Calculus Examples

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

Step 1

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

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Step 1.1

Divide each term in $2 x y \frac{d y}{d x} = x^{2} + y^{2}$ by $2 x y$ and simplify.

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Step 1.1.1

Divide each term in $2 x y \frac{d y}{d x} = x^{2} + y^{2}$ by $2 x y$ .

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

Step 1.1.2

Simplify the left side.

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Step 1.1.2.1

Cancel the common factor of $2$ .

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Step 1.1.2.1.1

Cancel the common factor.

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

Step 1.1.2.1.2

Rewrite the expression.

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

Step 1.1.2.2

Cancel the common factor of $x$ .

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Step 1.1.2.2.1

Cancel the common factor.

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

Step 1.1.2.2.2

Rewrite the expression.

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

Step 1.1.2.3

Cancel the common factor of $y$ .

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Step 1.1.2.3.1

Cancel the common factor.

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

Step 1.1.2.3.2

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

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

Step 1.1.3

Simplify the right side.

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Step 1.1.3.1

Simplify each term.

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Step 1.1.3.1.1

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

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Step 1.1.3.1.1.1

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

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

Step 1.1.3.1.1.2

Cancel the common factors.

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Step 1.1.3.1.1.2.1

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

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

Step 1.1.3.1.1.2.2

Cancel the common factor.

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

Step 1.1.3.1.1.2.3

Rewrite the expression.

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

Step 1.1.3.1.2

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

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Step 1.1.3.1.2.1

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

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

Step 1.1.3.1.2.2

Cancel the common factors.

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Step 1.1.3.1.2.2.1

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

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

Step 1.1.3.1.2.2.2

Cancel the common factor.

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

Step 1.1.3.1.2.2.3

Rewrite the expression.

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

Step 1.2

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

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Step 1.2.1

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

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Step 1.2.1.1

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

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

Step 1.2.1.2

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

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

Step 1.2.2

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

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

Step 1.3

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

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Step 1.3.1

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

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

Step 1.3.2

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

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

Step 2

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

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

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

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

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

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

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

Step 6.1.1.1.2

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

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

Step 6.1.1.1.3

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

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

Step 6.1.1.2

Subtract $V$ from both sides of the equation.

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

Step 6.1.1.3

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

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Step 6.1.1.3.1

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

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

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.

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

Step 6.1.1.3.2.1.2

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

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

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.

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

Step 6.1.1.3.3.2

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

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

Step 6.1.1.3.3.3

Simplify terms.

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Step 6.1.1.3.3.3.1

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

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

Step 6.1.1.3.3.3.2

Combine the numerators over the common denominator.

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

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 $V - V \cdot 2$ .

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Step 6.1.1.3.3.3.3.1.1.1

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

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

Step 6.1.1.3.3.3.3.1.1.2

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

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

Step 6.1.1.3.3.3.3.1.1.3

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

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

Step 6.1.1.3.3.3.3.1.1.4

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

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

Step 6.1.1.3.3.3.3.1.2

Multiply $- 1$ by $2$ .

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

Step 6.1.1.3.3.3.3.1.3

Subtract $2$ from $1$ .

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

Step 6.1.1.3.3.3.3.2

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

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

Step 6.1.1.3.3.3.3.3

Move the negative in front of the fraction.

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

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

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

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

Step 6.1.1.3.3.4.4.2

Rewrite $V \cdot V$ as $V^{2}$ .

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

Step 6.1.1.3.3.4.4.3

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 $b = V$ .

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

Step 6.1.1.3.3.5

Multiply the numerator by the reciprocal of the denominator.

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

Step 6.1.1.3.3.6

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

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

Step 6.1.2

Regroup factors.

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

Step 6.1.3

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

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

Step 6.1.4

Simplify.

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Step 6.1.4.1

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

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

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 $2 V x$ .

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

Step 6.1.4.2.2

Cancel the common factor.

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

Step 6.1.4.2.3

Rewrite the expression.

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

Step 6.1.4.3

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

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Step 6.1.4.3.1

Cancel the common factor.

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

Step 6.1.4.3.2

Rewrite the expression.

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

Step 6.1.5

Rewrite the equation.

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

Step 6.2

Integrate both sides.

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Step 6.2.1

Set up an integral on each side.

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

Step 6.2.2

Integrate the left side.

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Step 6.2.2.1

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

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

Step 6.2.2.2

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

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Step 6.2.2.2.1.1

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

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

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 $f (V) = 1 + V$ and $g (V) = 1 - V$ .

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

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 $1 - V$ with respect to $V$ is $\frac{d}{d V} [1] + \frac{d}{d V} [- V]$ .

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

Step 6.2.2.2.1.3.2

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

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

Step 6.2.2.2.1.3.3

Add $0$ and $\frac{d}{d V} [- V]$ .

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

Step 6.2.2.2.1.3.4

Since $- 1$ is constant with respect to $V$ , the derivative of $- V$ with respect to $V$ is $- \frac{d}{d V} [V]$ .

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

Step 6.2.2.2.1.3.5

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

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

Step 6.2.2.2.1.3.6

Simplify the expression.

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Step 6.2.2.2.1.3.6.1

Multiply $- 1$ by $1$ .

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

Step 6.2.2.2.1.3.6.2

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

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

Step 6.2.2.2.1.3.6.3

Rewrite $- 1 (1 + V)$ as $- (1 + V)$ .

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

Step 6.2.2.2.1.3.7

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

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

Step 6.2.2.2.1.3.8

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

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

Step 6.2.2.2.1.3.9

Add $0$ and $\frac{d}{d V} [V]$ .

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

Step 6.2.2.2.1.3.10

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

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

Step 6.2.2.2.1.3.11

Multiply $1 - V$ by $1$ .

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

Step 6.2.2.2.1.4

Simplify.

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Step 6.2.2.2.1.4.1

Apply the distributive property.

$- 1 \cdot 1 - V + 1 - V$

Step 6.2.2.2.1.4.2

Combine terms.

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Step 6.2.2.2.1.4.2.1

Multiply $- 1$ by $1$ .

$- 1 - V + 1 - V$

Step 6.2.2.2.1.4.2.2

Add $- 1$ and $1$ .

$- V + 0 - V$

Step 6.2.2.2.1.4.2.3

Add $- V$ and $0$ .

$- V - V$

Step 6.2.2.2.1.4.2.4

Subtract $V$ from $- V$ .

$- 2 V$

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

Move the negative in front of the fraction.

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

Step 6.2.2.3.2

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

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

Step 6.2.2.3.3

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

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

Step 6.2.2.5

Multiply $- 1$ by $2$ .

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

Step 6.2.2.6

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 6.2.2.7

Simplify.

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Step 6.2.2.7.1

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

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

Step 6.2.2.7.2

Cancel the common factor of $- 2$ and $2$ .

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Step 6.2.2.7.2.1

Factor $2$ out of $- 2$ .

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

Step 6.2.2.7.2.2

Cancel the common factors.

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Step 6.2.2.7.2.2.1

Factor $2$ out of $2$ .

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

Step 6.2.2.7.2.2.2

Cancel the common factor.

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

Step 6.2.2.7.2.2.3

Rewrite the expression.

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

Step 6.2.2.7.2.2.4

Divide $- 1$ by $1$ .

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

Step 6.2.2.8

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

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

Step 6.2.2.9

Simplify.

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

Step 6.2.2.10

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

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

Step 6.2.3

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

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

Step 6.2.4

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

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

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.

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

Step 6.3.2

Simplify each term.

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Step 6.3.2.1

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

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Step 6.3.2.1.1

Apply the distributive property.

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

Step 6.3.2.1.2

Apply the distributive property.

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

Step 6.3.2.1.3

Apply the distributive property.

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

Step 6.3.2.2

Simplify and combine like terms.

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Step 6.3.2.2.1

Simplify each term.

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Step 6.3.2.2.1.1

Multiply $1$ by $1$ .

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

Step 6.3.2.2.1.2

Multiply $- | x |$ by $1$ .

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

Step 6.3.2.2.1.3

Multiply $| x |$ by $1$ .

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

Step 6.3.2.2.1.4

Rewrite using the commutative property of multiplication.

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

Step 6.3.2.2.1.5

Multiply $| x |$ by $| x |$ by adding the exponents.

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Step 6.3.2.2.1.5.1

Move $| x |$ .

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

Step 6.3.2.2.1.5.2

Multiply $| x |$ by $| x |$ .

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

Step 6.3.2.2.2

Add $- | x |$ and $| x |$ .

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

Step 6.3.2.2.3

Add $1$ and $0$ .

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

Step 6.3.3

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

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

Step 6.3.4

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

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Step 6.3.4.1

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

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

Step 6.3.4.2

Simplify the left side.

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Step 6.3.4.2.1

Dividing two negative values results in a positive value.

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

Step 6.3.4.2.2

Divide $\ln (| 1 - V^{2} |)$ by $1$ .

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

Step 6.3.4.3

Simplify the right side.

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Step 6.3.4.3.1

Simplify each term.

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Step 6.3.4.3.1.1

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

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

Step 6.3.4.3.1.2

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

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

Step 6.3.4.3.1.3

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

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

Step 6.3.4.3.1.4

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

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

Step 6.3.5

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

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

Step 6.3.6

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

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

Step 6.3.7

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

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

Step 6.3.8

Apply the distributive property.

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

Step 6.3.9

Multiply $x$ by $1$ .

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

Step 6.3.10

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

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

Step 6.3.11

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

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

Step 6.3.12

Solve for $V$ .

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Step 6.3.12.1

Rewrite the equation as $| x - V^{2} x | = e^{- C}$ .

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

Step 6.3.12.2

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

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

Step 6.3.12.3

Subtract $x$ from both sides of the equation.

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

Step 6.3.12.4

Divide each term in $- V^{2} x = \pm e^{- C} - x$ by $- x$ and simplify.

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Step 6.3.12.4.1

Divide each term in $- V^{2} x = \pm e^{- C} - x$ by $- x$ .

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

Step 6.3.12.4.2

Simplify the left side.

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Step 6.3.12.4.2.1

Dividing two negative values results in a positive value.

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

Step 6.3.12.4.2.2

Cancel the common factor of $x$ .

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Step 6.3.12.4.2.2.1

Cancel the common factor.

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

Step 6.3.12.4.2.2.2

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

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

Step 6.3.12.4.3

Simplify the right side.

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Step 6.3.12.4.3.1

Simplify each term.

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Step 6.3.12.4.3.1.1

Simplify $\frac{\pm e^{- C}}{- x}$ .

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

Step 6.3.12.4.3.1.2

Dividing two negative values results in a positive value.

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

Step 6.3.12.4.3.1.3

Cancel the common factor of $x$ .

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Step 6.3.12.4.3.1.3.1

Cancel the common factor.

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

Step 6.3.12.4.3.1.3.2

Rewrite the expression.

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

Step 6.3.12.5

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

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

Step 6.3.12.6

Simplify $\pm \sqrt{\frac{\pm e^{- C}}{x} + 1}$ .

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Step 6.3.12.6.1

Write $1$ as a fraction with a common denominator.

$V = \pm \sqrt{\frac{\pm e^{- C}}{x} + \frac{x}{x}}$

Step 6.3.12.6.2

Combine the numerators over the common denominator.

$V = \pm \sqrt{\frac{\pm e^{- C} + x}{x}}$

Step 6.3.12.6.3

Rewrite $\sqrt{\frac{\pm e^{- C} + x}{x}}$ as $\frac{\sqrt{\pm e^{- C} + x}}{\sqrt{x}}$ .

$V = \pm \frac{\sqrt{\pm e^{- C} + x}}{\sqrt{x}}$

Step 6.3.12.6.4

Multiply $\frac{\sqrt{\pm e^{- C} + x}}{\sqrt{x}}$ by $\frac{\sqrt{x}}{\sqrt{x}}$ .

$V = \pm \frac{\sqrt{\pm e^{- C} + x}}{\sqrt{x}} \cdot \frac{\sqrt{x}}{\sqrt{x}}$

Step 6.3.12.6.5

Combine and simplify the denominator.

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Step 6.3.12.6.5.1

Multiply $\frac{\sqrt{\pm e^{- C} + x}}{\sqrt{x}}$ by $\frac{\sqrt{x}}{\sqrt{x}}$ .

$V = \pm \frac{\sqrt{\pm e^{- C} + x} \sqrt{x}}{\sqrt{x} \sqrt{x}}$

Step 6.3.12.6.5.2

Raise $\sqrt{x}$ to the power of $1$ .

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

Step 6.3.12.6.5.3

Raise $\sqrt{x}$ to the power of $1$ .

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

Step 6.3.12.6.5.4

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

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

Step 6.3.12.6.5.5

Add $1$ and $1$ .

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

Step 6.3.12.6.5.6

Rewrite ${\sqrt{x}}^{2}$ as $x$ .

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Step 6.3.12.6.5.6.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{x}$ as $x^{\frac{1}{2}}$ .

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

Step 6.3.12.6.5.6.2

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

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

Step 6.3.12.6.5.6.3

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

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

Step 6.3.12.6.5.6.4

Cancel the common factor of $2$ .

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Step 6.3.12.6.5.6.4.1

Cancel the common factor.

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

Step 6.3.12.6.5.6.4.2

Rewrite the expression.

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

Step 6.3.12.6.5.6.5

Simplify.

$V = \pm \frac{\sqrt{\pm e^{- C} + x} \sqrt{x}}{x}$

Step 6.3.12.6.6

Combine using the product rule for radicals.

$V = \pm \frac{\sqrt{(\pm e^{- C} + x) x}}{x}$

Step 6.3.12.6.7

Reorder factors in $\pm \frac{\sqrt{(\pm e^{- C} + x) x}}{x}$ .

$V = \pm \frac{\sqrt{x (\pm e^{- C} + x)}}{x}$

Step 6.4

Simplify the constant of integration.

$V = \pm \frac{\sqrt{x (C + x)}}{x}$

Step 7

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

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

Step 8

Solve $\frac{y}{x} = \pm \frac{\sqrt{x (C + x)}}{x}$ for $y$ .

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Step 8.1

Multiply both sides by $x$ .

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

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.

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

Step 8.2.1.2

Rewrite the expression.

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