac{\partial M}{\partial y} = - (\frac{y^{1} y}{x} \frac{d}{d y} [\frac{x}{y}] + \ln (\frac{x}{y}) \frac{d}{d y} [y])$

Step 2.10

Raise $y$ to the power of $1$ .

$\frac{\partial M}{\partial y} = - (\frac{y^{1} y^{1}}{x} \frac{d}{d y} [\frac{x}{y}] + \ln (\frac{x}{y}) \frac{d}{d y} [y])$

Step 2.11

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$\frac{\partial M}{\partial y} = - (\frac{y^{1 + 1}}{x} \frac{d}{d y} [\frac{x}{y}] + \ln (\frac{x}{y}) \frac{d}{d y} [y])$

Step 2.12

Add $1$ and $1$ .

$\frac{\partial M}{\partial y} = - (\frac{y^{2}}{x} \frac{d}{d y} [\frac{x}{y}] + \ln (\frac{x}{y}) \frac{d}{d y} [y])$

Step 2.13

Since $x$ is constant with respect to $y$ , the derivative of $\frac{x}{y}$ with respect to $y$ is $x \frac{d}{d y} [\frac{1}{y}]$ .

$\frac{\partial M}{\partial y} = - (\frac{y^{2}}{x} (x \frac{d}{d y} [\frac{1}{y}]) + \ln (\frac{x}{y}) \frac{d}{d y} [y])$

Step 2.14

Simplify terms.

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Step 2.14.1

Combine $x$ and $\frac{y^{2}}{x}$ .

$\frac{\partial M}{\partial y} = - (\frac{x y^{2}}{x} \frac{d}{d y} [\frac{1}{y}] + \ln (\frac{x}{y}) \frac{d}{d y} [y])$

Step 2.14.2

Cancel the common factor of $x$ .

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Step 2.14.2.1

Cancel the common factor.

$\frac{\partial M}{\partial y} = - (\frac{x y^{2}}{x} \frac{d}{d y} [\frac{1}{y}] + \ln (\frac{x}{y}) \frac{d}{d y} [y])$

Step 2.14.2.2

Divide $y^{2}$ by $1$ .

$\frac{\partial M}{\partial y} = - (y^{2} \frac{d}{d y} [\frac{1}{y}] + \ln (\frac{x}{y}) \frac{d}{d y} [y])$

Step 2.14.3

Rewrite $\frac{1}{y}$ as $y^{- 1}$ .

$\frac{\partial M}{\partial y} = - (y^{2} \frac{d}{d y} [y^{- 1}] + \ln (\frac{x}{y}) \frac{d}{d y} [y])$

Step 2.15

Differentiate using the Power Rule which states that $\frac{d}{d y} [y^{n}]$ is $n y^{n - 1}$ where $n = - 1$ .

$\frac{\partial M}{\partial y} = - (y^{2} (- y^{- 2}) + \ln (\frac{x}{y}) \frac{d}{d y} [y])$

Step 2.16

Multiply $y^{2}$ by $y^{- 2}$ by adding the exponents.

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Step 2.16.1

Move $y^{- 2}$ .

$\frac{\partial M}{\partial y} = - (y^{- 2} y^{2} \cdot - 1 + \ln (\frac{x}{y}) \frac{d}{d y} [y])$

Step 2.16.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$\frac{\partial M}{\partial y} = - (y^{- 2 + 2} \cdot - 1 + \ln (\frac{x}{y}) \frac{d}{d y} [y])$

Step 2.16.3

Add $- 2$ and $2$ .

$\frac{\partial M}{\partial y} = - (y^{0} \cdot - 1 + \ln (\frac{x}{y}) \frac{d}{d y} [y])$

Step 2.17

Simplify $y^{0} \cdot - 1$ .

$\frac{\partial M}{\partial y} = - (- 1 + \ln (\frac{x}{y}) \frac{d}{d y} [y])$

Step 2.18

Differentiate using the Power Rule which states that $\frac{d}{d y} [y^{n}]$ is $n y^{n - 1}$ where $n = 1$ .

$\frac{\partial M}{\partial y} = - (- 1 + \ln (\frac{x}{y}) \cdot 1)$

Step 2.19

Multiply $\ln (\frac{x}{y})$ by $1$ .

$\frac{\partial M}{\partial y} = - (- 1 + \ln (\frac{x}{y}))$

Step 2.20

Simplify.

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Step 2.20.1

Apply the distributive property.

$\frac{\partial M}{\partial y} = - - 1 - \ln (\frac{x}{y})$

Step 2.20.2

Multiply $- 1$ by $- 1$ .

$\frac{\partial M}{\partial y} = 1 - \ln (\frac{x}{y})$

Step 3

Find $\frac{\partial N}{\partial x}$ where $N (x, y) = x$ .

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Step 3.1

Differentiate $N$ with respect to $x$ .

$\frac{\partial N}{\partial x} = \frac{d}{d x} [x]$

Step 3.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

$\frac{\partial N}{\partial x} = 1$

Step 4

Check that $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$ .

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Step 4.1

Substitute $1 - \ln (\frac{x}{y})$ for $\frac{\partial M}{\partial y}$ and $1$ for $\frac{\partial N}{\partial x}$ .

$1 - \ln (\frac{x}{y}) = 1$

Step 4.2

Since the left side does not equal the right side, the equation is not an identity.

$1 - \ln (\frac{x}{y}) = 1$ is not an identity.

Step 5

Find the integration factor $μ (x, y) = e^{\int \frac{\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y}}{M} d y}$ .

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Step 5.1

Substitute $1$ for $\frac{\partial N}{\partial x}$ .

$\frac{1 - \frac{\partial M}{\partial y}}{M}$

Step 5.2

Substitute $1 - \ln (\frac{x}{y})$ for $\frac{\partial M}{\partial y}$ .

$\frac{1 - (1 - \ln (\frac{x}{y}))}{M}$

Step 5.3

Substitute $- y (\ln (x) - \ln (y))$ for $M$ .

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Step 5.3.1

Substitute $- y (\ln (x) - \ln (y))$ for $M$ .

$\frac{1 - (1 - \ln (\frac{x}{y}))}{- y (\ln (x) - \ln (y))}$

Step 5.3.2

Simplify the numerator.

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Step 5.3.2.1

Apply the distributive property.

$\frac{1 - 1 \cdot 1 - - \ln (\frac{x}{y})}{- y (\ln (x) - \ln (y))}$

Step 5.3.2.2

Multiply $- 1$ by $1$ .

$\frac{1 - 1 - - \ln (\frac{x}{y})}{- y (\ln (x) - \ln (y))}$

Step 5.3.2.3

Multiply $- - \ln (\frac{x}{y})$ .

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Step 5.3.2.3.1

Multiply $- 1$ by $- 1$ .

$\frac{1 - 1 + 1 \ln (\frac{x}{y})}{- y (\ln (x) - \ln (y))}$

Step 5.3.2.3.2

Multiply $\ln (\frac{x}{y})$ by $1$ .

$\frac{1 - 1 + \ln (\frac{x}{y})}{- y (\ln (x) - \ln (y))}$

Step 5.3.2.4

Subtract $1$ from $1$ .

$\frac{0 + \ln (\frac{x}{y})}{- y (\ln (x) - \ln (y))}$

Step 5.3.2.5

Add $0$ and $\ln (\frac{x}{y})$ .

$\frac{\ln (\frac{x}{y})}{- y (\ln (x) - \ln (y))}$

Step 5.3.3

Use the quotient property of logarithms, $\log_{b} (x) - \log_{b} (y) = \log_{b} (\frac{x}{y})$ .

$\frac{\ln (\frac{x}{y})}{- y \ln (\frac{x}{y})}$

Step 5.3.4

Cancel the common factor of $\ln (\frac{x}{y})$ .

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Step 5.3.4.1

Cancel the common factor.

$\frac{\ln (\frac{x}{y})}{- y \ln (\frac{x}{y})}$

Step 5.3.4.2

Rewrite the expression.

$\frac{1}{- y}$

Step 5.3.5

Substitute $- y (\ln (x) - \ln (y))$ for $M$ .

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Step 5.3.5.1

Rewrite $1$ as $- 1 (- 1)$ .

$\frac{- 1 (- 1)}{- y}$

Step 5.3.5.2

Move the negative in front of the fraction.

$- \frac{1}{y}$

Step 5.4

Find the integration factor $μ (x, y) = e^{\int \frac{\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y}}{M} d y}$ .

$μ (x, y) = e^{\int - \frac{1}{y} d y}$

Step 6

Evaluate the integral $e^{\int - \frac{1}{y} d y}$ .

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Step 6.1

Since $- 1$ is constant with respect to $y$ , move $- 1$ out of the integral.

$μ (x, y) = e^{- \int \frac{1}{y} d y}$

Step 6.2

The integral of $\frac{1}{y}$ with respect to $y$ is $\ln (| y |)$ .

$μ (x, y) = e^{- (\ln (| y |) + C)}$

Step 6.3

Simplify.

$μ (x, y) = e^{- \ln (| y |)} + C$

Step 6.4

Simplify each term.

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Step 6.4.1

Simplify $- \ln (| y |)$ by moving $- 1$ inside the logarithm.

$μ (x, y) = e^{\ln ({| y |}^{- 1})} + C$

Step 6.4.2

Exponentiation and log are inverse functions.

$μ (x, y) = {| y |}^{- 1} + C$

Step 6.4.3

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$μ (x, y) = \frac{1}{| y |} + C$

Step 7

Multiply both sides of $- y (\ln (x) - \ln (y)) d x + x d y = 0$ by the integration factor $\frac{1}{y}$ .

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Step 7.1

Multiply $- y (\ln (x) - \ln (y))$ by $\frac{1}{y}$ .

$- y (\ln (x) - \ln (y)) \frac{1}{y} d x + x d y = 0$

Step 7.2

Cancel the common factor of $y$ .

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Step 7.2.1

Factor $y$ out of $- y (\ln (x) - \ln (y))$ .

$y (- 1 (\ln (x) - \ln (y))) \frac{1}{y} d x + x d y = 0$

Step 7.2.2

Cancel the common factor.

$y (- 1 (\ln (x) - \ln (y))) \frac{1}{y} d x + x d y = 0$

Step 7.2.3

Rewrite the expression.

$- 1 (\ln (x) - \ln (y)) d x + x d y = 0$

Step 7.3

Use the quotient property of logarithms, $\log_{b} (x) - \log_{b} (y) = \log_{b} (\frac{x}{y})$ .

$- 1 \ln (\frac{x}{y}) d x + x d y = 0$

Step 7.4

Rewrite $- 1 \ln (\frac{x}{y})$ as $- \ln (\frac{x}{y})$ .

$- \ln (\frac{x}{y}) d x + x d y = 0$

Step 7.5

Multiply $x$ by $\frac{1}{y}$ .

$- \ln (\frac{x}{y}) d x + x \frac{1}{y} d y = 0$

Step 7.6

Combine $x$ and $\frac{1}{y}$ .

$- \ln (\frac{x}{y}) d x + \frac{x}{y} d y = 0$

Step 8

Set $f (x, y)$ equal to the integral of $N (x, y)$ .

$f (x, y) = \int \frac{x}{y} d y$

Step 9

Integrate $N (x, y) = \frac{x}{y}$ to find $f (x, y)$ .

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Step 9.1

Since $x$ is constant with respect to $y$ , move $x$ out of the integral.

$f (x, y) = x \int \frac{1}{y} d y$

Step 9.2

The integral of $\frac{1}{y}$ with respect to $y$ is $\ln (| y |)$ .

$f (x, y) = x (\ln (| y |) + C)$

Step 9.3

Simplify.

$f (x, y) = x \ln (| y |) + C$

Step 10

Since the integral of $g (x)$ will contain an integration constant, we can replace $C$ with $g (x)$ .

$f (x, y) = x \ln (| y |) + g (x)$

Step 11

Set $\frac{\partial f}{\partial x} = M (x, y)$ .

$\frac{\partial f}{\partial x} = - \ln (\frac{x}{y})$

Step 12

Differentiate $f$ with respect to $x$ .

$\frac{d}{d x} [x \ln (| y |) + g (x)] = - \ln (\frac{x}{y})$

Step 13

Solve for $\frac{d g}{d x}$ .

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Step 13.1

Solve for $\frac{d g}{d x}$ .

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Step 13.1.1

Rewrite.

$x d y = y (\ln (x) - \ln (y)) d x$

Step 13.1.2

Rewrite the differential equation to fit the Exact differential equation technique.

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Step 13.1.2.1

Subtract $y (\ln (x) - \ln (y)) d x$ from both sides of the equation.

$x d y - y (\ln (x) - \ln (y)) d x = 0$

Step 13.1.2.2

Rewrite.

$- y (\ln (x) - \ln (y)) d x + x d y = 0$

Step 13.1.3

Find $\frac{\partial M}{\partial y}$ where $M (x, y) = - y (\ln (x) - \ln (y))$ .

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Step 13.1.3.1

Differentiate $M$ with respect to $y$ .

$\frac{\partial M}{\partial y} = \frac{d}{d y} [- y (\ln (x) - \ln (y))]$

Step 13.1.3.2

Use the quotient property of logarithms, $\log_{b} (x) - \log_{b} (y) = \log_{b} (\frac{x}{y})$ .

$\frac{\partial M}{\partial y} = \frac{d}{d y} [- y \ln (\frac{x}{y})]$

Step 13.1.3.3

Since $- 1$ is constant with respect to $y$ , the derivative of $- y \ln (\frac{x}{y})$ with respect to $y$ is $- \frac{d}{d y} [y \ln (\frac{x}{y})]$ .

$\frac{\partial M}{\partial y} = - \frac{d}{d y} [y \ln (\frac{x}{y})]$

Step 13.1.3.4

Differentiate using the Product Rule which states that $\frac{d}{d y} [f (y) g (y)]$ is $f (y) \frac{d}{d y} [g (y)] + g (y) \frac{d}{d y} [f (y)]$ where $f (y) = y$ and $g (y) = \ln (\frac{x}{y})$ .

$\frac{\partial M}{\partial y} = - (y \frac{d}{d y} [\ln (\frac{x}{y})] + \ln (\frac{x}{y}) \frac{d}{d y} [y])$

Step 13.1.3.5

Differentiate using the chain rule, which states that $\frac{d}{d y} [f (g (y))]$ is $f' (g (y)) g' (y)$ where $f (y) = \ln (y)$ and $g (y) = \frac{x}{y}$ .

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Step 13.1.3.5.1

To apply the Chain Rule, set $u$ as $\frac{x}{y}$ .

$\frac{\partial M}{\partial y} = - (y (\frac{d}{d u} [\ln (u)] \frac{d}{d y} [\frac{x}{y}]) + \ln (\frac{x}{y}) \frac{d}{d y} [y])$

Step 13.1.3.5.2

The derivative of $\ln (u)$ with respect to $u$ is $\frac{1}{u}$ .

$\frac{\partial M}{\partial y} = - (y (\frac{1}{u} \frac{d}{d y} [\frac{x}{y}]) + \ln (\frac{x}{y}) \frac{d}{d y} [y])$

Step 13.1.3.5.3

Replace all occurrences of $u$ with $\frac{x}{y}$ .

$\frac{\partial M}{\partial y} = - (y (\frac{1}{\frac{x}{y}} \frac{d}{d y} [\frac{x}{y}]) + \ln (\frac{x}{y}) \frac{d}{d y} [y])$

Step 13.1.3.6

Multiply by the reciprocal of the fraction to divide by $\frac{x}{y}$ .

$\frac{\partial M}{\partial y} = - (y (1 \frac{y}{x} \frac{d}{d y} [\frac{x}{y}]) + \ln (\frac{x}{y}) \frac{d}{d y} [y])$

Step 13.1.3.7

Multiply $\frac{y}{x}$ by $1$ .

$\frac{\partial M}{\partial y} = - (y (\frac{y}{x} \frac{d}{d y} [\frac{x}{y}]) + \ln (\frac{x}{y}) \frac{d}{d y} [y])$

Step 13.1.3.8

Combine $\frac{y}{x}$ and $y$ .

$\frac{\partial M}{\partial y} = - (\frac{y \cdot y}{x} \frac{d}{d y} [\frac{x}{y}] + \ln (\frac{x}{y}) \frac{d}{d y} [y])$

Step 13.1.3.9

Raise $y$ to the power of $1$ .

$\frac{\partial M}{\partial y} = - (\frac{y^{1} y}{x} \frac{d}{d y} [\frac{x}{y}] + \ln (\frac{x}{y}) \frac{d}{d y} [y])$

Step 13.1.3.10

Raise $y$ to the power of $1$ .

$\frac{\partial M}{\partial y} = - (\frac{y^{1} y^{1}}{x} \frac{d}{d y} [\frac{x}{y}] + \ln (\frac{x}{y}) \frac{d}{d y} [y])$

Step 13.1.3.11

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$\frac{\partial M}{\partial y} = - (\frac{y^{1 + 1}}{x} \frac{d}{d y} [\frac{x}{y}] + \ln (\frac{x}{y}) \frac{d}{d y} [y])$

Step 13.1.3.12

Add $1$ and $1$ .

$\frac{\partial M}{\partial y} = - (\frac{y^{2}}{x} \frac{d}{d y} [\frac{x}{y}] + \ln (\frac{x}{y}) \frac{d}{d y} [y])$

Step 13.1.3.13

Since $x$ is constant with respect to $y$ , the derivative of $\frac{x}{y}$ with respect to $y$ is $x \frac{d}{d y} [\frac{1}{y}]$ .

$\frac{\partial M}{\partial y} = - (\frac{y^{2}}{x} (x \frac{d}{d y} [\frac{1}{y}]) + \ln (\frac{x}{y}) \frac{d}{d y} [y])$

Step 13.1.3.14

Simplify terms.

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Step 13.1.3.14.1

Combine $x$ and $\frac{y^{2}}{x}$ .

$\frac{\partial M}{\partial y} = - (\frac{x y^{2}}{x} \frac{d}{d y} [\frac{1}{y}] + \ln (\frac{x}{y}) \frac{d}{d y} [y])$

Step 13.1.3.14.2

Cancel the common factor of $x$ .

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Step 13.1.3.14.2.1

Cancel the common factor.

$\frac{\partial M}{\partial y} = - (\frac{x y^{2}}{x} \frac{d}{d y} [\frac{1}{y}] + \ln (\frac{x}{y}) \frac{d}{d y} [y])$

Step 13.1.3.14.2.2

Divide $y^{2}$ by $1$ .

$\frac{\partial M}{\partial y} = - (y^{2} \frac{d}{d y} [\frac{1}{y}] + \ln (\frac{x}{y}) \frac{d}{d y} [y])$

Step 13.1.3.14.3

Rewrite $\frac{1}{y}$ as $y^{- 1}$ .

$\frac{\partial M}{\partial y} = - (y^{2} \frac{d}{d y} [y^{- 1}] + \ln (\frac{x}{y}) \frac{d}{d y} [y])$

Step 13.1.3.15

Differentiate using the Power Rule which states that $\frac{d}{d y} [y^{n}]$ is $n y^{n - 1}$ where $n = - 1$ .

$\frac{\partial M}{\partial y} = - (y^{2} (- y^{- 2}) + \ln (\frac{x}{y}) \frac{d}{d y} [y])$

Step 13.1.3.16

Multiply $y^{2}$ by $y^{- 2}$ by adding the exponents.

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Step 13.1.3.16.1

Move $y^{- 2}$ .

$\frac{\partial M}{\partial y} = - (y^{- 2} y^{2} \cdot - 1 + \ln (\frac{x}{y}) \frac{d}{d y} [y])$

Step 13.1.3.16.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$\frac{\partial M}{\partial y} = - (y^{- 2 + 2} \cdot - 1 + \ln (\frac{x}{y}) \frac{d}{d y} [y])$

Step 13.1.3.16.3

Add $- 2$ and $2$ .

$\frac{\partial M}{\partial y} = - (y^{0} \cdot - 1 + \ln (\frac{x}{y}) \frac{d}{d y} [y])$

Step 13.1.3.17

Simplify $y^{0} \cdot - 1$ .

$\frac{\partial M}{\partial y} = - (- 1 + \ln (\frac{x}{y}) \frac{d}{d y} [y])$

Step 13.1.3.18

Differentiate using the Power Rule which states that $\frac{d}{d y} [y^{n}]$ is $n y^{n - 1}$ where $n = 1$ .

$\frac{\partial M}{\partial y} = - (- 1 + \ln (\frac{x}{y}) \cdot 1)$

Step 13.1.3.19

Multiply $\ln (\frac{x}{y})$ by $1$ .

$\frac{\partial M}{\partial y} = - (- 1 + \ln (\frac{x}{y}))$

Step 13.1.3.20

Simplify.

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Step 13.1.3.20.1

Apply the distributive property.

$\frac{\partial M}{\partial y} = - - 1 - \ln (\frac{x}{y})$

Step 13.1.3.20.2

Multiply $- 1$ by $- 1$ .

$\frac{\partial M}{\partial y} = 1 - \ln (\frac{x}{y})$

Step 13.1.4

Find $\frac{\partial N}{\partial x}$ where $N (x, y) = x$ .

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Step 13.1.4.1

Differentiate $N$ with respect to $x$ .

$\frac{\partial N}{\partial x} = \frac{d}{d x} [x]$

Step 13.1.4.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

$\frac{\partial N}{\partial x} = 1$

Step 13.1.5

Check that $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$ .

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Step 13.1.5.1

Substitute $1 - \ln (\frac{x}{y})$ for $\frac{\partial M}{\partial y}$ and $1$ for $\frac{\partial N}{\partial x}$ .

$1 - \ln (\frac{x}{y}) = 1$

Step 13.1.5.2

Since the left side does not equal the right side, the equation is not an identity.

$1 - \ln (\frac{x}{y}) = 1$ is not an identity.

Step 13.1.6

Find the integration factor $μ (x, y) = e^{\int \frac{\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y}}{M} d y}$ .

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Step 13.1.6.1

Substitute $1$ for $\frac{\partial N}{\partial x}$ .

$\frac{1 - \frac{\partial M}{\partial y}}{M}$

Step 13.1.6.2

Substitute $1 - \ln (\frac{x}{y})$ for $\frac{\partial M}{\partial y}$ .

$\frac{1 - (1 - \ln (\frac{x}{y}))}{M}$

Step 13.1.6.3

Substitute $- y (\ln (x) - \ln (y))$ for $M$ .

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Step 13.1.6.3.1

Substitute $- y (\ln (x) - \ln (y))$ for $M$ .

$\frac{1 - (1 - \ln (\frac{x}{y}))}{- y (\ln (x) - \ln (y))}$

Step 13.1.6.3.2

Simplify the numerator.

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Step 13.1.6.3.2.1

Apply the distributive property.

$\frac{1 - 1 \cdot 1 - - \ln (\frac{x}{y})}{- y (\ln (x) - \ln (y))}$

Step 13.1.6.3.2.2

Multiply $- 1$ by $1$ .

$\frac{1 - 1 - - \ln (\frac{x}{y})}{- y (\ln (x) - \ln (y))}$

Step 13.1.6.3.2.3

Multiply $- - \ln (\frac{x}{y})$ .

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Step 13.1.6.3.2.3.1

Multiply $- 1$ by $- 1$ .

$\frac{1 - 1 + 1 \ln (\frac{x}{y})}{- y (\ln (x) - \ln (y))}$

Step 13.1.6.3.2.3.2

Multiply $\ln (\frac{x}{y})$ by $1$ .

$\frac{1 - 1 + \ln (\frac{x}{y})}{- y (\ln (x) - \ln (y))}$

Step 13.1.6.3.2.4

Subtract $1$ from $1$ .

$\frac{0 + \ln (\frac{x}{y})}{- y (\ln (x) - \ln (y))}$

Step 13.1.6.3.2.5

Add $0$ and $\ln (\frac{x}{y})$ .

$\frac{\ln (\frac{x}{y})}{- y (\ln (x) - \ln (y))}$

Step 13.1.6.3.3

Use the quotient property of logarithms, $\log_{b} (x) - \log_{b} (y) = \log_{b} (\frac{x}{y})$ .

$\frac{\ln (\frac{x}{y})}{- y \ln (\frac{x}{y})}$

Step 13.1.6.3.4

Cancel the common factor of $\ln (\frac{x}{y})$ .

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Step 13.1.6.3.4.1

Cancel the common factor.

$\frac{\ln (\frac{x}{y})}{- y \ln (\frac{x}{y})}$

Step 13.1.6.3.4.2

Rewrite the expression.

$\frac{1}{- y}$

Step 13.1.6.3.5

Substitute $- y (\ln (x) - \ln (y))$ for $M$ .

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Step 13.1.6.3.5.1

Rewrite $1$ as $- 1 (- 1)$ .

$\frac{- 1 (- 1)}{- y}$

Step 13.1.6.3.5.2

Move the negative in front of the fraction.

$- \frac{1}{y}$

Step 13.1.6.4

Find the integration factor $μ (x, y) = e^{\int \frac{\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y}}{M} d y}$ .

$μ (x, y) = e^{\int - \frac{1}{y} d y}$

Step 13.1.7

Evaluate the integral $e^{\int - \frac{1}{y} d y}$ .

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Step 13.1.7.1

Since $- 1$ is constant with respect to $y$ , move $- 1$ out of the integral.

$μ (x, y) = e^{- \int \frac{1}{y} d y}$

Step 13.1.7.2

The integral of $\frac{1}{y}$ with respect to $y$ is $\ln (| y |)$ .

$μ (x, y) = e^{- (\ln (| y |) + C)}$

Step 13.1.7.3

Simplify.

$μ (x, y) = e^{- \ln (| y |)} + C$

Step 13.1.7.4

Simplify each term.

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Step 13.1.7.4.1

Simplify $- \ln (| y |)$ by moving $- 1$ inside the logarithm.

$μ (x, y) = e^{\ln ({| y |}^{- 1})} + C$

Step 13.1.7.4.2

Exponentiation and log are inverse functions.

$μ (x, y) = {| y |}^{- 1} + C$

Step 13.1.7.4.3

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$μ (x, y) = \frac{1}{| y |} + C$

Step 13.1.8

Multiply both sides of $- y (\ln (x) - \ln (y)) d x + x d y = 0$ by the integration factor $\frac{1}{y}$ .

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Step 13.1.8.1

Multiply $- y (\ln (x) - \ln (y))$ by $\frac{1}{y}$ .

$- y (\ln (x) - \ln (y)) \frac{1}{y} d x + x d y = 0$

Step 13.1.8.2

Cancel the common factor of $y$ .

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Step 13.1.8.2.1

Factor $y$ out of $- y (\ln (x) - \ln (y))$ .

$y (- 1 (\ln (x) - \ln (y))) \frac{1}{y} d x + x d y = 0$

Step 13.1.8.2.2

Cancel the common factor.

$y (- 1 (\ln (x) - \ln (y))) \frac{1}{y} d x + x d y = 0$

Step 13.1.8.2.3

Rewrite the expression.

$- 1 (\ln (x) - \ln (y)) d x + x d y = 0$

Step 13.1.8.3

Use the quotient property of logarithms, $\log_{b} (x) - \log_{b} (y) = \log_{b} (\frac{x}{y})$ .

$- 1 \ln (\frac{x}{y}) d x + x d y = 0$

Step 13.1.8.4

Rewrite $- 1 \ln (\frac{x}{y})$ as $- \ln (\frac{x}{y})$ .

$- \ln (\frac{x}{y}) d x + x d y = 0$

Step 13.1.8.5

Multiply $x$ by $\frac{1}{y}$ .

$- \ln (\frac{x}{y}) d x + x \frac{1}{y} d y = 0$

Step 13.1.8.6

Combine $x$ and $\frac{1}{y}$ .

$- \ln (\frac{x}{y}) d x + \frac{x}{y} d y = 0$

Step 13.1.9

Set $f (x, y)$ equal to the integral of $N (x, y)$ .

$f (x, y) = \int \frac{x}{y} d y$

Step 13.1.10

Integrate $N (x, y) = \frac{x}{y}$ to find $f (x, y)$ .

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Step 13.1.10.1

Since $x$ is constant with respect to $y$ , move $x$ out of the integral.

$f (x, y) = x \int \frac{1}{y} d y$

Step 13.1.10.2

The integral of $\frac{1}{y}$ with respect to $y$ is $\ln (| y |)$ .

$f (x, y) = x (\ln (| y |) + C)$

Step 13.1.10.3

Simplify.

$f (x, y) = x \ln (| y |) + C$

Step 13.1.11

Since the integral of $g (x)$ will contain an integration constant, we can replace $C$ with $g (x)$ .

$f (x, y) = x \ln (| y |) + g (x)$

Step 13.1.12

Set $\frac{\partial f}{\partial x} = M (x, y)$ .

$\frac{\partial f}{\partial x} = - \ln (\frac{x}{y})$

Step 13.1.13

Simplify the left side.

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Step 13.1.13.1

Simplify $\frac{d}{d x} \cdot (x \ln (| y |) + g (x))$ .

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Step 13.1.13.1.1

Cancel the common factor of $d$ .

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Step 13.1.13.1.1.1

Cancel the common factor.

$\frac{d}{d x} \cdot (x \ln (| y |) + g (x)) = - \ln (\frac{x}{y})$

Step 13.1.13.1.1.2

Rewrite the expression.

$\frac{1}{x} \cdot (x \ln (| y |) + g (x)) = - \ln (\frac{x}{y})$

Step 13.1.13.1.2

Multiply $\frac{1}{x}$ by $x \ln (| y |) + g x$ .

$\frac{x \ln (| y |) + g x}{x} = - \ln (\frac{x}{y})$

Step 13.1.13.1.3

Factor $x$ out of $x \ln (| y |) + g x$ .

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Step 13.1.13.1.3.1

Factor $x$ out of $x \ln (| y |)$ .

$\frac{x \ln (| y |) + g x}{x} = - \ln (\frac{x}{y})$

Step 13.1.13.1.3.2

Factor $x$ out of $g x$ .

$\frac{x \ln (| y |) + x g}{x} = - \ln (\frac{x}{y})$

Step 13.1.13.1.3.3

Factor $x$ out of $x \ln (| y |) + x g$ .

$\frac{x (\ln (| y |) + g)}{x} = - \ln (\frac{x}{y})$

Step 13.1.13.1.4

Cancel the common factor of $x$ .

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Step 13.1.13.1.4.1

Cancel the common factor.

$\frac{x (\ln (| y |) + g)}{x} = - \ln (\frac{x}{y})$

Step 13.1.13.1.4.2

Divide $\ln (| y |) + g$ by $1$ .

$\ln (| y |) + g = - \ln (\frac{x}{y})$

Step 13.1.14

Move all the terms containing a logarithm to the left side of the equation.

$\ln (| y |) + \ln (\frac{x}{y}) = - g$

Step 13.1.15

Use the product property of logarithms, $\log_{b} (x) + \log_{b} (y) = \log_{b} (x y)$ .

$\ln (| y | (\frac{x}{y})) = - g$

Step 13.1.16

Combine $| y |$ and $\frac{x}{y}$ .

$\ln (\frac{| y | x}{y}) = - g$

Step 13.1.17

Reorder factors in $\ln (\frac{| y | x}{y})$ .

$\ln (\frac{x | y |}{y}) = - g$

Step 14

Find the antiderivative of $- g$ to find $g (x)$ .

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Step 14.1

Integrate both sides of $\ln (\frac{x | y |}{y}) = - g$ .

$\int \ln (\frac{x | y |}{y}) d x = \int - g d x$

Step 14.2

Evaluate $\int \ln (\frac{x | y |}{y}) d x$ .

$g (x) = \int - g d x$

Step 14.3

Apply the constant rule.

$g (x) = - g x + C$

Step 15

Substitute for $g (x)$ in $f (x, y) = x \ln (| y |) + g (x)$ .

$f (x, y) = x \ln (| y |) - g x + C$