Calculus Examples

$\frac{y}{x} d x + (y^{3} - \ln (x)) d y = 0$

Step 1

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

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Step 1.1

Differentiate $M$ with respect to $y$ .

$\frac{\partial M}{\partial y} = \frac{d}{d y} [\frac{y}{x}]$

Step 1.2

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

$\frac{\partial M}{\partial y} = \frac{1}{x} \frac{d}{d y} [y]$

Step 1.3

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

$\frac{\partial M}{\partial y} = \frac{1}{x} \cdot 1$

Step 1.4

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

$\frac{\partial M}{\partial y} = \frac{1}{x}$

Step 2

Find $\frac{\partial N}{\partial x}$ where $N (x, y) = y^{3} - \ln (x)$ .

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Step 2.1

Differentiate $N$ with respect to $x$ .

$\frac{\partial N}{\partial x} = \frac{d}{d x} [y^{3} - \ln (x)]$

Step 2.2

Differentiate.

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Step 2.2.1

By the Sum Rule, the derivative of $y^{3} - \ln (x)$ with respect to $x$ is $\frac{d}{d x} [y^{3}] + \frac{d}{d x} [- \ln (x)]$ .

$\frac{\partial N}{\partial x} = \frac{d}{d x} [y^{3}] + \frac{d}{d x} [- \ln (x)]$

Step 2.2.2

Since $y^{3}$ is constant with respect to $x$ , the derivative of $y^{3}$ with respect to $x$ is $0$ .

$\frac{\partial N}{\partial x} = 0 + \frac{d}{d x} [- \ln (x)]$

Step 2.3

Evaluate $\frac{d}{d x} [- \ln (x)]$ .

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Step 2.3.1

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

$\frac{\partial N}{\partial x} = 0 - \frac{d}{d x} [\ln (x)]$

Step 2.3.2

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

$\frac{\partial N}{\partial x} = 0 - \frac{1}{x}$

Step 2.4

Subtract $\frac{1}{x}$ from $0$ .

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

Step 3

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

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Step 3.1

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

$\frac{1}{x} = - \frac{1}{x}$

Step 3.2

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

$\frac{1}{x} = - \frac{1}{x}$ is not an identity.

Step 4

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 4.1

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

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

Step 4.2

Substitute $\frac{1}{x}$ for $\frac{\partial M}{\partial y}$ .

$\frac{- \frac{1}{x} - \frac{1}{x}}{M}$

Step 4.3

Substitute $\frac{y}{x}$ for $M$ .

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Step 4.3.1

Substitute $\frac{y}{x}$ for $M$ .

$\frac{- \frac{1}{x} - \frac{1}{x}}{\frac{y}{x}}$

Step 4.3.2

Multiply the numerator by the reciprocal of the denominator.

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

Step 4.3.3

Combine the numerators over the common denominator.

$\frac{- 1 - 1}{x} \cdot \frac{x}{y}$

Step 4.3.4

Subtract $1$ from $- 1$ .

$\frac{- 2}{x} \cdot \frac{x}{y}$

Step 4.3.5

Cancel the common factor of $x$ .

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Step 4.3.5.1

Cancel the common factor.

$\frac{- 2}{x} \cdot \frac{x}{y}$

Step 4.3.5.2

Rewrite the expression.

$- 2 \frac{1}{y}$

Step 4.3.6

Combine $- 2$ and $\frac{1}{y}$ .

$\frac{- 2}{y}$

Step 4.3.7

Substitute $\frac{y}{x}$ for $M$ .

$- \frac{2}{y}$

Step 4.4

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

$μ (x, y) = e^{\int - \frac{2}{y} d y}$

Step 5

Evaluate the integral $e^{\int - \frac{2}{y} d y}$ .

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Step 5.1

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

$μ (x, y) = e^{- \int \frac{2}{y} d y}$

Step 5.2

Since $2$ is constant with respect to $y$ , move $2$ out of the integral.

$μ (x, y) = e^{- (2 \int \frac{1}{y} d y)}$

Step 5.3

Multiply $2$ by $- 1$ .

$μ (x, y) = e^{- 2 \int \frac{1}{y} d y}$

Step 5.4

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

$μ (x, y) = e^{- 2 (\ln (| y |) + C)}$

Step 5.5

Simplify.

$μ (x, y) = e^{- 2 \ln (| y |)} + C$

Step 5.6

Simplify each term.

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Step 5.6.1

Simplify $- 2 \ln (| y |)$ by moving $- 2$ inside the logarithm.

$μ (x, y) = e^{\ln ({| y |}^{- 2})} + C$

Step 5.6.2

Exponentiation and log are inverse functions.

$μ (x, y) = {| y |}^{- 2} + C$

Step 5.6.3

Remove the absolute value in ${| y |}^{- 2}$ because exponentiations with even powers are always positive.

$μ (x, y) = y^{- 2} + C$

Step 5.6.4

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

$μ (x, y) = \frac{1}{y^{2}} + C$

Step 6

Multiply both sides of $\frac{y}{x} d x + (y^{3} - \ln (x)) d y = 0$ by the integration factor $\frac{1}{y^{2}}$ .

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Step 6.1

Multiply $\frac{y}{x}$ by $\frac{1}{y^{2}}$ .

$\frac{y}{x} \cdot \frac{1}{y^{2}} d x + (y^{3} - \ln (x)) d y = 0$

Step 6.2

Combine.

$\frac{y \cdot 1}{x y^{2}} d x + (y^{3} - \ln (x)) d y = 0$

Step 6.3

Cancel the common factor of $y$ and $y^{2}$ .

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Step 6.3.1

Factor $y$ out of $y \cdot 1$ .

$\frac{y (1)}{x y^{2}} d x + (y^{3} - \ln (x)) d y = 0$

Step 6.3.2

Cancel the common factors.

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Step 6.3.2.1

Factor $y$ out of $x y^{2}$ .

$\frac{y (1)}{y (x y)} d x + (y^{3} - \ln (x)) d y = 0$

Step 6.3.2.2

Cancel the common factor.

$\frac{y \cdot 1}{y (x y)} d x + (y^{3} - \ln (x)) d y = 0$

Step 6.3.2.3

Rewrite the expression.

$\frac{1}{x y} d x + (y^{3} - \ln (x)) d y = 0$

Step 6.4

Multiply $y^{3} - \ln (x)$ by $\frac{1}{y^{2}}$ .

$\frac{1}{x y} d x + (y^{3} - \ln (x)) \frac{1}{y^{2}} d y = 0$

Step 6.5

Multiply $y^{3} - \ln (x)$ by $\frac{1}{y^{2}}$ .

$\frac{1}{x y} d x + \frac{y^{3} - \ln (x)}{y^{2}} d y = 0$

Step 7

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

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

Step 8

Integrate $M (x, y) = \frac{1}{x y}$ to find $f (x, y)$ .

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Step 8.1

Since $\frac{1}{y}$ is constant with respect to $x$ , move $\frac{1}{y}$ out of the integral.

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

Step 8.2

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

$f (x, y) = \frac{1}{y} (\ln (| x |) + C)$

Step 8.3

Simplify the answer.

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Step 8.3.1

Simplify.

$f (x, y) = \frac{1}{y} \ln (| x |) + C$

Step 8.3.2

Combine $\frac{1}{y}$ and $\ln (| x |)$ .

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

Step 9

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

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

Step 10

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

$\frac{\partial f}{\partial y} = \frac{y^{3} - \ln (x)}{y^{2}}$

Step 11

Differentiate $f$ with respect to $y$ .

$\frac{d}{d y} [\frac{\ln (| x |)}{y} + g (y)] = \frac{y^{3} - \ln (x)}{y^{2}}$

Step 12

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

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Step 12.1

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

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Step 12.1.1

Rewrite.

$\frac{y}{x} d x + (y^{3} - \ln (x)) d y = 0$

Step 12.1.2

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

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Step 12.1.2.1

Differentiate $M$ with respect to $y$ .

$\frac{\partial M}{\partial y} = \frac{d}{d y} [\frac{y}{x}]$

Step 12.1.2.2

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

$\frac{\partial M}{\partial y} = \frac{1}{x} \frac{d}{d y} [y]$

Step 12.1.2.3

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

$\frac{\partial M}{\partial y} = \frac{1}{x} \cdot 1$

Step 12.1.2.4

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

$\frac{\partial M}{\partial y} = \frac{1}{x}$

Step 12.1.3

Find $\frac{\partial N}{\partial x}$ where $N (x, y) = y^{3} - \ln (x)$ .

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Step 12.1.3.1

Differentiate $N$ with respect to $x$ .

$\frac{\partial N}{\partial x} = \frac{d}{d x} [y^{3} - \ln (x)]$

Step 12.1.3.2

Differentiate.

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Step 12.1.3.2.1

By the Sum Rule, the derivative of $y^{3} - \ln (x)$ with respect to $x$ is $\frac{d}{d x} [y^{3}] + \frac{d}{d x} [- \ln (x)]$ .

$\frac{\partial N}{\partial x} = \frac{d}{d x} [y^{3}] + \frac{d}{d x} [- \ln (x)]$

Step 12.1.3.2.2

Since $y^{3}$ is constant with respect to $x$ , the derivative of $y^{3}$ with respect to $x$ is $0$ .

$\frac{\partial N}{\partial x} = 0 + \frac{d}{d x} [- \ln (x)]$

Step 12.1.3.3

Evaluate $\frac{d}{d x} [- \ln (x)]$ .

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Step 12.1.3.3.1

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

$\frac{\partial N}{\partial x} = 0 - \frac{d}{d x} [\ln (x)]$

Step 12.1.3.3.2

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

$\frac{\partial N}{\partial x} = 0 - \frac{1}{x}$

Step 12.1.3.4

Subtract $\frac{1}{x}$ from $0$ .

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

Step 12.1.4

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

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Step 12.1.4.1

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

$\frac{1}{x} = - \frac{1}{x}$

Step 12.1.4.2

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

$\frac{1}{x} = - \frac{1}{x}$ is not an identity.

Step 12.1.5

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 12.1.5.1

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

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

Step 12.1.5.2

Substitute $\frac{1}{x}$ for $\frac{\partial M}{\partial y}$ .

$\frac{- \frac{1}{x} - \frac{1}{x}}{M}$

Step 12.1.5.3

Substitute $\frac{y}{x}$ for $M$ .

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Step 12.1.5.3.1

Substitute $\frac{y}{x}$ for $M$ .

$\frac{- \frac{1}{x} - \frac{1}{x}}{\frac{y}{x}}$

Step 12.1.5.3.2

Multiply the numerator by the reciprocal of the denominator.

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

Step 12.1.5.3.3

Combine the numerators over the common denominator.

$\frac{- 1 - 1}{x} \cdot \frac{x}{y}$

Step 12.1.5.3.4

Subtract $1$ from $- 1$ .

$\frac{- 2}{x} \cdot \frac{x}{y}$

Step 12.1.5.3.5

Cancel the common factor of $x$ .

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Step 12.1.5.3.5.1

Cancel the common factor.

$\frac{- 2}{x} \cdot \frac{x}{y}$

Step 12.1.5.3.5.2

Rewrite the expression.

$- 2 \frac{1}{y}$

Step 12.1.5.3.6

Combine $- 2$ and $\frac{1}{y}$ .

$\frac{- 2}{y}$

Step 12.1.5.3.7

Substitute $\frac{y}{x}$ for $M$ .

$- \frac{2}{y}$

Step 12.1.5.4

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

$μ (x, y) = e^{\int - \frac{2}{y} d y}$

Step 12.1.6

Evaluate the integral $e^{\int - \frac{2}{y} d y}$ .

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Step 12.1.6.1

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

$μ (x, y) = e^{- \int \frac{2}{y} d y}$

Step 12.1.6.2

Since $2$ is constant with respect to $y$ , move $2$ out of the integral.

$μ (x, y) = e^{- (2 \int \frac{1}{y} d y)}$

Step 12.1.6.3

Multiply $2$ by $- 1$ .

$μ (x, y) = e^{- 2 \int \frac{1}{y} d y}$

Step 12.1.6.4

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

$μ (x, y) = e^{- 2 (\ln (| y |) + C)}$

Step 12.1.6.5

Simplify.

$μ (x, y) = e^{- 2 \ln (| y |)} + C$

Step 12.1.6.6

Simplify each term.

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Step 12.1.6.6.1

Simplify $- 2 \ln (| y |)$ by moving $- 2$ inside the logarithm.

$μ (x, y) = e^{\ln ({| y |}^{- 2})} + C$

Step 12.1.6.6.2

Exponentiation and log are inverse functions.

$μ (x, y) = {| y |}^{- 2} + C$

Step 12.1.6.6.3

Remove the absolute value in ${| y |}^{- 2}$ because exponentiations with even powers are always positive.

$μ (x, y) = y^{- 2} + C$

Step 12.1.6.6.4

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

$μ (x, y) = \frac{1}{y^{2}} + C$

Step 12.1.7

Multiply both sides of $\frac{y}{x} d x + (y^{3} - \ln (x)) d y = 0$ by the integration factor $\frac{1}{y^{2}}$ .

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Step 12.1.7.1

Multiply $\frac{y}{x}$ by $\frac{1}{y^{2}}$ .

$\frac{y}{x} \cdot \frac{1}{y^{2}} d x + (y^{3} - \ln (x)) d y = 0$

Step 12.1.7.2

Combine.

$\frac{y \cdot 1}{x y^{2}} d x + (y^{3} - \ln (x)) d y = 0$

Step 12.1.7.3

Cancel the common factor of $y$ and $y^{2}$ .

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Step 12.1.7.3.1

Factor $y$ out of $y \cdot 1$ .

$\frac{y (1)}{x y^{2}} d x + (y^{3} - \ln (x)) d y = 0$

Step 12.1.7.3.2

Cancel the common factors.

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Step 12.1.7.3.2.1

Factor $y$ out of $x y^{2}$ .

$\frac{y (1)}{y (x y)} d x + (y^{3} - \ln (x)) d y = 0$

Step 12.1.7.3.2.2

Cancel the common factor.

$\frac{y \cdot 1}{y (x y)} d x + (y^{3} - \ln (x)) d y = 0$

Step 12.1.7.3.2.3

Rewrite the expression.

$\frac{1}{x y} d x + (y^{3} - \ln (x)) d y = 0$

Step 12.1.7.4

Multiply $y^{3} - \ln (x)$ by $\frac{1}{y^{2}}$ .

$\frac{1}{x y} d x + (y^{3} - \ln (x)) \frac{1}{y^{2}} d y = 0$

Step 12.1.7.5

Multiply $y^{3} - \ln (x)$ by $\frac{1}{y^{2}}$ .

$\frac{1}{x y} d x + \frac{y^{3} - \ln (x)}{y^{2}} d y = 0$

Step 12.1.8

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

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

Step 12.1.9

Integrate $M (x, y) = \frac{1}{x y}$ to find $f (x, y)$ .

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Step 12.1.9.1

Since $\frac{1}{y}$ is constant with respect to $x$ , move $\frac{1}{y}$ out of the integral.

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

Step 12.1.9.2

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

$f (x, y) = \frac{1}{y} (\ln (| x |) + C)$

Step 12.1.9.3

Simplify the answer.

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Step 12.1.9.3.1

Simplify.

$f (x, y) = \frac{1}{y} \ln (| x |) + C$

Step 12.1.9.3.2

Combine $\frac{1}{y}$ and $\ln (| x |)$ .

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

Step 12.1.10

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

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

Step 12.1.11

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

$\frac{\partial f}{\partial y} = \frac{y^{3} - \ln (x)}{y^{2}}$

Step 12.1.12

Simplify the left side.

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Step 12.1.12.1

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

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Step 12.1.12.1.1

Simplify terms.

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Step 12.1.12.1.1.1

Cancel the common factor of $d$ .

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Step 12.1.12.1.1.1.1

Cancel the common factor.

$\frac{d}{d y} \cdot (\frac{\ln (| x |)}{y} + g (y)) = \frac{y^{3} - \ln (x)}{y^{2}}$

Step 12.1.12.1.1.1.2

Rewrite the expression.

$\frac{1}{y} \cdot (\frac{\ln (| x |)}{y} + g (y)) = \frac{y^{3} - \ln (x)}{y^{2}}$

Step 12.1.12.1.1.2

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

$\frac{\frac{\ln (| x |)}{y} + g y}{y} = \frac{y^{3} - \ln (x)}{y^{2}}$

Step 12.1.12.1.2

Simplify the numerator.

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Step 12.1.12.1.2.1

To write $g y$ as a fraction with a common denominator, multiply by $\frac{y}{y}$ .

$\frac{\frac{\ln (| x |)}{y} + \frac{g y \cdot y}{y}}{y} = \frac{y^{3} - \ln (x)}{y^{2}}$

Step 12.1.12.1.2.2

Combine the numerators over the common denominator.

$\frac{\frac{\ln (| x |) + g y \cdot y}{y}}{y} = \frac{y^{3} - \ln (x)}{y^{2}}$

Step 12.1.12.1.2.3

Multiply $y$ by $y$ by adding the exponents.

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Step 12.1.12.1.2.3.1

Move $y$ .

$\frac{\frac{\ln (| x |) + g (y \cdot y)}{y}}{y} = \frac{y^{3} - \ln (x)}{y^{2}}$

Step 12.1.12.1.2.3.2

Multiply $y$ by $y$ .

$\frac{\frac{\ln (| x |) + g y^{2}}{y}}{y} = \frac{y^{3} - \ln (x)}{y^{2}}$

Step 12.1.12.1.3

Multiply the numerator by the reciprocal of the denominator.

$\frac{\ln (| x |) + g y^{2}}{y} \cdot \frac{1}{y} = \frac{y^{3} - \ln (x)}{y^{2}}$

Step 12.1.12.1.4

Multiply $\frac{\ln (| x |) + g y^{2}}{y} \cdot \frac{1}{y}$ .

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Step 12.1.12.1.4.1

Multiply $\frac{\ln (| x |) + g y^{2}}{y}$ by $\frac{1}{y}$ .

$\frac{\ln (| x |) + g y^{2}}{y \cdot y} = \frac{y^{3} - \ln (x)}{y^{2}}$

Step 12.1.12.1.4.2

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

$\frac{\ln (| x |) + g y^{2}}{y^{1} y} = \frac{y^{3} - \ln (x)}{y^{2}}$

Step 12.1.12.1.4.3

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

$\frac{\ln (| x |) + g y^{2}}{y^{1} y^{1}} = \frac{y^{3} - \ln (x)}{y^{2}}$

Step 12.1.12.1.4.4

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

$\frac{\ln (| x |) + g y^{2}}{y^{1 + 1}} = \frac{y^{3} - \ln (x)}{y^{2}}$

Step 12.1.12.1.4.5

Add $1$ and $1$ .

$\frac{\ln (| x |) + g y^{2}}{y^{2}} = \frac{y^{3} - \ln (x)}{y^{2}}$

Step 12.1.13

Multiply each term in $\frac{\ln (| x |) + g y^{2}}{y^{2}} = \frac{y^{3} - \ln (x)}{y^{2}}$ by $y^{2}$ to eliminate the fractions.

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Step 12.1.13.1

Multiply each term in $\frac{\ln (| x |) + g y^{2}}{y^{2}} = \frac{y^{3} - \ln (x)}{y^{2}}$ by $y^{2}$ .

$\frac{\ln (| x |) + g y^{2}}{y^{2}} y^{2} = \frac{y^{3} - \ln (x)}{y^{2}} y^{2}$

Step 12.1.13.2

Simplify the left side.

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Step 12.1.13.2.1

Cancel the common factor of $y^{2}$ .

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Step 12.1.13.2.1.1

Cancel the common factor.

$\frac{\ln (| x |) + g y^{2}}{y^{2}} y^{2} = \frac{y^{3} - \ln (x)}{y^{2}} y^{2}$

Step 12.1.13.2.1.2

Rewrite the expression.

$\ln (| x |) + g y^{2} = \frac{y^{3} - \ln (x)}{y^{2}} y^{2}$

Step 12.1.13.3

Simplify the right side.

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Step 12.1.13.3.1

Cancel the common factor of $y^{2}$ .

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Step 12.1.13.3.1.1

Cancel the common factor.

$\ln (| x |) + g y^{2} = \frac{y^{3} - \ln (x)}{y^{2}} y^{2}$

Step 12.1.13.3.1.2

Rewrite the expression.

$\ln (| x |) + g y^{2} = y^{3} - \ln (x)$

Step 12.1.14

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

$\ln (| x |) + \ln (x) = - g y^{2} + y^{3}$

Step 12.1.15

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

$\ln (| x | x) = - g y^{2} + y^{3}$

Step 12.1.16

Reorder factors in $\ln (| x | x)$ .

$\ln (x | x |) = - g y^{2} + y^{3}$

Step 13

Find the antiderivative of $- g y^{2} + y^{3}$ to find $g (y)$ .

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Step 13.1

Integrate both sides of $\ln (x | x |) = - g y^{2} + y^{3}$ .

$\int \ln (x | x |) d y = \int - g y^{2} + y^{3} d y$

Step 13.2

Rewrite.

$0 + 0 + g (y) = \int - g y^{2} + y^{3} d y$

Step 13.3

Add $0$ and $0$ .

$0 + g (y) = \int - g y^{2} + y^{3} d y$

Step 13.4

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

$g (y) = \int - g y^{2} + y^{3} d y$

Step 13.5

Split the single integral into multiple integrals.

$g (y) = \int - g y^{2} d y + \int y^{3} d y$

Step 13.6

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

$g (y) = - g \int y^{2} d y + \int y^{3} d y$

Step 13.7

By the Power Rule, the integral of $y^{2}$ with respect to $y$ is $\frac{1}{3} y^{3}$ .

$g (y) = - g (\frac{1}{3} y^{3} + C) + \int y^{3} d y$

Step 13.8

By the Power Rule, the integral of $y^{3}$ with respect to $y$ is $\frac{1}{4} y^{4}$ .

$g (y) = - g (\frac{1}{3} y^{3} + C) + \frac{1}{4} y^{4} + C$

Step 13.9

Combine $\frac{1}{3}$ and $y^{3}$ .

$g (y) = - g (\frac{y^{3}}{3} + C) + \frac{1}{4} y^{4} + C$

Step 13.10

Simplify.

$g (y) = - \frac{y^{3} g}{3} + \frac{1}{4} y^{4} + C$

Step 13.11

Simplify.

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Step 13.11.1

Reorder terms.

$g (y) = - (\frac{1}{3} y^{3} g) + \frac{1}{4} y^{4} + C$

Step 13.11.2

Remove parentheses.

$g (y) = - (\frac{1}{3} y^{3}) g + \frac{1}{4} y^{4} + C$

Step 13.11.3

Remove parentheses.

$g (y) = - \frac{1}{3} y^{3} g + \frac{1}{4} y^{4} + C$

Step 14

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

$f (x, y) = \frac{\ln (| x |)}{y} - \frac{1}{3} y^{3} g + \frac{1}{4} y^{4} + C$

Step 15

Simplify each term.

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Step 15.1

Combine $y^{3}$ and $\frac{1}{3}$ .

$f (x, y) = \frac{\ln (| x |)}{y} - \frac{y^{3}}{3} g + \frac{1}{4} y^{4} + C$

Step 15.2

Combine $g$ and $\frac{y^{3}}{3}$ .

$f (x, y) = \frac{\ln (| x |)}{y} - \frac{g y^{3}}{3} + \frac{1}{4} y^{4} + C$

Step 15.3

Combine $\frac{1}{4}$ and $y^{4}$ .

$f (x, y) = \frac{\ln (| x |)}{y} - \frac{g y^{3}}{3} + \frac{y^{4}}{4} + C$