Calculus Examples

$9 x y \frac{d y}{d x} = 8 y^{2} + 5 x^{2}$

Step 1

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

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Step 1.1

Divide each term in $9 x y \frac{d y}{d x} = 8 y^{2} + 5 x^{2}$ by $9 x y$ and simplify.

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Step 1.1.1

Divide each term in $9 x y \frac{d y}{d x} = 8 y^{2} + 5 x^{2}$ by $9 x y$ .

$\frac{9 x y \frac{d y}{d x}}{9 x y} = \frac{8 y^{2}}{9 x y} + \frac{5 x^{2}}{9 x y}$

Step 1.1.2

Simplify the left side.

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Step 1.1.2.1

Cancel the common factor of $9$ .

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Step 1.1.2.1.1

Cancel the common factor.

$\frac{9 x y \frac{d y}{d x}}{9 x y} = \frac{8 y^{2}}{9 x y} + \frac{5 x^{2}}{9 x y}$

Step 1.1.2.1.2

Rewrite the expression.

$\frac{x y \frac{d y}{d x}}{x y} = \frac{8 y^{2}}{9 x y} + \frac{5 x^{2}}{9 x y}$

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.

$\frac{x y \frac{d y}{d x}}{x y} = \frac{8 y^{2}}{9 x y} + \frac{5 x^{2}}{9 x y}$

Step 1.1.2.2.2

Rewrite the expression.

$\frac{y \frac{d y}{d x}}{y} = \frac{8 y^{2}}{9 x y} + \frac{5 x^{2}}{9 x y}$

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.

$\frac{y \frac{d y}{d x}}{y} = \frac{8 y^{2}}{9 x y} + \frac{5 x^{2}}{9 x y}$

Step 1.1.2.3.2

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

$\frac{d y}{d x} = \frac{8 y^{2}}{9 x y} + \frac{5 x^{2}}{9 x y}$

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 $y^{2}$ and $y$ .

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Step 1.1.3.1.1.1

Factor $y$ out of $8 y^{2}$ .

$\frac{d y}{d x} = \frac{y (8 y)}{9 x y} + \frac{5 x^{2}}{9 x y}$

Step 1.1.3.1.1.2

Cancel the common factors.

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Step 1.1.3.1.1.2.1

Factor $y$ out of $9 x y$ .

$\frac{d y}{d x} = \frac{y (8 y)}{y (9 x)} + \frac{5 x^{2}}{9 x y}$

Step 1.1.3.1.1.2.2

Cancel the common factor.

$\frac{d y}{d x} = \frac{y (8 y)}{y (9 x)} + \frac{5 x^{2}}{9 x y}$

Step 1.1.3.1.1.2.3

Rewrite the expression.

$\frac{d y}{d x} = \frac{8 y}{9 x} + \frac{5 x^{2}}{9 x y}$

Step 1.1.3.1.2

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

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Step 1.1.3.1.2.1

Factor $x$ out of $5 x^{2}$ .

$\frac{d y}{d x} = \frac{8 y}{9 x} + \frac{x (5 x)}{9 x y}$

Step 1.1.3.1.2.2

Cancel the common factors.

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Step 1.1.3.1.2.2.1

Factor $x$ out of $9 x y$ .

$\frac{d y}{d x} = \frac{8 y}{9 x} + \frac{x (5 x)}{x (9 y)}$

Step 1.1.3.1.2.2.2

Cancel the common factor.

$\frac{d y}{d x} = \frac{8 y}{9 x} + \frac{x (5 x)}{x (9 y)}$

Step 1.1.3.1.2.2.3

Rewrite the expression.

$\frac{d y}{d x} = \frac{8 y}{9 x} + \frac{5 x}{9 y}$

Step 1.2

Factor out $\frac{y}{x}$ from $\frac{8 y}{9 x}$ .

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Step 1.2.1

Factor $\frac{y}{x}$ out of $\frac{8 y}{9 x}$ .

$\frac{d y}{d x} = \frac{y}{x} \cdot \frac{8}{9} + \frac{5 x}{9 y}$

Step 1.2.2

Reorder $\frac{y}{x}$ and $\frac{8}{9}$ .

$\frac{d y}{d x} = \frac{8}{9} \cdot \frac{y}{x} + \frac{5 x}{9 y}$

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 $\frac{5 x}{9 y}$ .

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Step 1.3.1.1

Factor $\frac{x}{y}$ out of $\frac{5 x}{9 y}$ .

$\frac{d y}{d x} = \frac{8}{9} \cdot \frac{y}{x} + \frac{x}{y} \cdot \frac{5}{9}$

Step 1.3.1.2

Reorder $\frac{x}{y}$ and $\frac{5}{9}$ .

$\frac{d y}{d x} = \frac{8}{9} \cdot \frac{y}{x} + \frac{5}{9} \cdot \frac{x}{y}$

Step 1.3.2

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

$\frac{d y}{d x} = \frac{8}{9} \cdot \frac{y}{x} + \frac{5}{9} {(\frac{y}{x})}^{- 1}$

Step 2

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

$\frac{d y}{d x} = \frac{8}{9} V + \frac{5}{9} V^{- 1}$

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

$x \frac{d V}{d x} + V = \frac{8}{9} V + \frac{5}{9} V^{- 1}$

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 $\frac{8}{9}$ and $V$ .

$x \frac{d V}{d x} + V = \frac{8 V}{9} + \frac{5}{9} V^{- 1}$

Step 6.1.1.1.2

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

$x \frac{d V}{d x} + V = \frac{8 V}{9} + \frac{5}{9} \cdot \frac{1}{V}$

Step 6.1.1.1.3

Multiply $\frac{5}{9}$ by $\frac{1}{V}$ .

$x \frac{d V}{d x} + V = \frac{8 V}{9} + \frac{5}{9 V}$

Step 6.1.1.2

Subtract $V$ from both sides of the equation.

$x \frac{d V}{d x} = \frac{8 V}{9} + \frac{5}{9 V} - V$

Step 6.1.1.3

Divide each term in $x \frac{d V}{d x} = \frac{8 V}{9} + \frac{5}{9 V} - V$ by $x$ and simplify.

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Step 6.1.1.3.1

Divide each term in $x \frac{d V}{d x} = \frac{8 V}{9} + \frac{5}{9 V} - V$ by $x$ .

$\frac{x \frac{d V}{d x}}{x} = \frac{\frac{8 V}{9}}{x} + \frac{\frac{5}{9 V}}{x} + \frac{- V}{x}$

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.

$\frac{x \frac{d V}{d x}}{x} = \frac{\frac{8 V}{9}}{x} + \frac{\frac{5}{9 V}}{x} + \frac{- V}{x}$

Step 6.1.1.3.2.1.2

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

$\frac{d V}{d x} = \frac{\frac{8 V}{9}}{x} + \frac{\frac{5}{9 V}}{x} + \frac{- V}{x}$

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.

$\frac{d V}{d x} = \frac{\frac{8 V}{9} + \frac{5}{9 V} - V}{x}$

Step 6.1.1.3.3.2

To write $- V$ as a fraction with a common denominator, multiply by $\frac{9}{9}$ .

$\frac{d V}{d x} = \frac{\frac{5}{9 V} + \frac{8 V}{9} - V \cdot \frac{9}{9}}{x}$

Step 6.1.1.3.3.3

Simplify terms.

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Step 6.1.1.3.3.3.1

Combine $- V$ and $\frac{9}{9}$ .

$\frac{d V}{d x} = \frac{\frac{5}{9 V} + \frac{8 V}{9} + \frac{- V \cdot 9}{9}}{x}$

Step 6.1.1.3.3.3.2

Combine the numerators over the common denominator.

$\frac{d V}{d x} = \frac{\frac{5}{9 V} + \frac{8 V - V \cdot 9}{9}}{x}$

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 $8 V - V \cdot 9$ .

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Step 6.1.1.3.3.3.3.1.1.1

Factor $V$ out of $8 V$ .

$\frac{d V}{d x} = \frac{\frac{5}{9 V} + \frac{V \cdot 8 - V \cdot 9}{9}}{x}$

Step 6.1.1.3.3.3.3.1.1.2

Factor $V$ out of $- V \cdot 9$ .

$\frac{d V}{d x} = \frac{\frac{5}{9 V} + \frac{V \cdot 8 + V (- 1 \cdot 9)}{9}}{x}$

Step 6.1.1.3.3.3.3.1.1.3

Factor $V$ out of $V \cdot 8 + V (- 1 \cdot 9)$ .

$\frac{d V}{d x} = \frac{\frac{5}{9 V} + \frac{V (8 - 1 \cdot 9)}{9}}{x}$

Step 6.1.1.3.3.3.3.1.2

Multiply $- 1$ by $9$ .

$\frac{d V}{d x} = \frac{\frac{5}{9 V} + \frac{V (8 - 9)}{9}}{x}$

Step 6.1.1.3.3.3.3.1.3

Subtract $9$ from $8$ .

$\frac{d V}{d x} = \frac{\frac{5}{9 V} + \frac{V \cdot - 1}{9}}{x}$

Step 6.1.1.3.3.3.3.2

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

$\frac{d V}{d x} = \frac{\frac{5}{9 V} + \frac{- 1 \cdot V}{9}}{x}$

Step 6.1.1.3.3.3.3.3

Move the negative in front of the fraction.

$\frac{d V}{d x} = \frac{\frac{5}{9 V} - \frac{V}{9}}{x}$

Step 6.1.1.3.3.4

Simplify the numerator.

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Step 6.1.1.3.3.4.1

To write $- \frac{V}{9}$ as a fraction with a common denominator, multiply by $\frac{V}{V}$ .

$\frac{d V}{d x} = \frac{\frac{5}{9 V} - \frac{V}{9} \cdot \frac{V}{V}}{x}$

Step 6.1.1.3.3.4.2

Multiply $\frac{V}{9}$ by $\frac{V}{V}$ .

$\frac{d V}{d x} = \frac{\frac{5}{9 V} - \frac{V \cdot V}{9 V}}{x}$

Step 6.1.1.3.3.4.3

Combine the numerators over the common denominator.

$\frac{d V}{d x} = \frac{\frac{5 - V \cdot V}{9 V}}{x}$

Step 6.1.1.3.3.4.4

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

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Step 6.1.1.3.3.4.4.1

Move $V$ .

$\frac{d V}{d x} = \frac{\frac{5 - (V \cdot V)}{9 V}}{x}$

Step 6.1.1.3.3.4.4.2

Multiply $V$ by $V$ .

$\frac{d V}{d x} = \frac{\frac{5 - V^{2}}{9 V}}{x}$

Step 6.1.1.3.3.5

Multiply the numerator by the reciprocal of the denominator.

$\frac{d V}{d x} = \frac{5 - V^{2}}{9 V} \cdot \frac{1}{x}$

Step 6.1.1.3.3.6

Multiply $\frac{5 - V^{2}}{9 V}$ by $\frac{1}{x}$ .

$\frac{d V}{d x} = \frac{5 - V^{2}}{9 V x}$

Step 6.1.2

Regroup factors.

$\frac{d V}{d x} = \frac{5 - V^{2}}{9 V} \cdot \frac{1}{x}$

Step 6.1.3

Multiply both sides by $\frac{9 V}{5 - V^{2}}$ .

$\frac{9 V}{5 - V^{2}} \frac{d V}{d x} = \frac{9 V}{5 - V^{2}} (\frac{5 - V^{2}}{9 V} \cdot \frac{1}{x})$

Step 6.1.4

Simplify.

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Step 6.1.4.1

Multiply $\frac{5 - V^{2}}{9 V}$ by $\frac{1}{x}$ .

$\frac{9 V}{5 - V^{2}} \frac{d V}{d x} = \frac{9 V}{5 - V^{2}} \cdot \frac{5 - V^{2}}{9 V x}$

Step 6.1.4.2

Cancel the common factor of $9 V$ .

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Step 6.1.4.2.1

Factor $9 V$ out of $9 V x$ .

$\frac{9 V}{5 - V^{2}} \frac{d V}{d x} = \frac{9 V}{5 - V^{2}} \cdot \frac{5 - V^{2}}{9 V (x)}$

Step 6.1.4.2.2

Cancel the common factor.

$\frac{9 V}{5 - V^{2}} \frac{d V}{d x} = \frac{9 V}{5 - V^{2}} \cdot \frac{5 - V^{2}}{9 V x}$

Step 6.1.4.2.3

Rewrite the expression.

$\frac{9 V}{5 - V^{2}} \frac{d V}{d x} = \frac{1}{5 - V^{2}} \cdot \frac{5 - V^{2}}{x}$

Step 6.1.4.3

Cancel the common factor of $5 - V^{2}$ .

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Step 6.1.4.3.1

Cancel the common factor.

$\frac{9 V}{5 - V^{2}} \frac{d V}{d x} = \frac{1}{5 - V^{2}} \cdot \frac{5 - V^{2}}{x}$

Step 6.1.4.3.2

Rewrite the expression.

$\frac{9 V}{5 - V^{2}} \frac{d V}{d x} = \frac{1}{x}$

Step 6.1.5

Rewrite the equation.

$\frac{9 V}{5 - V^{2}} d V = \frac{1}{x} d x$

Step 6.2

Integrate both sides.

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Step 6.2.1

Set up an integral on each side.

$\int \frac{9 V}{5 - V^{2}} d V = \int \frac{1}{x} d x$

Step 6.2.2

Integrate the left side.

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Step 6.2.2.1

Since $9$ is constant with respect to $V$ , move $9$ out of the integral.

$9 \int \frac{V}{5 - V^{2}} d V = \int \frac{1}{x} d x$

Step 6.2.2.2

Let $u = 5 - V^{2}$ . 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 $u = 5 - V^{2}$ . Find $\frac{d u}{d V}$ .

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Step 6.2.2.2.1.1

Differentiate $5 - V^{2}$ .

$\frac{d}{d V} [5 - V^{2}]$

Step 6.2.2.2.1.2

Differentiate.

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Step 6.2.2.2.1.2.1

By the Sum Rule, the derivative of $5 - V^{2}$ with respect to $V$ is $\frac{d}{d V} [5] + \frac{d}{d V} [- V^{2}]$ .

$\frac{d}{d V} [5] + \frac{d}{d V} [- V^{2}]$

Step 6.2.2.2.1.2.2

Since $5$ is constant with respect to $V$ , the derivative of $5$ with respect to $V$ is $0$ .

$0 + \frac{d}{d V} [- V^{2}]$

Step 6.2.2.2.1.3

Evaluate $\frac{d}{d V} [- V^{2}]$ .

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Step 6.2.2.2.1.3.1

Since $- 1$ is constant with respect to $V$ , the derivative of $- V^{2}$ with respect to $V$ is $- \frac{d}{d V} [V^{2}]$ .

$0 - \frac{d}{d V} [V^{2}]$

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 $n = 2$ .

$0 - (2 V)$

Step 6.2.2.2.1.3.3

Multiply $2$ by $- 1$ .

$0 - 2 V$

Step 6.2.2.2.1.4

Subtract $2 V$ from $0$ .

$- 2 V$

Step 6.2.2.2.2

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

$9 \int \frac{1}{u} \cdot \frac{1}{- 2} d u = \int \frac{1}{x} d x$

Step 6.2.2.3

Simplify.

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Step 6.2.2.3.1

Move the negative in front of the fraction.

$9 \int \frac{1}{u} (- \frac{1}{2}) d u = \int \frac{1}{x} d x$

Step 6.2.2.3.2

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

$9 \int - \frac{1}{u \cdot 2} d u = \int \frac{1}{x} d x$

Step 6.2.2.3.3

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

$9 \int - \frac{1}{2 u} d u = \int \frac{1}{x} d x$

Step 6.2.2.4

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

$9 (- \int \frac{1}{2 u} d u) = \int \frac{1}{x} d x$

Step 6.2.2.5

Multiply $- 1$ by $9$ .

$- 9 \int \frac{1}{2 u} d u = \int \frac{1}{x} d x$

Step 6.2.2.6

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

$- 9 (\frac{1}{2} \int \frac{1}{u} d u) = \int \frac{1}{x} d x$

Step 6.2.2.7

Simplify.

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Step 6.2.2.7.1

Combine $\frac{1}{2}$ and $- 9$ .

$\frac{- 9}{2} \int \frac{1}{u} d u = \int \frac{1}{x} d x$

Step 6.2.2.7.2

Move the negative in front of the fraction.

$- \frac{9}{2} \int \frac{1}{u} d u = \int \frac{1}{x} d x$

Step 6.2.2.8

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

$- \frac{9}{2} (\ln (| u |) + C_{1}) = \int \frac{1}{x} d x$

Step 6.2.2.9

Simplify.

$- \frac{9}{2} \ln (| u |) + C_{1} = \int \frac{1}{x} d x$

Step 6.2.2.10

Replace all occurrences of $u$ with $5 - V^{2}$ .

$- \frac{9}{2} \ln (| 5 - V^{2} |) + C_{1} = \int \frac{1}{x} d x$

Step 6.2.3

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

$- \frac{9}{2} \ln (| 5 - V^{2} |) + C_{1} = \ln (| x |) + C_{2}$

Step 6.2.4

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

$- \frac{9}{2} \ln (| 5 - V^{2} |) = \ln (| x |) + C$

Step 6.3

Solve for $V$ .

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Step 6.3.1

Multiply both sides of the equation by $- \frac{2}{9}$ .

$- \frac{2}{9} (- \frac{9}{2} \ln (| 5 - V^{2} |)) = - \frac{2}{9} (\ln (| x |) + C)$

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 $- \frac{2}{9} (- \frac{9}{2} \ln (| 5 - V^{2} |))$ .

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Step 6.3.2.1.1.1

Combine $\ln (| 5 - V^{2} |)$ and $\frac{9}{2}$ .

$- \frac{2}{9} (- \frac{\ln (| 5 - V^{2} |) \cdot 9}{2}) = - \frac{2}{9} (\ln (| x |) + C)$

Step 6.3.2.1.1.2

Cancel the common factor of $2$ .

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Step 6.3.2.1.1.2.1

Move the leading negative in $- \frac{2}{9}$ into the numerator.

$\frac{- 2}{9} (- \frac{\ln (| 5 - V^{2} |) \cdot 9}{2}) = - \frac{2}{9} (\ln (| x |) + C)$

Step 6.3.2.1.1.2.2

Move the leading negative in $- \frac{\ln (| 5 - V^{2} |) \cdot 9}{2}$ into the numerator.

$\frac{- 2}{9} \cdot \frac{- \ln (| 5 - V^{2} |) \cdot 9}{2} = - \frac{2}{9} (\ln (| x |) + C)$

Step 6.3.2.1.1.2.3

Factor $2$ out of $- 2$ .

$\frac{2 (- 1)}{9} \cdot \frac{- \ln (| 5 - V^{2} |) \cdot 9}{2} = - \frac{2}{9} (\ln (| x |) + C)$

Step 6.3.2.1.1.2.4

Cancel the common factor.

$\frac{2 \cdot - 1}{9} \cdot \frac{- \ln (| 5 - V^{2} |) \cdot 9}{2} = - \frac{2}{9} (\ln (| x |) + C)$

Step 6.3.2.1.1.2.5

Rewrite the expression.

$\frac{- 1}{9} (- \ln (| 5 - V^{2} |) \cdot 9) = - \frac{2}{9} (\ln (| x |) + C)$

Step 6.3.2.1.1.3

Cancel the common factor of $9$ .

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Step 6.3.2.1.1.3.1

Factor $9$ out of $- \ln (| 5 - V^{2} |) \cdot 9$ .

$\frac{- 1}{9} (9 \cdot (- \ln (| 5 - V^{2} |))) = - \frac{2}{9} (\ln (| x |) + C)$

Step 6.3.2.1.1.3.2

Cancel the common factor.

$\frac{- 1}{9} (9 \cdot (- \ln (| 5 - V^{2} |))) = - \frac{2}{9} (\ln (| x |) + C)$

Step 6.3.2.1.1.3.3

Rewrite the expression.

$- - \ln (| 5 - V^{2} |) = - \frac{2}{9} (\ln (| x |) + C)$

Step 6.3.2.1.1.4

Multiply.

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Step 6.3.2.1.1.4.1

Multiply $- 1$ by $- 1$ .

$1 \ln (| 5 - V^{2} |) = - \frac{2}{9} (\ln (| x |) + C)$

Step 6.3.2.1.1.4.2

Multiply $\ln (| 5 - V^{2} |)$ by $1$ .

$\ln (| 5 - V^{2} |) = - \frac{2}{9} (\ln (| x |) + C)$

Step 6.3.2.2

Simplify the right side.

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Step 6.3.2.2.1

Simplify $- \frac{2}{9} (\ln (| x |) + C)$ .

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Step 6.3.2.2.1.1

Simplify terms.

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Step 6.3.2.2.1.1.1

Apply the distributive property.

$\ln (| 5 - V^{2} |) = - \frac{2}{9} \ln (| x |) - \frac{2}{9} C$

Step 6.3.2.2.1.1.2

Combine $\ln (| x |)$ and $\frac{2}{9}$ .

$\ln (| 5 - V^{2} |) = - \frac{\ln (| x |) \cdot 2}{9} - \frac{2}{9} C$

Step 6.3.2.2.1.1.3

Combine $C$ and $\frac{2}{9}$ .

$\ln (| 5 - V^{2} |) = - \frac{\ln (| x |) \cdot 2}{9} - \frac{C \cdot 2}{9}$

Step 6.3.2.2.1.2

Simplify each term.

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Step 6.3.2.2.1.2.1

Move $2$ to the left of $\ln (| x |)$ .

$\ln (| 5 - V^{2} |) = - \frac{2 \cdot \ln (| x |)}{9} - \frac{C \cdot 2}{9}$

Step 6.3.2.2.1.2.2

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

$\ln (| 5 - V^{2} |) = - \frac{2 \ln (| x |)}{9} - \frac{2 C}{9}$

Step 6.3.3

Multiply each term in $\ln (| 5 - V^{2} |) = - \frac{2 \ln (| x |)}{9} - \frac{2 C}{9}$ by $9$ to eliminate the fractions.

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Step 6.3.3.1

Multiply each term in $\ln (| 5 - V^{2} |) = - \frac{2 \ln (| x |)}{9} - \frac{2 C}{9}$ by $9$ .

$\ln (| 5 - V^{2} |) \cdot 9 = - \frac{2 \ln (| x |)}{9} \cdot 9 - \frac{2 C}{9} \cdot 9$

Step 6.3.3.2

Simplify the left side.

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Step 6.3.3.2.1

Move $9$ to the left of $\ln (| 5 - V^{2} |)$ .

$9 \ln (| 5 - V^{2} |) = - \frac{2 \ln (| x |)}{9} \cdot 9 - \frac{2 C}{9} \cdot 9$

Step 6.3.3.3

Simplify the right side.

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Step 6.3.3.3.1

Simplify each term.

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Step 6.3.3.3.1.1

Cancel the common factor of $9$ .

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Step 6.3.3.3.1.1.1

Move the leading negative in $- \frac{2 \ln (| x |)}{9}$ into the numerator.

$9 \ln (| 5 - V^{2} |) = \frac{- 2 \ln (| x |)}{9} \cdot 9 - \frac{2 C}{9} \cdot 9$

Step 6.3.3.3.1.1.2

Cancel the common factor.

$9 \ln (| 5 - V^{2} |) = \frac{- 2 \ln (| x |)}{9} \cdot 9 - \frac{2 C}{9} \cdot 9$

Step 6.3.3.3.1.1.3

Rewrite the expression.

$9 \ln (| 5 - V^{2} |) = - 2 \ln (| x |) - \frac{2 C}{9} \cdot 9$

Step 6.3.3.3.1.2

Cancel the common factor of $9$ .

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Step 6.3.3.3.1.2.1

Move the leading negative in $- \frac{2 C}{9}$ into the numerator.

$9 \ln (| 5 - V^{2} |) = - 2 \ln (| x |) + \frac{- 2 C}{9} \cdot 9$

Step 6.3.3.3.1.2.2

Cancel the common factor.

$9 \ln (| 5 - V^{2} |) = - 2 \ln (| x |) + \frac{- 2 C}{9} \cdot 9$

Step 6.3.3.3.1.2.3

Rewrite the expression.

$9 \ln (| 5 - V^{2} |) = - 2 \ln (| x |) - 2 C$

Step 6.3.4

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

$9 \ln (| 5 - V^{2} |) + 2 \ln (| x |) = - 2 C$

Step 6.3.5

Simplify the left side.

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Step 6.3.5.1

Simplify $9 \ln (| 5 - V^{2} |) + 2 \ln (| x |)$ .

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Step 6.3.5.1.1

Simplify each term.

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Step 6.3.5.1.1.1

Simplify $9 \ln (| 5 - V^{2} |)$ by moving $9$ inside the logarithm.

$\ln ({| 5 - V^{2} |}^{9}) + 2 \ln (| x |) = - 2 C$

Step 6.3.5.1.1.2

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

$\ln ({| 5 - V^{2} |}^{9}) + \ln ({| x |}^{2}) = - 2 C$

Step 6.3.5.1.1.3

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

$\ln ({| 5 - V^{2} |}^{9}) + \ln (x^{2}) = - 2 C$

Step 6.3.5.1.2

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

$\ln ({| 5 - V^{2} |}^{9} x^{2}) = - 2 C$

Step 6.3.5.1.3

Reorder factors in $\ln ({| 5 - V^{2} |}^{9} x^{2})$ .

$\ln (x^{2} {| 5 - V^{2} |}^{9}) = - 2 C$

Step 6.3.6

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

$e^{\ln (x^{2} {| 5 - V^{2} |}^{9})} = e^{- 2 C}$

Step 6.3.7

Rewrite $\ln (x^{2} {| 5 - V^{2} |}^{9}) = - 2 C$ 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 $b^{y} = x$ .

$e^{- 2 C} = x^{2} {| 5 - V^{2} |}^{9}$

Step 6.3.8

Solve for $V$ .

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Step 6.3.8.1

Rewrite the equation as $x^{2} {| 5 - V^{2} |}^{9} = e^{- 2 C}$ .

$x^{2} {| 5 - V^{2} |}^{9} = e^{- 2 C}$

Step 6.3.8.2

Divide each term in $x^{2} {| 5 - V^{2} |}^{9} = e^{- 2 C}$ by $x^{2}$ and simplify.

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Step 6.3.8.2.1

Divide each term in $x^{2} {| 5 - V^{2} |}^{9} = e^{- 2 C}$ by $x^{2}$ .

$\frac{x^{2} {| 5 - V^{2} |}^{9}}{x^{2}} = \frac{e^{- 2 C}}{x^{2}}$

Step 6.3.8.2.2

Simplify the left side.

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Step 6.3.8.2.2.1

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

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Step 6.3.8.2.2.1.1

Cancel the common factor.

$\frac{x^{2} {| 5 - V^{2} |}^{9}}{x^{2}} = \frac{e^{- 2 C}}{x^{2}}$

Step 6.3.8.2.2.1.2

Divide ${| 5 - V^{2} |}^{9}$ by $1$ .

${| 5 - V^{2} |}^{9} = \frac{e^{- 2 C}}{x^{2}}$

Step 6.3.8.3

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

$| 5 - V^{2} | = \sqrt[9]{\frac{e^{- 2 C}}{x^{2}}}$

Step 6.3.8.4

Simplify $\sqrt[9]{\frac{e^{- 2 C}}{x^{2}}}$ .

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Step 6.3.8.4.1

Rewrite $\sqrt[9]{\frac{e^{- 2 C}}{x^{2}}}$ as $\frac{\sqrt[9]{e^{- 2 C}}}{\sqrt[9]{x^{2}}}$ .

$| 5 - V^{2} | = \frac{\sqrt[9]{e^{- 2 C}}}{\sqrt[9]{x^{2}}}$

Step 6.3.8.4.2

Multiply $\frac{\sqrt[9]{e^{- 2 C}}}{\sqrt[9]{x^{2}}}$ by $\frac{{\sqrt[9]{x^{2}}}^{8}}{{\sqrt[9]{x^{2}}}^{8}}$ .

$| 5 - V^{2} | = \frac{\sqrt[9]{e^{- 2 C}}}{\sqrt[9]{x^{2}}} \cdot \frac{{\sqrt[9]{x^{2}}}^{8}}{{\sqrt[9]{x^{2}}}^{8}}$

Step 6.3.8.4.3

Combine and simplify the denominator.

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Step 6.3.8.4.3.1

Multiply $\frac{\sqrt[9]{e^{- 2 C}}}{\sqrt[9]{x^{2}}}$ by $\frac{{\sqrt[9]{x^{2}}}^{8}}{{\sqrt[9]{x^{2}}}^{8}}$ .

$| 5 - V^{2} | = \frac{\sqrt[9]{e^{- 2 C}} {\sqrt[9]{x^{2}}}^{8}}{\sqrt[9]{x^{2}} {\sqrt[9]{x^{2}}}^{8}}$

Step 6.3.8.4.3.2

Raise $\sqrt[9]{x^{2}}$ to the power of $1$ .

$| 5 - V^{2} | = \frac{\sqrt[9]{e^{- 2 C}} {\sqrt[9]{x^{2}}}^{8}}{{\sqrt[9]{x^{2}}}^{1} {\sqrt[9]{x^{2}}}^{8}}$

Step 6.3.8.4.3.3

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

$| 5 - V^{2} | = \frac{\sqrt[9]{e^{- 2 C}} {\sqrt[9]{x^{2}}}^{8}}{{\sqrt[9]{x^{2}}}^{1 + 8}}$

Step 6.3.8.4.3.4

Add $1$ and $8$ .

$| 5 - V^{2} | = \frac{\sqrt[9]{e^{- 2 C}} {\sqrt[9]{x^{2}}}^{8}}{{\sqrt[9]{x^{2}}}^{9}}$

Step 6.3.8.4.3.5

Rewrite ${\sqrt[9]{x^{2}}}^{9}$ as $x^{2}$ .

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Step 6.3.8.4.3.5.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt[9]{x^{2}}$ as $x^{\frac{2}{9}}$ .

$| 5 - V^{2} | = \frac{\sqrt[9]{e^{- 2 C}} {\sqrt[9]{x^{2}}}^{8}}{{(x^{\frac{2}{9}})}^{9}}$

Step 6.3.8.4.3.5.2

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

$| 5 - V^{2} | = \frac{\sqrt[9]{e^{- 2 C}} {\sqrt[9]{x^{2}}}^{8}}{x^{\frac{2}{9} \cdot 9}}$

Step 6.3.8.4.3.5.3

Combine $\frac{2}{9}$ and $9$ .

$| 5 - V^{2} | = \frac{\sqrt[9]{e^{- 2 C}} {\sqrt[9]{x^{2}}}^{8}}{x^{\frac{2 \cdot 9}{9}}}$

Step 6.3.8.4.3.5.4

Multiply $2$ by $9$ .

$| 5 - V^{2} | = \frac{\sqrt[9]{e^{- 2 C}} {\sqrt[9]{x^{2}}}^{8}}{x^{\frac{18}{9}}}$

Step 6.3.8.4.3.5.5

Cancel the common factor of $18$ and $9$ .

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Step 6.3.8.4.3.5.5.1

Factor $9$ out of $18$ .

$| 5 - V^{2} | = \frac{\sqrt[9]{e^{- 2 C}} {\sqrt[9]{x^{2}}}^{8}}{x^{\frac{9 \cdot 2}{9}}}$

Step 6.3.8.4.3.5.5.2

Cancel the common factors.

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Step 6.3.8.4.3.5.5.2.1

Factor $9$ out of $9$ .

$| 5 - V^{2} | = \frac{\sqrt[9]{e^{- 2 C}} {\sqrt[9]{x^{2}}}^{8}}{x^{\frac{9 \cdot 2}{9 (1)}}}$

Step 6.3.8.4.3.5.5.2.2

Cancel the common factor.

$| 5 - V^{2} | = \frac{\sqrt[9]{e^{- 2 C}} {\sqrt[9]{x^{2}}}^{8}}{x^{\frac{9 \cdot 2}{9 \cdot 1}}}$

Step 6.3.8.4.3.5.5.2.3

Rewrite the expression.

$| 5 - V^{2} | = \frac{\sqrt[9]{e^{- 2 C}} {\sqrt[9]{x^{2}}}^{8}}{x^{\frac{2}{1}}}$

Step 6.3.8.4.3.5.5.2.4

Divide $2$ by $1$ .

$| 5 - V^{2} | = \frac{\sqrt[9]{e^{- 2 C}} {\sqrt[9]{x^{2}}}^{8}}{x^{2}}$

Step 6.3.8.4.4

Simplify the numerator.

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Step 6.3.8.4.4.1

Rewrite ${\sqrt[9]{x^{2}}}^{8}$ as $\sqrt[9]{{(x^{2})}^{8}}$ .

$| 5 - V^{2} | = \frac{\sqrt[9]{e^{- 2 C}} \sqrt[9]{{(x^{2})}^{8}}}{x^{2}}$

Step 6.3.8.4.4.2

Multiply the exponents in ${(x^{2})}^{8}$ .

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Step 6.3.8.4.4.2.1

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

$| 5 - V^{2} | = \frac{\sqrt[9]{e^{- 2 C}} \sqrt[9]{x^{2 \cdot 8}}}{x^{2}}$

Step 6.3.8.4.4.2.2

Multiply $2$ by $8$ .

$| 5 - V^{2} | = \frac{\sqrt[9]{e^{- 2 C}} \sqrt[9]{x^{16}}}{x^{2}}$

Step 6.3.8.4.4.3

Factor out $x^{9}$ .

$| 5 - V^{2} | = \frac{\sqrt[9]{e^{- 2 C}} \sqrt[9]{x^{9} x^{7}}}{x^{2}}$

Step 6.3.8.4.4.4

Pull terms out from under the radical.

$| 5 - V^{2} | = \frac{\sqrt[9]{e^{- 2 C}} x \sqrt[9]{x^{7}}}{x^{2}}$

Step 6.3.8.4.4.5

Combine using the product rule for radicals.

$| 5 - V^{2} | = \frac{x \sqrt[9]{e^{- 2 C} x^{7}}}{x^{2}}$

Step 6.3.8.4.5

Reduce the expression by cancelling the common factors.

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Step 6.3.8.4.5.1

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

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Step 6.3.8.4.5.1.1

Factor $x$ out of $x \sqrt[9]{e^{- 2 C} x^{7}}$ .

$| 5 - V^{2} | = \frac{x \sqrt[9]{e^{- 2 C} x^{7}}}{x^{2}}$

Step 6.3.8.4.5.1.2

Cancel the common factors.

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Step 6.3.8.4.5.1.2.1

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

$| 5 - V^{2} | = \frac{x \sqrt[9]{e^{- 2 C} x^{7}}}{x \cdot x}$

Step 6.3.8.4.5.1.2.2

Cancel the common factor.

$| 5 - V^{2} | = \frac{x \sqrt[9]{e^{- 2 C} x^{7}}}{x \cdot x}$

Step 6.3.8.4.5.1.2.3

Rewrite the expression.

$| 5 - V^{2} | = \frac{\sqrt[9]{e^{- 2 C} x^{7}}}{x}$

Step 6.3.8.4.5.2

Reorder factors in $\frac{\sqrt[9]{e^{- 2 C} x^{7}}}{x}$ .

$| 5 - V^{2} | = \frac{\sqrt[9]{x^{7} e^{- 2 C}}}{x}$

Step 6.3.8.5

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

$5 - V^{2} = \pm \frac{\sqrt[9]{x^{7} e^{- 2 C}}}{x}$

Step 6.3.8.6

Subtract $5$ from both sides of the equation.

$- V^{2} = \pm \frac{\sqrt[9]{x^{7} e^{- 2 C}}}{x} - 5$

Step 6.3.8.7

Divide each term in $- V^{2} = \pm \frac{\sqrt[9]{x^{7} e^{- 2 C}}}{x} - 5$ by $- 1$ and simplify.

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Step 6.3.8.7.1

Divide each term in $- V^{2} = \pm \frac{\sqrt[9]{x^{7} e^{- 2 C}}}{x} - 5$ by $- 1$ .

$\frac{- V^{2}}{- 1} = \frac{\pm \frac{\sqrt[9]{x^{7} e^{- 2 C}}}{x}}{- 1} + \frac{- 5}{- 1}$

Step 6.3.8.7.2

Simplify the left side.

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Step 6.3.8.7.2.1

Dividing two negative values results in a positive value.

$\frac{V^{2}}{1} = \frac{\pm \frac{\sqrt[9]{x^{7} e^{- 2 C}}}{x}}{- 1} + \frac{- 5}{- 1}$

Step 6.3.8.7.2.2

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

$V^{2} = \frac{\pm \frac{\sqrt[9]{x^{7} e^{- 2 C}}}{x}}{- 1} + \frac{- 5}{- 1}$

Step 6.3.8.7.3

Simplify the right side.

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Step 6.3.8.7.3.1

Simplify each term.

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Step 6.3.8.7.3.1.1

Move the negative one from the denominator of $\frac{\pm \frac{\sqrt[9]{x^{7} e^{- 2 C}}}{x}}{- 1}$ .

$V^{2} = - 1 \cdot (\pm \frac{\sqrt[9]{x^{7} e^{- 2 C}}}{x}) + \frac{- 5}{- 1}$

Step 6.3.8.7.3.1.2

Rewrite $- 1 \cdot (\pm \frac{\sqrt[9]{x^{7} e^{- 2 C}}}{x})$ as $- (\pm \frac{\sqrt[9]{x^{7} e^{- 2 C}}}{x})$ .

$V^{2} = - (\pm \frac{\sqrt[9]{x^{7} e^{- 2 C}}}{x}) + \frac{- 5}{- 1}$

Step 6.3.8.7.3.1.3

Divide $- 5$ by $- 1$ .

$V^{2} = - (\pm \frac{\sqrt[9]{x^{7} e^{- 2 C}}}{x}) + 5$

Step 6.3.8.8

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

$V = \pm \sqrt{- (\pm \frac{\sqrt[9]{x^{7} e^{- 2 C}}}{x}) + 5}$

Step 6.4

Simplify the constant of integration.

$V = \pm \sqrt{- (\pm \frac{\sqrt[9]{x^{7} C}}{x}) + 5}$

Step 7

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

$\frac{y}{x} = \pm \sqrt{- (\pm \frac{\sqrt[9]{x^{7} C}}{x}) + 5}$

Step 8

Solve $\frac{y}{x} = \pm \sqrt{- (\pm \frac{\sqrt[9]{x^{7} C}}{x}) + 5}$ for $y$ .

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Step 8.1

Multiply both sides by $x$ .

$\frac{y}{x} x = (\pm \sqrt{- (\pm \frac{\sqrt[9]{x^{7} C}}{x}) + 5}) x$

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.

$\frac{y}{x} x = (\pm \sqrt{- (\pm \frac{\sqrt[9]{x^{7} C}}{x}) + 5}) x$

Step 8.2.1.2

Rewrite the expression.

$y = (\pm \sqrt{- (\pm \frac{\sqrt[9]{x^{7} C}}{x}) + 5}) x$