Calculus Examples

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

Step 1

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

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Step 1.1

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

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Step 1.1.1

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

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

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 $2 y^{2}$ .

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

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 $x y$ .

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

Step 1.1.2.1.2.2

Cancel the common factor.

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

Step 1.1.2.1.2.3

Rewrite the expression.

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

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 $- x^{2}$ .

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

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 $x y$ .

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

Step 1.1.2.2.2.2

Cancel the common factor.

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

Step 1.1.2.2.2.3

Rewrite the expression.

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

Step 1.1.2.3

Move the negative in front of the fraction.

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

Step 1.2

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

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

Step 1.3

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

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Step 1.3.1

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

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

Step 1.3.2

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

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

Step 2

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

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

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

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

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

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

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

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.

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

Step 6.1.1.2.2

Subtract $V$ from $2 V$ .

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

Step 6.1.1.3

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

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Step 6.1.1.3.1

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

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

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.

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

Step 6.1.1.3.2.1.2

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

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

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.

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

Step 6.1.1.3.3.1.2

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

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

Step 6.1.2

Factor.

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Step 6.1.2.1

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

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

Step 6.1.2.2

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

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

Step 6.1.2.3

Combine the numerators over the common denominator.

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

Step 6.1.2.4

Simplify the numerator.

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Step 6.1.2.4.1

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

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

Step 6.1.2.4.2

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

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

Step 6.1.2.4.3

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

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

Step 6.1.2.4.4

Add $1$ and $1$ .

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

Step 6.1.2.4.5

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

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

Step 6.1.2.4.6

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 $b = 1$ .

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

Step 6.1.3

Regroup factors.

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

Step 6.1.4

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

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

Step 6.1.5

Simplify.

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Step 6.1.5.1

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

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

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 $V x$ .

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

Step 6.1.5.2.2

Cancel the common factor.

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

Step 6.1.5.2.3

Rewrite the expression.

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

Step 6.1.5.3

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

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Step 6.1.5.3.1

Cancel the common factor.

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

Step 6.1.5.3.2

Rewrite the expression.

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

Step 6.1.6

Rewrite the equation.

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

Step 6.2

Integrate both sides.

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Step 6.2.1

Set up an integral on each side.

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

Step 6.2.2

Integrate the left side.

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Step 6.2.2.1

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 6.2.2.1.1

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

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Step 6.2.2.1.1.1

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

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

Step 6.2.2.1.1.2

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 6.2.2.1.1.3

Differentiate.

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Step 6.2.2.1.1.3.1

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 6.2.2.1.1.3.2

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 6.2.2.1.1.3.3

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 6.2.2.1.1.3.4

Simplify the expression.

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Step 6.2.2.1.1.3.4.1

Add $1$ and $0$ .

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

Step 6.2.2.1.1.3.4.2

Multiply $V + 1$ by $1$ .

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

Step 6.2.2.1.1.3.5

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 6.2.2.1.1.3.6

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 6.2.2.1.1.3.7

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 6.2.2.1.1.3.8

Simplify by adding terms.

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Step 6.2.2.1.1.3.8.1

Add $1$ and $0$ .

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

Step 6.2.2.1.1.3.8.2

Multiply $V - 1$ by $1$ .

$V + 1 + V - 1$

Step 6.2.2.1.1.3.8.3

Add $V$ and $V$ .

$2 V + 1 - 1$

Step 6.2.2.1.1.3.8.4

Simplify by subtracting numbers.

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Step 6.2.2.1.1.3.8.4.1

Subtract $1$ from $1$ .

$2 V + 0$

Step 6.2.2.1.1.3.8.4.2

Add $2 V$ and $0$ .

$2 V$

Step 6.2.2.1.2

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

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

Step 6.2.2.2

Simplify.

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Step 6.2.2.2.1

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

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

Step 6.2.2.2.2

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

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

Step 6.2.2.3

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

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

Step 6.2.2.4

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

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

Step 6.2.2.5

Simplify.

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

Step 6.2.2.6

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

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

Step 6.2.3

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

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

Step 6.2.4

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

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

Step 6.3

Solve for $V$ .

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Step 6.3.1

Multiply both sides of the equation by $2$ .

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

Step 6.3.2

Simplify both sides of the equation.

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Step 6.3.2.1

Simplify the left side.

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Step 6.3.2.1.1

Simplify $2 (\frac{1}{2} \ln (| (V + 1) (V - 1) |))$ .

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Step 6.3.2.1.1.1

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

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Step 6.3.2.1.1.1.1

Apply the distributive property.

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

Step 6.3.2.1.1.1.2

Apply the distributive property.

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

Step 6.3.2.1.1.1.3

Apply the distributive property.

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

Step 6.3.2.1.1.2

Simplify and combine like terms.

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Step 6.3.2.1.1.2.1

Simplify each term.

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Step 6.3.2.1.1.2.1.1

Multiply $V$ by $V$ .

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

Step 6.3.2.1.1.2.1.2

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

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

Step 6.3.2.1.1.2.1.3

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

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

Step 6.3.2.1.1.2.1.4

Multiply $V$ by $1$ .

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

Step 6.3.2.1.1.2.1.5

Multiply $- 1$ by $1$ .

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

Step 6.3.2.1.1.2.2

Add $- V$ and $V$ .

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

Step 6.3.2.1.1.2.3

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

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

Step 6.3.2.1.1.3

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

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

Step 6.3.2.1.1.4

Cancel the common factor of $2$ .

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Step 6.3.2.1.1.4.1

Cancel the common factor.

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

Step 6.3.2.1.1.4.2

Rewrite the expression.

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

Step 6.3.2.2

Simplify the right side.

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Step 6.3.2.2.1

Apply the distributive property.

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

Step 6.3.3

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

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

Step 6.3.4

Simplify the left side.

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Step 6.3.4.1

Simplify $\ln (| V^{2} - 1 |) - 2 \ln (| x |)$ .

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Step 6.3.4.1.1

Simplify each term.

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Step 6.3.4.1.1.1

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

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

Step 6.3.4.1.1.2

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

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

Step 6.3.4.1.2

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

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

Step 6.3.4.1.3

Simplify the numerator.

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Step 6.3.4.1.3.1

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

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

Step 6.3.4.1.3.2

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 $b = 1$ .

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

Step 6.3.4.1.3.3

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

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Step 6.3.4.1.3.3.1

Apply the distributive property.

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

Step 6.3.4.1.3.3.2

Apply the distributive property.

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

Step 6.3.4.1.3.3.3

Apply the distributive property.

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

Step 6.3.4.1.3.4

Simplify and combine like terms.

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Step 6.3.4.1.3.4.1

Simplify each term.

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Step 6.3.4.1.3.4.1.1

Multiply $V$ by $V$ .

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

Step 6.3.4.1.3.4.1.2

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

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

Step 6.3.4.1.3.4.1.3

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

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

Step 6.3.4.1.3.4.1.4

Multiply $V$ by $1$ .

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

Step 6.3.4.1.3.4.1.5

Multiply $- 1$ by $1$ .

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

Step 6.3.4.1.3.4.2

Add $- V$ and $V$ .

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

Step 6.3.4.1.3.4.3

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

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

Step 6.3.4.1.3.5

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

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

Step 6.3.4.1.3.6

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 $b = 1$ .

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

Step 6.3.5

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

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

Step 6.3.6

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

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

Step 6.3.7

Solve for $V$ .

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Step 6.3.7.1

Rewrite the equation as $\frac{| (V + 1) (V - 1) |}{x^{2}} = e^{2 C}$ .

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

Step 6.3.7.2

Multiply both sides by $x^{2}$ .

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

Step 6.3.7.3

Simplify.

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Step 6.3.7.3.1

Simplify the left side.

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Step 6.3.7.3.1.1

Simplify $\frac{| (V + 1) (V - 1) |}{x^{2}} x^{2}$ .

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Step 6.3.7.3.1.1.1

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

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Step 6.3.7.3.1.1.1.1

Cancel the common factor.

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

Step 6.3.7.3.1.1.1.2

Rewrite the expression.

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

Step 6.3.7.3.1.1.2

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

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Step 6.3.7.3.1.1.2.1

Apply the distributive property.

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

Step 6.3.7.3.1.1.2.2

Apply the distributive property.

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

Step 6.3.7.3.1.1.2.3

Apply the distributive property.

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

Step 6.3.7.3.1.1.3

Simplify and combine like terms.

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Step 6.3.7.3.1.1.3.1

Simplify each term.

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Step 6.3.7.3.1.1.3.1.1

Multiply $V$ by $V$ .

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

Step 6.3.7.3.1.1.3.1.2

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

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

Step 6.3.7.3.1.1.3.1.3

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

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

Step 6.3.7.3.1.1.3.1.4

Multiply $V$ by $1$ .

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

Step 6.3.7.3.1.1.3.1.5

Multiply $- 1$ by $1$ .

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

Step 6.3.7.3.1.1.3.2

Add $- V$ and $V$ .

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

Step 6.3.7.3.1.1.3.3

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

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

Step 6.3.7.3.2

Simplify the right side.

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Step 6.3.7.3.2.1

Reorder factors in $e^{2 C} x^{2}$ .

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

Step 6.3.7.4

Solve for $V$ .

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Step 6.3.7.4.1

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

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

Step 6.3.7.4.2

Add $1$ to both sides of the equation.

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

Step 6.3.7.4.3

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

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

Step 6.4

Group the constant terms together.

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Step 6.4.1

Simplify the constant of integration.

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

Step 6.4.2

Combine constants with the plus or minus.

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

Step 7

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

$\frac{y}{x} = \pm \sqrt{C x^{2} + 1}$

Step 8

Solve $\frac{y}{x} = \pm \sqrt{C x^{2} + 1}$ for $y$ .

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Step 8.1

Multiply both sides by $x$ .

$\frac{y}{x} x = (\pm \sqrt{C x^{2} + 1}) x$

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.

$\frac{y}{x} x = (\pm \sqrt{C x^{2} + 1}) x$

Step 8.2.1.2

Rewrite the expression.

$y = (\pm \sqrt{C x^{2} + 1}) x$