Calculus Examples

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

Step 1

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

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Step 1.1

Split $\frac{3 y^{2} - x^{2}}{2 x y}$ and simplify.

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Step 1.1.1

Split the fraction $\frac{3 y^{2} - x^{2}}{2 x y}$ into two fractions.

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

Step 1.1.2

Simplify each term.

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Step 1.1.2.1

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

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Step 1.1.2.1.1

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

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

Step 1.1.2.1.2

Cancel the common factors.

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Step 1.1.2.1.2.1

Factor $y$ out of $2 x y$ .

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

Step 1.1.2.1.2.2

Cancel the common factor.

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

Step 1.1.2.1.2.3

Rewrite the expression.

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

Step 1.1.2.2

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

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Step 1.1.2.2.1

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

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

Step 1.1.2.2.2

Cancel the common factors.

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Step 1.1.2.2.2.1

Factor $x$ out of $2 x y$ .

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

Step 1.1.2.2.2.2

Cancel the common factor.

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

Step 1.1.2.2.2.3

Rewrite the expression.

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

Step 1.1.2.3

Move the negative in front of the fraction.

$\frac{d y}{d x} = \frac{3 y}{2 x} - \frac{x}{2 y}$

Step 1.2

Factor out $\frac{y}{x}$ from $\frac{3 y}{2 x}$ .

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Step 1.2.1

Factor $\frac{y}{x}$ out of $\frac{3 y}{2 x}$ .

$\frac{d y}{d x} = \frac{y}{x} \cdot \frac{3}{2} - \frac{x}{2 y}$

Step 1.2.2

Reorder $\frac{y}{x}$ and $\frac{3}{2}$ .

$\frac{d y}{d x} = \frac{3}{2} \cdot \frac{y}{x} - \frac{x}{2 y}$

Step 1.3

Rewrite the differential equation as $f (\frac{y}{x})$ .

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Step 1.3.1

Factor out $\frac{x}{y}$ from $\frac{x}{2 y}$ .

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Step 1.3.1.1

Factor $\frac{x}{y}$ out of $\frac{x}{2 y}$ .

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

Step 1.3.1.2

Reorder $\frac{x}{y}$ and $\frac{1}{2}$ .

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

Step 1.3.2

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

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

Step 2

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

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

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

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

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 each term.

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Step 6.1.1.1.1

Combine $\frac{3}{2}$ and $V$ .

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

Step 6.1.1.1.2

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

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

Step 6.1.1.1.3

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

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

Step 6.1.1.1.4

Move $2$ to the left of $V$ .

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

Step 6.1.1.2

Subtract $V$ from both sides of the equation.

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

Step 6.1.1.3

Divide each term in $x \frac{d V}{d x} = \frac{3 V}{2} - \frac{1}{2 V} - V$ by $x$ and simplify.

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Step 6.1.1.3.1

Divide each term in $x \frac{d V}{d x} = \frac{3 V}{2} - \frac{1}{2 V} - V$ by $x$ .

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

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.

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

Step 6.1.1.3.2.1.2

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

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

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.

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

Step 6.1.1.3.3.2

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

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

Step 6.1.1.3.3.3

Simplify terms.

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Step 6.1.1.3.3.3.1

Combine $- V$ and $\frac{2}{2}$ .

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

Step 6.1.1.3.3.3.2

Combine the numerators over the common denominator.

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

Step 6.1.1.3.3.3.3

Simplify each term.

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Step 6.1.1.3.3.3.3.1

Simplify the numerator.

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Step 6.1.1.3.3.3.3.1.1

Factor $V$ out of $3 V - V \cdot 2$ .

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Step 6.1.1.3.3.3.3.1.1.1

Factor $V$ out of $3 V$ .

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

Step 6.1.1.3.3.3.3.1.1.2

Factor $V$ out of $- V \cdot 2$ .

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

Step 6.1.1.3.3.3.3.1.1.3

Factor $V$ out of $V \cdot 3 + V (- 1 \cdot 2)$ .

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

Step 6.1.1.3.3.3.3.1.2

Multiply $- 1$ by $2$ .

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

Step 6.1.1.3.3.3.3.1.3

Subtract $2$ from $3$ .

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

Step 6.1.1.3.3.3.3.2

Multiply $V$ by $1$ .

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

Step 6.1.1.3.3.4

Simplify the numerator.

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Step 6.1.1.3.3.4.1

To write $\frac{V}{2}$ as a fraction with a common denominator, multiply by $\frac{V}{V}$ .

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

Step 6.1.1.3.3.4.2

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

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

Step 6.1.1.3.3.4.3

Combine the numerators over the common denominator.

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

Step 6.1.1.3.3.4.4

Simplify the numerator.

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Step 6.1.1.3.3.4.4.1

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

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

Step 6.1.1.3.3.4.4.2

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

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

Step 6.1.1.3.3.4.4.3

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

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

Step 6.1.1.3.3.4.4.4

Add $1$ and $1$ .

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

Step 6.1.1.3.3.4.4.5

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

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

Step 6.1.1.3.3.4.4.6

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

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

Step 6.1.1.3.3.4.4.7

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

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

Step 6.1.1.3.3.5

Multiply the numerator by the reciprocal of the denominator.

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

Step 6.1.1.3.3.6

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

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

Step 6.1.2

Regroup factors.

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

Step 6.1.3

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

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

Step 6.1.4

Simplify.

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Step 6.1.4.1

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

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

Step 6.1.4.2

Cancel the common factor of $2 V$ .

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Step 6.1.4.2.1

Factor $2 V$ out of $2 V x$ .

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

Step 6.1.4.2.2

Cancel the common factor.

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

Step 6.1.4.2.3

Rewrite the expression.

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

Step 6.1.4.3

Cancel the common factor of $(V + 1) (V - 1)$ .

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Step 6.1.4.3.1

Cancel the common factor.

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

Step 6.1.4.3.2

Rewrite the expression.

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

Step 6.1.5

Rewrite the equation.

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

Step 6.2

Integrate both sides.

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Step 6.2.1

Set up an integral on each side.

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

Step 6.2.2

Integrate the left side.

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Step 6.2.2.1

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

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

Step 6.2.2.2

Let $u = (V + 1) (V - 1)$ . Then $d u = 2 V d V$ , so $\frac{1}{2} d u = V d V$ . Rewrite using $u$ and $d$ $u$ .

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Step 6.2.2.2.1

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

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Step 6.2.2.2.1.1

Differentiate $(V + 1) (V - 1)$ .

$\frac{d}{d V} [(V + 1) (V - 1)]$

Step 6.2.2.2.1.2

Differentiate using the Product Rule which states that $\frac{d}{d V} [f (V) g (V)]$ is $f (V) \frac{d}{d V} [g (V)] + g (V) \frac{d}{d V} [f (V)]$ where $f (V) = V + 1$ and $g (V) = V - 1$ .

$(V + 1) \frac{d}{d V} [V - 1] + (V - 1) \frac{d}{d V} [V + 1]$

Step 6.2.2.2.1.3

Differentiate.

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Step 6.2.2.2.1.3.1

By the Sum Rule, the derivative of $V - 1$ with respect to $V$ is $\frac{d}{d V} [V] + \frac{d}{d V} [- 1]$ .

$(V + 1) (\frac{d}{d V} [V] + \frac{d}{d V} [- 1]) + (V - 1) \frac{d}{d V} [V + 1]$

Step 6.2.2.2.1.3.2

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

$(V + 1) (1 + \frac{d}{d V} [- 1]) + (V - 1) \frac{d}{d V} [V + 1]$

Step 6.2.2.2.1.3.3

Since $- 1$ is constant with respect to $V$ , the derivative of $- 1$ with respect to $V$ is $0$ .

$(V + 1) (1 + 0) + (V - 1) \frac{d}{d V} [V + 1]$

Step 6.2.2.2.1.3.4

Simplify the expression.

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Step 6.2.2.2.1.3.4.1

Add $1$ and $0$ .

$(V + 1) \cdot 1 + (V - 1) \frac{d}{d V} [V + 1]$

Step 6.2.2.2.1.3.4.2

Multiply $V + 1$ by $1$ .

$V + 1 + (V - 1) \frac{d}{d V} [V + 1]$

Step 6.2.2.2.1.3.5

By the Sum Rule, the derivative of $V + 1$ with respect to $V$ is $\frac{d}{d V} [V] + \frac{d}{d V} [1]$ .

$V + 1 + (V - 1) (\frac{d}{d V} [V] + \frac{d}{d V} [1])$

Step 6.2.2.2.1.3.6

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

$V + 1 + (V - 1) (1 + \frac{d}{d V} [1])$

Step 6.2.2.2.1.3.7

Since $1$ is constant with respect to $V$ , the derivative of $1$ with respect to $V$ is $0$ .

$V + 1 + (V - 1) (1 + 0)$

Step 6.2.2.2.1.3.8

Simplify by adding terms.

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Step 6.2.2.2.1.3.8.1

Add $1$ and $0$ .

$V + 1 + (V - 1) \cdot 1$

Step 6.2.2.2.1.3.8.2

Multiply $V - 1$ by $1$ .

$V + 1 + V - 1$

Step 6.2.2.2.1.3.8.3

Add $V$ and $V$ .

$2 V + 1 - 1$

Step 6.2.2.2.1.3.8.4

Simplify by subtracting numbers.

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Step 6.2.2.2.1.3.8.4.1

Subtract $1$ from $1$ .

$2 V + 0$

Step 6.2.2.2.1.3.8.4.2

Add $2 V$ and $0$ .

$2 V$

Step 6.2.2.2.2

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

$2 \int \frac{1}{u} \cdot \frac{1}{2} d u = \int \frac{1}{x} d x$

Step 6.2.2.3

Simplify.

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Step 6.2.2.3.1

Multiply $\frac{1}{u}$ by $\frac{1}{2}$ .

$2 \int \frac{1}{u \cdot 2} d u = \int \frac{1}{x} d x$

Step 6.2.2.3.2

Move $2$ to the left of $u$ .

$2 \int \frac{1}{2 u} d u = \int \frac{1}{x} d x$

Step 6.2.2.4

Since $\frac{1}{2}$ is constant with respect to $u$ , move $\frac{1}{2}$ out of the integral.

$2 (\frac{1}{2} \int \frac{1}{u} d u) = \int \frac{1}{x} d x$

Step 6.2.2.5

Simplify.

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Step 6.2.2.5.1

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

$\frac{2}{2} \int \frac{1}{u} d u = \int \frac{1}{x} d x$

Step 6.2.2.5.2

Cancel the common factor of $2$ .

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Step 6.2.2.5.2.1

Cancel the common factor.

$\frac{2}{2} \int \frac{1}{u} d u = \int \frac{1}{x} d x$

Step 6.2.2.5.2.2

Rewrite the expression.

$1 \int \frac{1}{u} d u = \int \frac{1}{x} d x$

Step 6.2.2.5.3

Multiply $\int \frac{1}{u} d u$ by $1$ .

$\int \frac{1}{u} d u = \int \frac{1}{x} d x$

Step 6.2.2.6

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

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

Step 6.2.2.7

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

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

Step 6.2.3

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

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

Step 6.2.4

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

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

Step 6.3

Solve for $V$ .

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Step 6.3.1

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

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

Step 6.3.2

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

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

Step 6.3.3

To solve for $V$ , rewrite the equation using properties of logarithms.

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

Step 6.3.4

Rewrite $\ln (\frac{| (V + 1) (V - 1) |}{| x |}) = C$ in exponential form using the definition of a logarithm. If $x$ and $b$ are positive real numbers and $b \neq 1$ , then $\log_{b} (x) = y$ is equivalent to $b^{y} = x$ .

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

Step 6.3.5

Solve for $V$ .

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Step 6.3.5.1

Rewrite the equation as $\frac{| (V + 1) (V - 1) |}{| x |} = e^{C}$ .

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

Step 6.3.5.2

Multiply both sides by $| x |$ .

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

Step 6.3.5.3

Simplify the left side.

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Step 6.3.5.3.1

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

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Step 6.3.5.3.1.1

Cancel the common factor of $| x |$ .

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Step 6.3.5.3.1.1.1

Cancel the common factor.

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

Step 6.3.5.3.1.1.2

Rewrite the expression.

$| (V + 1) (V - 1) | = e^{C} | x |$

Step 6.3.5.3.1.2

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

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Step 6.3.5.3.1.2.1

Apply the distributive property.

$| V (V - 1) + 1 (V - 1) | = e^{C} | x |$

Step 6.3.5.3.1.2.2

Apply the distributive property.

$| V \cdot V + V \cdot - 1 + 1 (V - 1) | = e^{C} | x |$

Step 6.3.5.3.1.2.3

Apply the distributive property.

$| V \cdot V + V \cdot - 1 + 1 V + 1 \cdot - 1 | = e^{C} | x |$

Step 6.3.5.3.1.3

Simplify and combine like terms.

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Step 6.3.5.3.1.3.1

Simplify each term.

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Step 6.3.5.3.1.3.1.1

Multiply $V$ by $V$ .

$| V^{2} + V \cdot - 1 + 1 V + 1 \cdot - 1 | = e^{C} | x |$

Step 6.3.5.3.1.3.1.2

Move $- 1$ to the left of $V$ .

$| V^{2} - 1 \cdot V + 1 V + 1 \cdot - 1 | = e^{C} | x |$

Step 6.3.5.3.1.3.1.3

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

$| V^{2} - V + 1 V + 1 \cdot - 1 | = e^{C} | x |$

Step 6.3.5.3.1.3.1.4

Multiply $V$ by $1$ .

$| V^{2} - V + V + 1 \cdot - 1 | = e^{C} | x |$

Step 6.3.5.3.1.3.1.5

Multiply $- 1$ by $1$ .

$| V^{2} - V + V - 1 | = e^{C} | x |$

Step 6.3.5.3.1.3.2

Add $- V$ and $V$ .

$| V^{2} + 0 - 1 | = e^{C} | x |$

Step 6.3.5.3.1.3.3

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

$| V^{2} - 1 | = e^{C} | x |$

Step 6.3.5.4

Solve for $V$ .

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Step 6.3.5.4.1

Reorder factors in $e^{C} | x |$ .

$| V^{2} - 1 | = | x | e^{C}$

Step 6.3.5.4.2

Remove the absolute value term. This creates a $\pm$ on the right side of the equation because $| x | = \pm x$ .

$V^{2} - 1 = \pm | x | e^{C}$

Step 6.3.5.4.3

Reorder factors in $\pm | x | e^{C}$ .

$V^{2} - 1 = \pm e^{C} | x |$

Step 6.3.5.4.4

Add $1$ to both sides of the equation.

$V^{2} = \pm e^{C} | x | + 1$

Step 6.3.5.4.5

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$V = \pm \sqrt{\pm e^{C} | x | + 1}$

Step 6.4

Group the constant terms together.

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Step 6.4.1

Simplify the constant of integration.

$V = \pm \sqrt{\pm C | x | + 1}$

Step 6.4.2

Combine constants with the plus or minus.

$V = \pm \sqrt{C | x | + 1}$

Step 7

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

$\frac{y}{x} = \pm \sqrt{C | x | + 1}$

Step 8

Solve $\frac{y}{x} = \pm \sqrt{C | x | + 1}$ for $y$ .

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Step 8.1

Multiply both sides by $x$ .

$\frac{y}{x} x = (\pm \sqrt{C | x | + 1}) x$

Step 8.2

Simplify the left side.

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Step 8.2.1

Cancel the common factor of $x$ .

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Step 8.2.1.1

Cancel the common factor.

$\frac{y}{x} x = (\pm \sqrt{C | x | + 1}) x$

Step 8.2.1.2

Rewrite the expression.

$y = (\pm \sqrt{C | x | + 1}) x$