Calculus Examples

$\frac{d y}{d x} = \frac{x y}{x^{2} - y^{2}}$

Step 1

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

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Step 1.1

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

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

Step 1.2

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

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

Step 1.3

Apply the distributive property.

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

Step 1.4

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

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Step 1.4.1

Cancel the common factor.

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

Step 1.4.2

Rewrite the expression.

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

Step 1.5

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

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

Step 1.6

Cancel the common factor of $x$ .

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Step 1.6.1

Factor $x$ out of $x y$ .

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

Step 1.6.2

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

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

Step 1.6.3

Cancel the common factor.

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

Step 1.6.4

Rewrite the expression.

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

Step 1.7

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

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

Step 1.8

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

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

Step 2

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

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

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

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

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 the denominator.

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Step 6.1.1.1.1

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

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

Step 6.1.1.1.2

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

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

Step 6.1.1.2

Subtract $V$ from both sides of the equation.

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

Step 6.1.1.3

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

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Step 6.1.1.3.1

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

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

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.

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

Step 6.1.1.3.2.1.2

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

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

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.

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

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 $\frac{V}{(1 + V) (1 - V)} - V$ .

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Step 6.1.1.3.3.2.1.1

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

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

Step 6.1.1.3.3.2.1.2

Factor $V$ out of $- V$ .

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

Step 6.1.1.3.3.2.1.3

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

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

Step 6.1.1.3.3.2.2

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

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

Step 6.1.1.3.3.2.3

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

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

Step 6.1.1.3.3.2.4

Combine the numerators over the common denominator.

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

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.

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

Step 6.1.1.3.3.2.5.2

Multiply $- 1$ by $1$ .

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

Step 6.1.1.3.3.2.5.3

Expand $(- 1 - V) (1 - V)$ using the FOIL Method.

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Step 6.1.1.3.3.2.5.3.1

Apply the distributive property.

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

Step 6.1.1.3.3.2.5.3.2

Apply the distributive property.

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

Step 6.1.1.3.3.2.5.3.3

Apply the distributive property.

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

Step 6.1.1.3.3.2.5.4

Simplify and combine like terms.

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Step 6.1.1.3.3.2.5.4.1

Simplify each term.

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Step 6.1.1.3.3.2.5.4.1.1

Multiply $- 1$ by $1$ .

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

Step 6.1.1.3.3.2.5.4.1.2

Multiply $- 1 (- V)$ .

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Step 6.1.1.3.3.2.5.4.1.2.1

Multiply $- 1$ by $- 1$ .

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

Step 6.1.1.3.3.2.5.4.1.2.2

Multiply $V$ by $1$ .

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

Step 6.1.1.3.3.2.5.4.1.3

Multiply $- 1$ by $1$ .

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

Step 6.1.1.3.3.2.5.4.1.4

Rewrite using the commutative property of multiplication.

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

Step 6.1.1.3.3.2.5.4.1.5

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

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Step 6.1.1.3.3.2.5.4.1.5.1

Move $V$ .

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

Step 6.1.1.3.3.2.5.4.1.5.2

Multiply $V$ by $V$ .

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

Step 6.1.1.3.3.2.5.4.1.6

Multiply $- 1$ by $- 1$ .

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

Step 6.1.1.3.3.2.5.4.1.7

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

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

Step 6.1.1.3.3.2.5.4.2

Subtract $V$ from $V$ .

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

Step 6.1.1.3.3.2.5.4.3

Add $- 1$ and $0$ .

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

Step 6.1.1.3.3.2.5.5

Subtract $1$ from $1$ .

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

Step 6.1.1.3.3.2.5.6

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

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

Step 6.1.1.3.3.3

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

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

Step 6.1.1.3.3.4

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

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Step 6.1.1.3.3.4.1

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

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Step 6.1.1.3.3.4.1.1

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

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

Step 6.1.1.3.3.4.1.2

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

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

Step 6.1.1.3.3.4.2

Add $1$ and $2$ .

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

Step 6.1.1.3.3.5

Multiply the numerator by the reciprocal of the denominator.

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

Step 6.1.1.3.3.6

Combine.

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

Step 6.1.1.3.3.7

Simplify the expression.

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Step 6.1.1.3.3.7.1

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

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

Step 6.1.1.3.3.7.2

Reorder factors in $\frac{V^{3}}{(1 + V) (1 - V) x}$ .

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

Step 6.1.2

Regroup factors.

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

Step 6.1.3

Multiply both sides by $\frac{(1 + V) (1 - V)}{V^{3}}$ .

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

Step 6.1.4

Simplify.

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Step 6.1.4.1

Combine.

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

Step 6.1.4.2

Combine.

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

Step 6.1.4.3

Cancel the common factor of $1 + V$ .

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Step 6.1.4.3.1

Cancel the common factor.

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

Step 6.1.4.3.2

Rewrite the expression.

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

Step 6.1.4.4

Cancel the common factor of $1 - V$ .

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Step 6.1.4.4.1

Cancel the common factor.

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

Step 6.1.4.4.2

Rewrite the expression.

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

Step 6.1.4.5

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

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Step 6.1.4.5.1

Cancel the common factor.

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

Step 6.1.4.5.2

Rewrite the expression.

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

Step 6.1.5

Rewrite the equation.

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

Step 6.2

Integrate both sides.

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Step 6.2.1

Set up an integral on each side.

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

Step 6.2.2

Integrate the left side.

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Step 6.2.2.1

Move $V^{3}$ out of the denominator by raising it to the $- 1$ power.

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

Step 6.2.2.2

Multiply the exponents in ${(V^{3})}^{- 1}$ .

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Step 6.2.2.2.1

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

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

Step 6.2.2.2.2

Multiply $3$ by $- 1$ .

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

Step 6.2.2.3

Expand $(1 + V) (1 - V) V^{- 3}$ .

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Step 6.2.2.3.1

Apply the distributive property.

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

Step 6.2.2.3.2

Apply the distributive property.

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

Step 6.2.2.3.3

Apply the distributive property.

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

Step 6.2.2.3.4

Apply the distributive property.

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

Step 6.2.2.3.5

Apply the distributive property.

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

Step 6.2.2.3.6

Apply the distributive property.

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

Step 6.2.2.3.7

Reorder $V$ and $1$ .

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

Step 6.2.2.3.8

Reorder $V$ and $- 1$ .

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

Step 6.2.2.3.9

Multiply $1$ by $1$ .

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

Step 6.2.2.3.10

Multiply $V^{- 3}$ by $1$ .

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

Step 6.2.2.3.11

Multiply $- 1$ by $1$ .

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

Step 6.2.2.3.12

Factor out negative.

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

Step 6.2.2.3.13

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

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

Step 6.2.2.3.14

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

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

Step 6.2.2.3.15

Subtract $3$ from $1$ .

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

Step 6.2.2.3.16

Multiply $V$ by $1$ .

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

Step 6.2.2.3.17

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

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

Step 6.2.2.3.18

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

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

Step 6.2.2.3.19

Subtract $3$ from $1$ .

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

Step 6.2.2.3.20

Factor out negative.

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

Step 6.2.2.3.21

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

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

Step 6.2.2.3.22

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

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

Step 6.2.2.3.23

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

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

Step 6.2.2.3.24

Add $1$ and $1$ .

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

Step 6.2.2.3.25

Factor out negative.

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

Step 6.2.2.3.26

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

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

Step 6.2.2.3.27

Subtract $3$ from $2$ .

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

Step 6.2.2.3.28

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

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

Step 6.2.2.3.29

Subtract $V^{- 1}$ from $0$ .

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

Step 6.2.2.3.30

Reorder $V^{- 3}$ and $- V^{- 1}$ .

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

Step 6.2.2.4

Split the single integral into multiple integrals.

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

Step 6.2.2.5

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

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

Step 6.2.2.6

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

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

Step 6.2.2.7

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

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

Step 6.2.2.8

Simplify.

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Step 6.2.2.8.1

Simplify.

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

Step 6.2.2.8.2

Simplify.

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Step 6.2.2.8.2.1

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

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

Step 6.2.2.8.2.2

Move $2$ to the left of $V^{2}$ .

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

Step 6.2.2.9

Reorder terms.

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

Step 6.2.3

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

$- \frac{1}{2 V^{2}} - \ln (| V |) + C_{3} = \ln (| x |) + C_{4}$

Step 6.2.4

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

$- \frac{1}{2 V^{2}} - \ln (| V |) = \ln (| x |) + C$

Step 7

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

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

Step 8

Solve $- \frac{1}{2 {(\frac{y}{x})}^{2}} - \ln (| \frac{y}{x} |) = \ln (| x |) + C$ for $y$ .

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Step 8.1

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

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

Step 8.2

Simplify each term.

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Step 8.2.1

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

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

Step 8.2.2

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

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

Step 8.2.3

Multiply the numerator by the reciprocal of the denominator.

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

Step 8.2.4

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

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

Step 8.3

Divide each term in $- \ln (| \frac{y}{x} |) = \frac{x^{2}}{2 y^{2}} + C + \ln (| x |)$ by $- 1$ and simplify.

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Step 8.3.1

Divide each term in $- \ln (| \frac{y}{x} |) = \frac{x^{2}}{2 y^{2}} + C + \ln (| x |)$ by $- 1$ .

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

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.

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

Step 8.3.2.2

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

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

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 $\frac{\frac{x^{2}}{2 y^{2}}}{- 1}$ .

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

Step 8.3.3.1.2

Rewrite $- 1 \cdot \frac{x^{2}}{2 y^{2}}$ as $- \frac{x^{2}}{2 y^{2}}$ .

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

Step 8.3.3.1.3

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

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

Step 8.3.3.1.4

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

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

Step 8.3.3.1.5

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

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

Step 8.3.3.1.6

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

$\ln (| \frac{y}{x} |) = - \frac{x^{2}}{2 y^{2}} - C - \ln (| x |)$

Step 8.4

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

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

Step 8.5

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

$\ln (| \frac{y}{x} | | x |) = - \frac{x^{2}}{2 y^{2}} - C$

Step 8.6

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

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Step 8.6.1

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

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

Step 8.6.2

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

$\ln (| \frac{y x}{x} |) = - \frac{x^{2}}{2 y^{2}} - C$

Step 8.7

Cancel the common factor of $x$ .

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Step 8.7.1

Cancel the common factor.

$\ln (| \frac{y x}{x} |) = - \frac{x^{2}}{2 y^{2}} - C$

Step 8.7.2

Divide $y$ by $1$ .

$\ln (| y |) = - \frac{x^{2}}{2 y^{2}} - C$