Calculus Examples

$\frac{d y}{d x} = \frac{x^{3} + y^{3}}{x^{2} y + x y^{2}}$

Step 1

Rewrite the differential equation as a function of $\frac{y}{x}$ .

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Step 1.1

Multiply $\frac{x^{3} + y^{3}}{x^{2} y + x y^{2}}$ by $\frac{\frac{1}{x^{3}}}{\frac{1}{x^{3}}}$ .

$\frac{d y}{d x} = \frac{x^{3} + y^{3}}{x^{2} y + x y^{2}} \cdot \frac{\frac{1}{x^{3}}}{\frac{1}{x^{3}}}$

Step 1.2

Multiply $\frac{x^{3} + y^{3}}{x^{2} y + x y^{2}}$ by $\frac{\frac{1}{x^{3}}}{\frac{1}{x^{3}}}$ .

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

Step 1.3

Apply the distributive property.

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

Step 1.4

Cancel the common factor of $x^{3}$ .

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Step 1.4.1

Cancel the common factor.

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

Step 1.4.2

Rewrite the expression.

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

Step 1.5

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

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

Step 1.6

Apply the distributive property.

$\frac{d y}{d x} = \frac{1 + \frac{y^{3}}{x^{3}}}{x^{2} y \frac{1}{x^{3}} + x y^{2} \frac{1}{x^{3}}}$

Step 1.7

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

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Step 1.7.1

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

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

Step 1.7.2

Factor $x^{2}$ out of $x^{3}$ .

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

Step 1.7.3

Cancel the common factor.

$\frac{d y}{d x} = \frac{1 + \frac{y^{3}}{x^{3}}}{x^{2} y \frac{1}{x^{2} x} + x y^{2} \frac{1}{x^{3}}}$

Step 1.7.4

Rewrite the expression.

$\frac{d y}{d x} = \frac{1 + \frac{y^{3}}{x^{3}}}{y \frac{1}{x} + x y^{2} \frac{1}{x^{3}}}$

Step 1.8

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

$\frac{d y}{d x} = \frac{1 + \frac{y^{3}}{x^{3}}}{\frac{y}{x} + x y^{2} \frac{1}{x^{3}}}$

Step 1.9

Cancel the common factor of $x$ .

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Step 1.9.1

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

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

Step 1.9.2

Factor $x$ out of $x^{3}$ .

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

Step 1.9.3

Cancel the common factor.

$\frac{d y}{d x} = \frac{1 + \frac{y^{3}}{x^{3}}}{\frac{y}{x} + x y^{2} \frac{1}{x \cdot x^{2}}}$

Step 1.9.4

Rewrite the expression.

$\frac{d y}{d x} = \frac{1 + \frac{y^{3}}{x^{3}}}{\frac{y}{x} + y^{2} \frac{1}{x^{2}}}$

Step 1.10

Combine $y^{2}$ and $\frac{1}{x^{2}}$ .

$\frac{d y}{d x} = \frac{1 + \frac{y^{3}}{x^{3}}}{\frac{y}{x} + \frac{y^{2}}{x^{2}}}$

Step 1.11

Use the power of quotient rule $\frac{a^{m}}{b^{m}} = {(\frac{a}{b})}^{m}$ .

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

Step 1.12

Use the power of quotient rule $\frac{a^{m}}{b^{m}} = {(\frac{a}{b})}^{m}$ .

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

Step 2

Let $V = \frac{y}{x}$ . Substitute $V$ for $\frac{y}{x}$ .

$\frac{d y}{d x} = \frac{1 + V^{3}}{V + V^{2}}$

Step 3

Solve $V = \frac{y}{x}$ for $y$ .

$y = V x$

Step 4

Use the product rule to find the derivative of $y = V x$ with respect to $x$ .

$\frac{d y}{d x} = x \frac{d V}{d x} + V$

Step 5

Substitute $x \frac{d V}{d x} + V$ for $\frac{d y}{d x}$ .

$x \frac{d V}{d x} + V = \frac{1 + V^{3}}{V + V^{2}}$

Step 6

Solve the substituted differential equation.

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Step 6.1

Separate the variables.

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Step 6.1.1

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

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Step 6.1.1.1

Simplify $\frac{1 + V^{3}}{V + V^{2}}$ .

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Step 6.1.1.1.1

Simplify the numerator.

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Step 6.1.1.1.1.1

Rewrite $1$ as $1^{3}$ .

$x \frac{d V}{d x} + V = \frac{1^{3} + V^{3}}{V + V^{2}}$

Step 6.1.1.1.1.2

Since both terms are perfect cubes, factor using the sum of cubes formula, $a^{3} + b^{3} = (a + b) (a^{2} - a b + b^{2})$ where $a = 1$ and $b = V$ .

$x \frac{d V}{d x} + V = \frac{(1 + V) (1^{2} - 1 V + V^{2})}{V + V^{2}}$

Step 6.1.1.1.1.3

Simplify.

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Step 6.1.1.1.1.3.1

One to any power is one.

$x \frac{d V}{d x} + V = \frac{(1 + V) (1 - 1 V + V^{2})}{V + V^{2}}$

Step 6.1.1.1.1.3.2

Rewrite $- 1 V$ as $- V$ .

$x \frac{d V}{d x} + V = \frac{(1 + V) (1 - V + V^{2})}{V + V^{2}}$

Step 6.1.1.1.2

Simplify terms.

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Step 6.1.1.1.2.1

Factor $V$ out of $V + V^{2}$ .

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Step 6.1.1.1.2.1.1

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

$x \frac{d V}{d x} + V = \frac{(1 + V) (1 - V + V^{2})}{V^{1} + V^{2}}$

Step 6.1.1.1.2.1.2

Factor $V$ out of $V^{1}$ .

$x \frac{d V}{d x} + V = \frac{(1 + V) (1 - V + V^{2})}{V \cdot 1 + V^{2}}$

Step 6.1.1.1.2.1.3

Factor $V$ out of $V^{2}$ .

$x \frac{d V}{d x} + V = \frac{(1 + V) (1 - V + V^{2})}{V \cdot 1 + V \cdot V}$

Step 6.1.1.1.2.1.4

Factor $V$ out of $V \cdot 1 + V \cdot V$ .

$x \frac{d V}{d x} + V = \frac{(1 + V) (1 - V + V^{2})}{V (1 + V)}$

Step 6.1.1.1.2.2

Cancel the common factor of $1 + V$ .

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Step 6.1.1.1.2.2.1

Cancel the common factor.

$x \frac{d V}{d x} + V = \frac{(1 + V) (1 - V + V^{2})}{V (1 + V)}$

Step 6.1.1.1.2.2.2

Rewrite the expression.

$x \frac{d V}{d x} + V = \frac{1 - V + V^{2}}{V}$

Step 6.1.1.2

Move all terms not containing $\frac{d V}{d x}$ to the right side of the equation.

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Step 6.1.1.2.1

Subtract $V$ from both sides of the equation.

$x \frac{d V}{d x} = \frac{1 - V + V^{2}}{V} - V$

Step 6.1.1.2.2

Simplify each term.

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Step 6.1.1.2.2.1

Split the fraction $\frac{1 - V + V^{2}}{V}$ into two fractions.

$x \frac{d V}{d x} = \frac{1 - V}{V} + \frac{V^{2}}{V} - V$

Step 6.1.1.2.2.2

Simplify each term.

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Step 6.1.1.2.2.2.1

Split the fraction $\frac{1 - V}{V}$ into two fractions.

$x \frac{d V}{d x} = \frac{1}{V} + \frac{- V}{V} + \frac{V^{2}}{V} - V$

Step 6.1.1.2.2.2.2

Cancel the common factor of $V$ .

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Step 6.1.1.2.2.2.2.1

Cancel the common factor.

$x \frac{d V}{d x} = \frac{1}{V} + \frac{- V}{V} + \frac{V^{2}}{V} - V$

Step 6.1.1.2.2.2.2.2

Divide $- 1$ by $1$ .

$x \frac{d V}{d x} = \frac{1}{V} - 1 + \frac{V^{2}}{V} - V$

Step 6.1.1.2.2.2.3

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

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Step 6.1.1.2.2.2.3.1

Factor $V$ out of $V^{2}$ .

$x \frac{d V}{d x} = \frac{1}{V} - 1 + \frac{V \cdot V}{V} - V$

Step 6.1.1.2.2.2.3.2

Cancel the common factors.

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Step 6.1.1.2.2.2.3.2.1

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

$x \frac{d V}{d x} = \frac{1}{V} - 1 + \frac{V \cdot V}{V^{1}} - V$

Step 6.1.1.2.2.2.3.2.2

Factor $V$ out of $V^{1}$ .

$x \frac{d V}{d x} = \frac{1}{V} - 1 + \frac{V \cdot V}{V \cdot 1} - V$

Step 6.1.1.2.2.2.3.2.3

Cancel the common factor.

$x \frac{d V}{d x} = \frac{1}{V} - 1 + \frac{V \cdot V}{V \cdot 1} - V$

Step 6.1.1.2.2.2.3.2.4

Rewrite the expression.

$x \frac{d V}{d x} = \frac{1}{V} - 1 + \frac{V}{1} - V$

Step 6.1.1.2.2.2.3.2.5

Divide $V$ by $1$ .

$x \frac{d V}{d x} = \frac{1}{V} - 1 + V - V$

Step 6.1.1.2.3

Combine the opposite terms in $\frac{1}{V} - 1 + V - V$ .

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Step 6.1.1.2.3.1

Subtract $V$ from $V$ .

$x \frac{d V}{d x} = \frac{1}{V} - 1 + 0$

Step 6.1.1.2.3.2

Add $\frac{1}{V} - 1$ and $0$ .

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

Step 6.1.1.3

Divide each term in $x \frac{d V}{d x} = \frac{1}{V} - 1$ by $x$ and simplify.

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Step 6.1.1.3.1

Divide each term in $x \frac{d V}{d x} = \frac{1}{V} - 1$ by $x$ .

$\frac{x \frac{d V}{d x}}{x} = \frac{\frac{1}{V}}{x} + \frac{- 1}{x}$

Step 6.1.1.3.2

Simplify the left side.

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Step 6.1.1.3.2.1

Cancel the common factor of $x$ .

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Step 6.1.1.3.2.1.1

Cancel the common factor.

$\frac{x \frac{d V}{d x}}{x} = \frac{\frac{1}{V}}{x} + \frac{- 1}{x}$

Step 6.1.1.3.2.1.2

Divide $\frac{d V}{d x}$ by $1$ .

$\frac{d V}{d x} = \frac{\frac{1}{V}}{x} + \frac{- 1}{x}$

Step 6.1.1.3.3

Simplify the right side.

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Step 6.1.1.3.3.1

Simplify each term.

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Step 6.1.1.3.3.1.1

Multiply the numerator by the reciprocal of the denominator.

$\frac{d V}{d x} = \frac{1}{V} \cdot \frac{1}{x} + \frac{- 1}{x}$

Step 6.1.1.3.3.1.2

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

$\frac{d V}{d x} = \frac{1}{V x} + \frac{- 1}{x}$

Step 6.1.1.3.3.1.3

Move the negative in front of the fraction.

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

Step 6.1.2

Factor.

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Step 6.1.2.1

To write $- \frac{1}{x}$ as a fraction with a common denominator, multiply by $\frac{V}{V}$ .

$\frac{d V}{d x} = \frac{1}{V x} - \frac{1}{x} \cdot \frac{V}{V}$

Step 6.1.2.2

Write each expression with a common denominator of $V x$ , by multiplying each by an appropriate factor of $1$ .

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Step 6.1.2.2.1

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

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

Step 6.1.2.2.2

Reorder the factors of $x V$ .

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

Step 6.1.2.3

Combine the numerators over the common denominator.

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

Step 6.1.3

Regroup factors.

$\frac{d V}{d x} = \frac{1 - V}{V} \cdot \frac{1}{x}$

Step 6.1.4

Multiply both sides by $\frac{V}{1 - V}$ .

$\frac{V}{1 - V} \frac{d V}{d x} = \frac{V}{1 - V} (\frac{1 - V}{V} \cdot \frac{1}{x})$

Step 6.1.5

Simplify.

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Step 6.1.5.1

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

$\frac{V}{1 - V} \frac{d V}{d x} = \frac{V}{1 - V} \cdot \frac{1 - V}{V x}$

Step 6.1.5.2

Cancel the common factor of $V$ .

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Step 6.1.5.2.1

Factor $V$ out of $V x$ .

$\frac{V}{1 - V} \frac{d V}{d x} = \frac{V}{1 - V} \cdot \frac{1 - V}{V (x)}$

Step 6.1.5.2.2

Cancel the common factor.

$\frac{V}{1 - V} \frac{d V}{d x} = \frac{V}{1 - V} \cdot \frac{1 - V}{V x}$

Step 6.1.5.2.3

Rewrite the expression.

$\frac{V}{1 - V} \frac{d V}{d x} = \frac{1}{1 - V} \cdot \frac{1 - V}{x}$

Step 6.1.5.3

Cancel the common factor of $1 - V$ .

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Step 6.1.5.3.1

Cancel the common factor.

$\frac{V}{1 - V} \frac{d V}{d x} = \frac{1}{1 - V} \cdot \frac{1 - V}{x}$

Step 6.1.5.3.2

Rewrite the expression.

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

Step 6.1.6

Rewrite the equation.

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

Step 6.2

Integrate both sides.

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Step 6.2.1

Set up an integral on each side.

$\int \frac{V}{1 - V} d V = \int \frac{1}{x} d x$

Step 6.2.2

Integrate the left side.

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Step 6.2.2.1

Reorder $1$ and $- V$ .

$\int \frac{V}{- V + 1} d V = \int \frac{1}{x} d x$

Step 6.2.2.2

Divide $V$ by $- V + 1$ .

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Step 6.2.2.2.1

Set up the polynomials to be divided. If there is not a term for every exponent, insert one with a value of $0$ .

$V$

$1$

$V$

$0$

Step 6.2.2.2.2

Divide the highest order term in the dividend $V$ by the highest order term in divisor $- V$ .

				-	$1$
-	$V$	+	$1$		$V$	+	$0$

Step 6.2.2.2.3

Multiply the new quotient term by the divisor.

				-	$1$
-	$V$	+	$1$		$V$	+	$0$
				+	$V$	-	$1$

Step 6.2.2.2.4

The expression needs to be subtracted from the dividend, so change all the signs in $V - 1$

				-	$1$
-	$V$	+	$1$		$V$	+	$0$
				-	$V$	+	$1$

Step 6.2.2.2.5

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

				-	$1$
-	$V$	+	$1$		$V$	+	$0$
				-	$V$	+	$1$
						+	$1$

Step 6.2.2.2.6

The final answer is the quotient plus the remainder over the divisor.

$\int - 1 + \frac{1}{- V + 1} d V = \int \frac{1}{x} d x$

Step 6.2.2.3

Split the single integral into multiple integrals.

$\int - 1 d V + \int \frac{1}{- V + 1} d V = \int \frac{1}{x} d x$

Step 6.2.2.4

Apply the constant rule.

$- V + C_{1} + \int \frac{1}{- V + 1} d V = \int \frac{1}{x} d x$

Step 6.2.2.5

Let $u = - V + 1$ . Then $d u = - d V$ , so $- d u = d V$ . Rewrite using $u$ and $d$ $u$ .

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Step 6.2.2.5.1

Let $u = - V + 1$ . Find $\frac{d u}{d V}$ .

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Step 6.2.2.5.1.1

Rewrite.

$\frac{- 1}{1}$

Step 6.2.2.5.1.2

Divide $- 1$ by $1$ .

$- 1$

Step 6.2.2.5.2

Rewrite the problem using $u$ and $d u$ .

$- V + C_{1} + \int \frac{- 1}{u} d u = \int \frac{1}{x} d x$

Step 6.2.2.6

Move the negative in front of the fraction.

$- V + C_{1} + \int - \frac{1}{u} d u = \int \frac{1}{x} d x$

Step 6.2.2.7

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

$- V + C_{1} - \int \frac{1}{u} d u = \int \frac{1}{x} d x$

Step 6.2.2.8

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

$- V + C_{1} - (\ln (| u |) + C_{2}) = \int \frac{1}{x} d x$

Step 6.2.2.9

Simplify.

$- V - \ln (| u |) + C_{3} = \int \frac{1}{x} d x$

Step 6.2.2.10

Replace all occurrences of $u$ with $- V + 1$ .

$- V - \ln (| - V + 1 |) + C_{3} = \int \frac{1}{x} d x$

Step 6.2.3

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

$- V - \ln (| - V + 1 |) + C_{3} = \ln (| x |) + C_{4}$

Step 6.2.4

Group the constant of integration on the right side as $C$ .

$- V - \ln (| - V + 1 |) = \ln (| x |) + C$

Step 7

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

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

Step 8

Solve $- \frac{y}{x} - \ln (| - \frac{y}{x} + 1 |) = \ln (| x |) + C$ for $y$ .

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Step 8.1

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

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

Step 8.2

Add $\ln (| x |)$ to both sides of the equation.

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

Step 8.3

Divide each term in $- \ln (| - \frac{y}{x} + 1 |) = \frac{y}{x} + C + \ln (| x |)$ by $- 1$ and simplify.

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Step 8.3.1

Divide each term in $- \ln (| - \frac{y}{x} + 1 |) = \frac{y}{x} + C + \ln (| x |)$ by $- 1$ .

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

Step 8.3.2

Simplify the left side.

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Step 8.3.2.1

Dividing two negative values results in a positive value.

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

Step 8.3.2.2

Divide $\ln (| - \frac{y}{x} + 1 |)$ by $1$ .

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

Step 8.3.3

Simplify the right side.

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Step 8.3.3.1

Simplify each term.

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Step 8.3.3.1.1

Move the negative one from the denominator of $\frac{\frac{y}{x}}{- 1}$ .

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

Step 8.3.3.1.2

Rewrite $- 1 \cdot \frac{y}{x}$ as $- \frac{y}{x}$ .

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

Step 8.3.3.1.3

Move the negative one from the denominator of $\frac{C}{- 1}$ .

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

Step 8.3.3.1.4

Rewrite $- 1 \cdot C$ as $- C$ .

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

Step 8.3.3.1.5

Move the negative one from the denominator of $\frac{\ln (| x |)}{- 1}$ .

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

Step 8.3.3.1.6

Rewrite $- 1 \cdot \ln (| x |)$ as $- \ln (| x |)$ .

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

Step 8.4

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

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

Step 8.5

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

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

Step 8.6

To multiply absolute values, multiply the terms inside each absolute value.

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

Step 8.7

Apply the distributive property.

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

Step 8.8

Cancel the common factor of $x$ .

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Step 8.8.1

Move the leading negative in $- \frac{y}{x}$ into the numerator.

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

Step 8.8.2

Cancel the common factor.

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

Step 8.8.3

Rewrite the expression.

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

Step 8.9

Multiply $x$ by $1$ .

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