Calculus Examples

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

Step 1

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

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Step 1.1

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

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

Step 1.2

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

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

Step 1.3

Apply the distributive property.

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

Step 1.4

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

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Step 1.4.1

Cancel the common factor.

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

Step 1.4.2

Rewrite the expression.

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

Step 1.5

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

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

Step 1.6

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

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Step 1.6.1

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

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

Step 1.6.2

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

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

Step 1.6.3

Cancel the common factor.

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

Step 1.6.4

Rewrite the expression.

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

Step 1.7

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

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

Step 1.8

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

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

Step 2

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

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

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 $\frac{d y}{d x}$ .

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

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 the denominator.

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Step 6.1.1.1.1

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

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

Step 6.1.1.1.2

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 $b = V$ .

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

Step 6.1.1.1.3

Simplify.

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Step 6.1.1.1.3.1

One to any power is one.

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

Step 6.1.1.1.3.2

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

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

Step 6.1.1.2

Subtract $V$ from both sides of the equation.

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

Step 6.1.1.3

Divide each term in $x \frac{d V}{d x} = \frac{V}{(1 + V) (1 - V + V^{2})} - V$ by $x$ and simplify.

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Step 6.1.1.3.1

Divide each term in $x \frac{d V}{d x} = \frac{V}{(1 + V) (1 - V + V^{2})} - V$ by $x$ .

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

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.

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

Step 6.1.1.3.2.1.2

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

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

Step 6.1.1.3.3

Simplify the right side.

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Step 6.1.1.3.3.1

Combine the numerators over the common denominator.

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

Step 6.1.1.3.3.2

Simplify the numerator.

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Step 6.1.1.3.3.2.1

Factor $V$ out of $\frac{V}{(1 + V) (1 - V + V^{2})} - V$ .

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Step 6.1.1.3.3.2.1.1

Factor $V$ out of $\frac{V}{(1 + V) (1 - V + V^{2})}$ .

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

Step 6.1.1.3.3.2.1.2

Factor $V$ out of $- V$ .

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

Step 6.1.1.3.3.2.1.3

Factor $V$ out of $V \frac{1}{(1 + V) (1 - V + V^{2})} + V \cdot - 1$ .

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

Step 6.1.1.3.3.2.2

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{(1 + V) (1 - V + V^{2})}{(1 + V) (1 - V + V^{2})}$ .

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

Step 6.1.1.3.3.2.3

Combine $- 1$ and $\frac{(1 + V) (1 - V + V^{2})}{(1 + V) (1 - V + V^{2})}$ .

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

Step 6.1.1.3.3.2.4

Combine the numerators over the common denominator.

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

Step 6.1.1.3.3.2.5

Simplify the numerator.

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Step 6.1.1.3.3.2.5.1

Apply the distributive property.

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

Step 6.1.1.3.3.2.5.2

Multiply $- 1$ by $1$ .

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

Step 6.1.1.3.3.2.5.3

Expand $(- 1 - V) (1 - V + V^{2})$ by multiplying each term in the first expression by each term in the second expression.

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

Step 6.1.1.3.3.2.5.4

Simplify each term.

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Step 6.1.1.3.3.2.5.4.1

Multiply $- 1$ by $1$ .

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

Step 6.1.1.3.3.2.5.4.2

Multiply $- 1 (- V)$ .

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Step 6.1.1.3.3.2.5.4.2.1

Multiply $- 1$ by $- 1$ .

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

Step 6.1.1.3.3.2.5.4.2.2

Multiply $V$ by $1$ .

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

Step 6.1.1.3.3.2.5.4.3

Rewrite $- 1 V^{2}$ as $- V^{2}$ .

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

Step 6.1.1.3.3.2.5.4.4

Multiply $- 1$ by $1$ .

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

Step 6.1.1.3.3.2.5.4.5

Rewrite using the commutative property of multiplication.

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

Step 6.1.1.3.3.2.5.4.6

Multiply $V$ by $V$ by adding the exponents.

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Step 6.1.1.3.3.2.5.4.6.1

Move $V$ .

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

Step 6.1.1.3.3.2.5.4.6.2

Multiply $V$ by $V$ .

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

Step 6.1.1.3.3.2.5.4.7

Multiply $- 1$ by $- 1$ .

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

Step 6.1.1.3.3.2.5.4.8

Multiply $V^{2}$ by $1$ .

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

Step 6.1.1.3.3.2.5.4.9

Multiply $V$ by $V^{2}$ by adding the exponents.

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Step 6.1.1.3.3.2.5.4.9.1

Move $V^{2}$ .

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

Step 6.1.1.3.3.2.5.4.9.2

Multiply $V^{2}$ by $V$ .

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Step 6.1.1.3.3.2.5.4.9.2.1

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

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

Step 6.1.1.3.3.2.5.4.9.2.2

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

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

Step 6.1.1.3.3.2.5.4.9.3

Add $2$ and $1$ .

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

Step 6.1.1.3.3.2.5.5

Combine the opposite terms in $- 1 + V - V^{2} - V + V^{2} - V^{3}$ .

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Step 6.1.1.3.3.2.5.5.1

Subtract $V$ from $V$ .

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

Step 6.1.1.3.3.2.5.5.2

Add $- 1 - V^{2}$ and $0$ .

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

Step 6.1.1.3.3.2.5.5.3

Add $- V^{2}$ and $V^{2}$ .

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

Step 6.1.1.3.3.2.5.5.4

Add $- 1$ and $0$ .

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

Step 6.1.1.3.3.2.5.6

Subtract $1$ from $1$ .

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

Step 6.1.1.3.3.2.5.7

Subtract $V^{3}$ from $0$ .

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

Step 6.1.1.3.3.2.6

Move the negative in front of the fraction.

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

Step 6.1.1.3.3.2.7

Combine exponents.

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Step 6.1.1.3.3.2.7.1

Factor out negative.

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

Step 6.1.1.3.3.2.7.2

Combine $V$ and $\frac{V^{3}}{(1 + V) (1 - V + V^{2})}$ .

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

Step 6.1.1.3.3.2.7.3

Multiply $V$ by $V^{3}$ by adding the exponents.

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Step 6.1.1.3.3.2.7.3.1

Multiply $V$ by $V^{3}$ .

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Step 6.1.1.3.3.2.7.3.1.1

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

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

Step 6.1.1.3.3.2.7.3.1.2

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

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

Step 6.1.1.3.3.2.7.3.2

Add $1$ and $3$ .

$\frac{d V}{d x} = \frac{- \frac{V^{4}}{(1 + V) (1 - V + V^{2})}}{x}$

Step 6.1.1.3.3.3

Multiply the numerator by the reciprocal of the denominator.

$\frac{d V}{d x} = - \frac{V^{4}}{(1 + V) (1 - V + V^{2})} \cdot \frac{1}{x}$

Step 6.1.1.3.3.4

Multiply $\frac{1}{x}$ by $\frac{V^{4}}{(1 + V) (1 - V + V^{2})}$ .

$\frac{d V}{d x} = - \frac{V^{4}}{x (1 + V) (1 - V + V^{2})}$

Step 6.1.2

Regroup factors.

$\frac{d V}{d x} = - \frac{V^{4}}{(1 + V) (1 - V + V^{2})} \cdot \frac{1}{x}$

Step 6.1.3

Multiply both sides by $\frac{(1 + V) (1 - V + V^{2})}{V^{4}}$ .

$\frac{(1 + V) (1 - V + V^{2})}{V^{4}} \frac{d V}{d x} = \frac{(1 + V) (1 - V + V^{2})}{V^{4}} (- \frac{V^{4}}{(1 + V) (1 - V + V^{2})} \cdot \frac{1}{x})$

Step 6.1.4

Simplify.

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Step 6.1.4.1

Rewrite using the commutative property of multiplication.

$\frac{(1 + V) (1 - V + V^{2})}{V^{4}} \frac{d V}{d x} = - \frac{(1 + V) (1 - V + V^{2})}{V^{4}} (\frac{V^{4}}{(1 + V) (1 - V + V^{2})} \cdot \frac{1}{x})$

Step 6.1.4.2

Combine.

$\frac{(1 + V) (1 - V + V^{2})}{V^{4}} \frac{d V}{d x} = - \frac{(1 + V) (1 - V + V^{2})}{V^{4}} \cdot \frac{V^{4} \cdot 1}{(1 + V) (1 - V + V^{2}) x}$

Step 6.1.4.3

Cancel the common factor of $(1 + V) (1 - V + V^{2})$ .

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Step 6.1.4.3.1

Move the leading negative in $- \frac{(1 + V) (1 - V + V^{2})}{V^{4}}$ into the numerator.

$\frac{(1 + V) (1 - V + V^{2})}{V^{4}} \frac{d V}{d x} = \frac{- (1 + V) (1 - V + V^{2})}{V^{4}} \cdot \frac{V^{4} \cdot 1}{(1 + V) (1 - V + V^{2}) x}$

Step 6.1.4.3.2

Factor $(1 + V) (1 - V + V^{2})$ out of $- (1 + V) (1 - V + V^{2})$ .

$\frac{(1 + V) (1 - V + V^{2})}{V^{4}} \frac{d V}{d x} = \frac{(1 + V) (1 - V + V^{2}) (- 1)}{V^{4}} \cdot \frac{V^{4} \cdot 1}{(1 + V) (1 - V + V^{2}) x}$

Step 6.1.4.3.3

Factor $(1 + V) (1 - V + V^{2})$ out of $(1 + V) (1 - V + V^{2}) x$ .

$\frac{(1 + V) (1 - V + V^{2})}{V^{4}} \frac{d V}{d x} = \frac{(1 + V) (1 - V + V^{2}) (- 1)}{V^{4}} \cdot \frac{V^{4} \cdot 1}{(1 + V) (1 - V + V^{2}) (x)}$

Step 6.1.4.3.4

Cancel the common factor.

$\frac{(1 + V) (1 - V + V^{2})}{V^{4}} \frac{d V}{d x} = \frac{(1 + V) (1 - V + V^{2}) \cdot - 1}{V^{4}} \cdot \frac{V^{4} \cdot 1}{(1 + V) (1 - V + V^{2}) x}$

Step 6.1.4.3.5

Rewrite the expression.

$\frac{(1 + V) (1 - V + V^{2})}{V^{4}} \frac{d V}{d x} = \frac{- 1}{V^{4}} \cdot \frac{V^{4} \cdot 1}{x}$

Step 6.1.4.4

Cancel the common factor of $V^{4}$ .

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Step 6.1.4.4.1

Factor $V^{4}$ out of $V^{4} \cdot 1$ .

$\frac{(1 + V) (1 - V + V^{2})}{V^{4}} \frac{d V}{d x} = \frac{- 1}{V^{4}} \cdot \frac{V^{4} (1)}{x}$

Step 6.1.4.4.2

Cancel the common factor.

$\frac{(1 + V) (1 - V + V^{2})}{V^{4}} \frac{d V}{d x} = \frac{- 1}{V^{4}} \cdot \frac{V^{4} \cdot 1}{x}$

Step 6.1.4.4.3

Rewrite the expression.

$\frac{(1 + V) (1 - V + V^{2})}{V^{4}} \frac{d V}{d x} = - \frac{1}{x}$

Step 6.1.5

Rewrite the equation.

$\frac{(1 + V) (1 - V + V^{2})}{V^{4}} d V = - \frac{1}{x} d x$

Step 6.2

Integrate both sides.

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Step 6.2.1

Set up an integral on each side.

$\int \frac{(1 + V) (1 - V + V^{2})}{V^{4}} d V = \int - \frac{1}{x} d x$

Step 6.2.2

Integrate the left side.

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Step 6.2.2.1

Move $V^{4}$ out of the denominator by raising it to the $- 1$ power.

$\int (1 + V) (1 - V + V^{2}) {(V^{4})}^{- 1} d V = \int - \frac{1}{x} d x$

Step 6.2.2.2

Multiply the exponents in ${(V^{4})}^{- 1}$ .

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Step 6.2.2.2.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$\int (1 + V) (1 - V + V^{2}) V^{4 \cdot - 1} d V = \int - \frac{1}{x} d x$

Step 6.2.2.2.2

Multiply $4$ by $- 1$ .

$\int (1 + V) (1 - V + V^{2}) V^{- 4} d V = \int - \frac{1}{x} d x$

Step 6.2.2.3

Expand $(1 + V) (1 - V + V^{2}) V^{- 4}$ .

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Step 6.2.2.3.1

Apply the distributive property.

$\int (1 (1 - V + V^{2}) + V (1 - V + V^{2})) V^{- 4} d V = \int - \frac{1}{x} d x$

Step 6.2.2.3.2

Apply the distributive property.

$\int (1 (1 - V) + 1 V^{2} + V (1 - V + V^{2})) V^{- 4} d V = \int - \frac{1}{x} d x$

Step 6.2.2.3.3

Apply the distributive property.

$\int (1 \cdot 1 + 1 (- V) + 1 V^{2} + V (1 - V + V^{2})) V^{- 4} d V = \int - \frac{1}{x} d x$

Step 6.2.2.3.4

Apply the distributive property.

$\int (1 \cdot 1 + 1 (- V) + 1 V^{2} + V (1 - V) + V \cdot V^{2}) V^{- 4} d V = \int - \frac{1}{x} d x$

Step 6.2.2.3.5

Apply the distributive property.

$\int (1 \cdot 1 + 1 (- V) + 1 V^{2} + V \cdot 1 + V (- V) + V \cdot V^{2}) V^{- 4} d V = \int - \frac{1}{x} d x$

Step 6.2.2.3.6

Apply the distributive property.

$\int (1 \cdot 1 + 1 (- V) + 1 V^{2}) V^{- 4} + (V \cdot 1 + V (- V) + V \cdot V^{2}) V^{- 4} d V = \int - \frac{1}{x} d x$

Step 6.2.2.3.7

Apply the distributive property.

$\int (1 \cdot 1 + 1 (- V)) V^{- 4} + 1 V^{2} V^{- 4} + (V \cdot 1 + V (- V) + V \cdot V^{2}) V^{- 4} d V = \int - \frac{1}{x} d x$

Step 6.2.2.3.8

Apply the distributive property.

$\int 1 \cdot 1 V^{- 4} + 1 (- V) V^{- 4} + 1 V^{2} V^{- 4} + (V \cdot 1 + V (- V) + V \cdot V^{2}) V^{- 4} d V = \int - \frac{1}{x} d x$

Step 6.2.2.3.9

Apply the distributive property.

$\int 1 \cdot 1 V^{- 4} + 1 (- V) V^{- 4} + 1 V^{2} V^{- 4} + (V \cdot 1 + V (- V)) V^{- 4} + V \cdot V^{2} V^{- 4} d V = \int - \frac{1}{x} d x$

Step 6.2.2.3.10

Apply the distributive property.

$\int 1 \cdot 1 V^{- 4} + 1 (- V) V^{- 4} + 1 V^{2} V^{- 4} + V \cdot 1 V^{- 4} + V (- V) V^{- 4} + V \cdot V^{2} V^{- 4} d V = \int - \frac{1}{x} d x$

Step 6.2.2.3.11

Reorder $V$ and $1$ .

$\int 1 \cdot 1 V^{- 4} + 1 \cdot - 1 V \cdot V^{- 4} + 1 V^{2} V^{- 4} + 1 \cdot V \cdot V^{- 4} + V (- V) V^{- 4} + V \cdot V^{2} V^{- 4} d V = \int - \frac{1}{x} d x$

Step 6.2.2.3.12

Reorder $V$ and $- 1$ .

$\int 1 \cdot 1 V^{- 4} + 1 \cdot - 1 V \cdot V^{- 4} + 1 V^{2} V^{- 4} + 1 \cdot V \cdot V^{- 4} - 1 \cdot V \cdot V \cdot V^{- 4} + V \cdot V^{2} V^{- 4} d V = \int - \frac{1}{x} d x$

Step 6.2.2.3.13

Multiply $1$ by $1$ .

$\int 1 V^{- 4} + 1 \cdot - 1 V \cdot V^{- 4} + 1 V^{2} V^{- 4} + 1 V \cdot V^{- 4} - 1 V \cdot V \cdot V^{- 4} + V \cdot V^{2} V^{- 4} d V = \int - \frac{1}{x} d x$

Step 6.2.2.3.14

Multiply $V^{- 4}$ by $1$ .

$\int V^{- 4} + 1 \cdot - 1 V \cdot V^{- 4} + 1 V^{2} V^{- 4} + 1 V \cdot V^{- 4} - 1 V \cdot V \cdot V^{- 4} + V \cdot V^{2} V^{- 4} d V = \int - \frac{1}{x} d x$

Step 6.2.2.3.15

Multiply $- 1$ by $1$ .

$\int V^{- 4} - V \cdot V^{- 4} + 1 V^{2} V^{- 4} + 1 V \cdot V^{- 4} - 1 V \cdot V \cdot V^{- 4} + V \cdot V^{2} V^{- 4} d V = \int - \frac{1}{x} d x$

Step 6.2.2.3.16

Factor out negative.

$\int V^{- 4} - (V \cdot V^{- 4}) + 1 V^{2} V^{- 4} + 1 V \cdot V^{- 4} - 1 V \cdot V \cdot V^{- 4} + V \cdot V^{2} V^{- 4} d V = \int - \frac{1}{x} d x$

Step 6.2.2.3.17

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

$\int V^{- 4} - (V^{1} V^{- 4}) + 1 V^{2} V^{- 4} + 1 V \cdot V^{- 4} - 1 V \cdot V \cdot V^{- 4} + V \cdot V^{2} V^{- 4} d V = \int - \frac{1}{x} d x$

Step 6.2.2.3.18

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

$\int V^{- 4} - V^{1 - 4} + 1 V^{2} V^{- 4} + 1 V \cdot V^{- 4} - 1 V \cdot V \cdot V^{- 4} + V \cdot V^{2} V^{- 4} d V = \int - \frac{1}{x} d x$

Step 6.2.2.3.19

Subtract $4$ from $1$ .

$\int V^{- 4} - V^{- 3} + 1 V^{2} V^{- 4} + 1 V \cdot V^{- 4} - 1 V \cdot V \cdot V^{- 4} + V \cdot V^{2} V^{- 4} d V = \int - \frac{1}{x} d x$

Step 6.2.2.3.20

Multiply $V^{2}$ by $1$ .

$\int V^{- 4} - V^{- 3} + V^{2} V^{- 4} + 1 V \cdot V^{- 4} - 1 V \cdot V \cdot V^{- 4} + V \cdot V^{2} V^{- 4} d V = \int - \frac{1}{x} d x$

Step 6.2.2.3.21

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

$\int V^{- 4} - V^{- 3} + V^{2 - 4} + 1 V \cdot V^{- 4} - 1 V \cdot V \cdot V^{- 4} + V \cdot V^{2} V^{- 4} d V = \int - \frac{1}{x} d x$

Step 6.2.2.3.22

Subtract $4$ from $2$ .

$\int V^{- 4} - V^{- 3} + V^{- 2} + 1 V \cdot V^{- 4} - 1 V \cdot V \cdot V^{- 4} + V \cdot V^{2} V^{- 4} d V = \int - \frac{1}{x} d x$

Step 6.2.2.3.23

Multiply $V$ by $1$ .

$\int V^{- 4} - V^{- 3} + V^{- 2} + V \cdot V^{- 4} - 1 V \cdot V \cdot V^{- 4} + V \cdot V^{2} V^{- 4} d V = \int - \frac{1}{x} d x$

Step 6.2.2.3.24

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

$\int V^{- 4} - V^{- 3} + V^{- 2} + V^{1} V^{- 4} - 1 V \cdot V \cdot V^{- 4} + V \cdot V^{2} V^{- 4} d V = \int - \frac{1}{x} d x$

Step 6.2.2.3.25

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

$\int V^{- 4} - V^{- 3} + V^{- 2} + V^{1 - 4} - 1 V \cdot V \cdot V^{- 4} + V \cdot V^{2} V^{- 4} d V = \int - \frac{1}{x} d x$

Step 6.2.2.3.26

Subtract $4$ from $1$ .

$\int V^{- 4} - V^{- 3} + V^{- 2} + V^{- 3} - 1 V \cdot V \cdot V^{- 4} + V \cdot V^{2} V^{- 4} d V = \int - \frac{1}{x} d x$

Step 6.2.2.3.27

Factor out negative.

$\int V^{- 4} - V^{- 3} + V^{- 2} + V^{- 3} - (V \cdot V) V^{- 4} + V \cdot V^{2} V^{- 4} d V = \int - \frac{1}{x} d x$

Step 6.2.2.3.28

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

$\int V^{- 4} - V^{- 3} + V^{- 2} + V^{- 3} - (V^{1} V) V^{- 4} + V \cdot V^{2} V^{- 4} d V = \int - \frac{1}{x} d x$

Step 6.2.2.3.29

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

$\int V^{- 4} - V^{- 3} + V^{- 2} + V^{- 3} - (V^{1} V^{1}) V^{- 4} + V \cdot V^{2} V^{- 4} d V = \int - \frac{1}{x} d x$

Step 6.2.2.3.30

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

$\int V^{- 4} - V^{- 3} + V^{- 2} + V^{- 3} - V^{1 + 1} V^{- 4} + V \cdot V^{2} V^{- 4} d V = \int - \frac{1}{x} d x$

Step 6.2.2.3.31

Add $1$ and $1$ .

$\int V^{- 4} - V^{- 3} + V^{- 2} + V^{- 3} - V^{2} V^{- 4} + V \cdot V^{2} V^{- 4} d V = \int - \frac{1}{x} d x$

Step 6.2.2.3.32

Factor out negative.

$\int V^{- 4} - V^{- 3} + V^{- 2} + V^{- 3} - (V^{2} V^{- 4}) + V \cdot V^{2} V^{- 4} d V = \int - \frac{1}{x} d x$

Step 6.2.2.3.33

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

$\int V^{- 4} - V^{- 3} + V^{- 2} + V^{- 3} - V^{2 - 4} + V \cdot V^{2} V^{- 4} d V = \int - \frac{1}{x} d x$

Step 6.2.2.3.34

Subtract $4$ from $2$ .

$\int V^{- 4} - V^{- 3} + V^{- 2} + V^{- 3} - V^{- 2} + V \cdot V^{2} V^{- 4} d V = \int - \frac{1}{x} d x$

Step 6.2.2.3.35

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

$\int V^{- 4} - V^{- 3} + V^{- 2} + V^{- 3} - V^{- 2} + V^{1} V^{2} V^{- 4} d V = \int - \frac{1}{x} d x$

Step 6.2.2.3.36

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

$\int V^{- 4} - V^{- 3} + V^{- 2} + V^{- 3} - V^{- 2} + V^{1 + 2} V^{- 4} d V = \int - \frac{1}{x} d x$

Step 6.2.2.3.37

Add $1$ and $2$ .

$\int V^{- 4} - V^{- 3} + V^{- 2} + V^{- 3} - V^{- 2} + V^{3} V^{- 4} d V = \int - \frac{1}{x} d x$

Step 6.2.2.3.38

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

$\int V^{- 4} - V^{- 3} + V^{- 2} + V^{- 3} - V^{- 2} + V^{3 - 4} d V = \int - \frac{1}{x} d x$

Step 6.2.2.3.39

Subtract $4$ from $3$ .

$\int V^{- 4} - V^{- 3} + V^{- 2} + V^{- 3} - V^{- 2} + V^{- 1} d V = \int - \frac{1}{x} d x$

Step 6.2.2.3.40

Reorder $V^{- 4}$ and $- V^{- 3}$ .

$\int - V^{- 3} + V^{- 4} + V^{- 2} + V^{- 3} - V^{- 2} + V^{- 1} d V = \int - \frac{1}{x} d x$

Step 6.2.2.3.41

Move $V^{- 4}$ .

$\int - V^{- 3} + V^{- 2} + V^{- 4} + V^{- 3} - V^{- 2} + V^{- 1} d V = \int - \frac{1}{x} d x$

Step 6.2.2.3.42

Reorder $- V^{- 3}$ and $V^{- 2}$ .

$\int V^{- 2} - V^{- 3} + V^{- 4} + V^{- 3} - V^{- 2} + V^{- 1} d V = \int - \frac{1}{x} d x$

Step 6.2.2.3.43

Reorder $V^{- 3}$ and $- V^{- 2}$ .

$\int V^{- 2} - V^{- 3} + V^{- 4} - V^{- 2} + V^{- 3} + V^{- 1} d V = \int - \frac{1}{x} d x$

Step 6.2.2.3.44

Move $V^{- 3}$ .

$\int V^{- 2} - V^{- 3} + V^{- 4} - V^{- 2} + V^{- 1} + V^{- 3} d V = \int - \frac{1}{x} d x$

Step 6.2.2.3.45

Reorder $- V^{- 2}$ and $V^{- 1}$ .

$\int V^{- 2} - V^{- 3} + V^{- 4} + V^{- 1} - V^{- 2} + V^{- 3} d V = \int - \frac{1}{x} d x$

Step 6.2.2.3.46

Move $V^{- 4}$ .

$\int V^{- 2} - V^{- 3} + V^{- 1} - V^{- 2} + V^{- 3} + V^{- 4} d V = \int - \frac{1}{x} d x$

Step 6.2.2.3.47

Move $- V^{- 3}$ .

$\int V^{- 2} + V^{- 1} - V^{- 2} + V^{- 3} - V^{- 3} + V^{- 4} d V = \int - \frac{1}{x} d x$

Step 6.2.2.3.48

Reorder $V^{- 2}$ and $V^{- 1}$ .

$\int V^{- 1} + V^{- 2} - V^{- 2} + V^{- 3} - V^{- 3} + V^{- 4} d V = \int - \frac{1}{x} d x$

Step 6.2.2.3.49

Subtract $V^{- 2}$ from $V^{- 2}$ .

$\int V^{- 1} + 0 + V^{- 3} - V^{- 3} + V^{- 4} d V = \int - \frac{1}{x} d x$

Step 6.2.2.3.50

Add $V^{- 1}$ and $0$ .

$\int V^{- 1} + V^{- 3} - V^{- 3} + V^{- 4} d V = \int - \frac{1}{x} d x$

Step 6.2.2.3.51

Subtract $V^{- 3}$ from $V^{- 3}$ .

$\int V^{- 1} + 0 + V^{- 4} d V = \int - \frac{1}{x} d x$

Step 6.2.2.3.52

Add $V^{- 1}$ and $0$ .

$\int V^{- 1} + V^{- 4} d V = \int - \frac{1}{x} d x$

Step 6.2.2.4

Split the single integral into multiple integrals.

$\int V^{- 1} d V + \int V^{- 4} d V = \int - \frac{1}{x} d x$

Step 6.2.2.5

The integral of $V^{- 1}$ with respect to $V$ is $\ln (| V |)$ .

$\ln (| V |) + C_{1} + \int V^{- 4} d V = \int - \frac{1}{x} d x$

Step 6.2.2.6

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

$\ln (| V |) + C_{1} - \frac{1}{3} V^{- 3} + C_{2} = \int - \frac{1}{x} d x$

Step 6.2.2.7

Simplify.

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Step 6.2.2.7.1

Simplify.

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

Step 6.2.2.7.2

Simplify.

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Step 6.2.2.7.2.1

Multiply $\frac{1}{V^{3}}$ by $\frac{1}{3}$ .

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

Step 6.2.2.7.2.2

Move $3$ to the left of $V^{3}$ .

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

Step 6.2.2.8

Reorder terms.

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

Step 6.2.3

Integrate the right side.

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Step 6.2.3.1

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

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

Step 6.2.3.2

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

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

Step 6.2.3.3

Simplify.

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

Step 6.2.4

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

$- \frac{1}{3 V^{3}} + \ln (| V |) = - \ln (| x |) + C$

Step 7

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

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

Step 8

Solve $- \frac{1}{3 {(\frac{y}{x})}^{3}} + \ln (| \frac{y}{x} |) = - \ln (| x |) + C$ for $y$ .

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Step 8.1

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

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

Step 8.2

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

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

Step 8.3

Multiply $| \frac{y}{x} | | x |$ .

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Step 8.3.1

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

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

Step 8.3.2

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

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

Step 8.4

Cancel the common factor of $x$ .

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Step 8.4.1

Cancel the common factor.

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

Step 8.4.2

Divide $y$ by $1$ .

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

Step 8.5

Simplify each term.

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Step 8.5.1

Apply the product rule to $\frac{y}{x}$ .

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

Step 8.5.2

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

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

Step 8.5.3

Multiply the numerator by the reciprocal of the denominator.

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

Step 8.5.4

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

$\ln (| y |) = \frac{x^{3}}{3 y^{3}} + C$