Calculus Examples

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

Step 1

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

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Step 1.1

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

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Step 1.1.1

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

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

Step 1.1.2

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

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Step 1.1.2.1

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

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

Step 1.1.2.2

Cancel the common factors.

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Step 1.1.2.2.1

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

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

Step 1.1.2.2.2

Cancel the common factor.

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

Step 1.1.2.2.3

Rewrite the expression.

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

Step 1.2

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

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Step 1.2.1

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

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

Step 1.2.2

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

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

Step 1.3

Rewrite $\frac{y^{2}}{x^{2}}$ as ${(\frac{y}{x})}^{2}$ .

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

Step 2

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

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

Step 3

Solve $V = \frac{y}{x}$ for $y$ .

$y = V x$

Step 4

Use the product rule to find the derivative of $y = V x$ with respect to $x$ .

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

Step 5

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

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

Step 6

Solve the substituted differential equation.

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Step 6.1

Separate the variables.

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Step 6.1.1

Solve for $\frac{d V}{d x}$ .

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Step 6.1.1.1

Move all terms not containing $\frac{d V}{d x}$ to the right side of the equation.

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Step 6.1.1.1.1

Subtract $V$ from both sides of the equation.

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

Step 6.1.1.1.2

Subtract $V$ from $2 V$ .

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

Step 6.1.1.2

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

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Step 6.1.1.2.1

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

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

Step 6.1.1.2.2

Simplify the left side.

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Step 6.1.1.2.2.1

Cancel the common factor of $x$ .

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Step 6.1.1.2.2.1.1

Cancel the common factor.

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

Step 6.1.1.2.2.1.2

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

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

Step 6.1.2

Factor.

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Step 6.1.2.1

Combine the numerators over the common denominator.

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

Step 6.1.2.2

Factor $V$ out of $V^{2} + V$ .

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Step 6.1.2.2.1

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

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

Step 6.1.2.2.2

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

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

Step 6.1.2.2.3

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

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

Step 6.1.2.2.4

Factor $V$ out of $V \cdot V + V \cdot 1$ .

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

Step 6.1.3

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

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

Step 6.1.4

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

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Step 6.1.4.1

Cancel the common factor.

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

Step 6.1.4.2

Rewrite the expression.

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

Step 6.1.5

Rewrite the equation.

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

Step 6.2

Integrate both sides.

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Step 6.2.1

Set up an integral on each side.

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

Step 6.2.2

Integrate the left side.

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Step 6.2.2.1

Write the fraction using partial fraction decomposition.

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Step 6.2.2.1.1

Decompose the fraction and multiply through by the common denominator.

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Step 6.2.2.1.1.1

For each factor in the denominator, create a new fraction using the factor as the denominator, and an unknown value as the numerator. Since the factor in the denominator is linear, put a single variable in its place $B$ .

$\frac{A}{V} + \frac{B}{V + 1}$

Step 6.2.2.1.1.2

Multiply each fraction in the equation by the denominator of the original expression. In this case, the denominator is $V (V + 1)$ .

$\frac{1 (V (V + 1))}{V (V + 1)} = \frac{A (V (V + 1))}{V} + \frac{(B) (V (V + 1))}{V + 1}$

Step 6.2.2.1.1.3

Cancel the common factor of $V$ .

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Step 6.2.2.1.1.3.1

Cancel the common factor.

$\frac{1 (V (V + 1))}{V (V + 1)} = \frac{A (V (V + 1))}{V} + \frac{(B) (V (V + 1))}{V + 1}$

Step 6.2.2.1.1.3.2

Rewrite the expression.

$\frac{1 (V + 1)}{V + 1} = \frac{A (V (V + 1))}{V} + \frac{(B) (V (V + 1))}{V + 1}$

Step 6.2.2.1.1.4

Cancel the common factor of $V + 1$ .

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Step 6.2.2.1.1.4.1

Cancel the common factor.

$\frac{1 (V + 1)}{V + 1} = \frac{A (V (V + 1))}{V} + \frac{(B) (V (V + 1))}{V + 1}$

Step 6.2.2.1.1.4.2

Rewrite the expression.

$1 = \frac{A (V (V + 1))}{V} + \frac{(B) (V (V + 1))}{V + 1}$

Step 6.2.2.1.1.5

Simplify each term.

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Step 6.2.2.1.1.5.1

Cancel the common factor of $V$ .

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Step 6.2.2.1.1.5.1.1

Cancel the common factor.

$1 = \frac{A (V (V + 1))}{V} + \frac{(B) (V (V + 1))}{V + 1}$

Step 6.2.2.1.1.5.1.2

Divide $A (V + 1)$ by $1$ .

$1 = A (V + 1) + \frac{(B) (V (V + 1))}{V + 1}$

Step 6.2.2.1.1.5.2

Apply the distributive property.

$1 = A V + A \cdot 1 + \frac{(B) (V (V + 1))}{V + 1}$

Step 6.2.2.1.1.5.3

Multiply $A$ by $1$ .

$1 = A V + A + \frac{(B) (V (V + 1))}{V + 1}$

Step 6.2.2.1.1.5.4

Cancel the common factor of $V + 1$ .

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Step 6.2.2.1.1.5.4.1

Cancel the common factor.

$1 = A V + A + \frac{B (V (V + 1))}{V + 1}$

Step 6.2.2.1.1.5.4.2

Divide $(B) (V)$ by $1$ .

$1 = A V + A + (B) (V)$

$1 = A V + A + B V$

Step 6.2.2.1.1.6

Move $A$ .

$1 = A V + B V + A$

Step 6.2.2.1.2

Create equations for the partial fraction variables and use them to set up a system of equations.

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Step 6.2.2.1.2.1

Create an equation for the partial fraction variables by equating the coefficients of $V$ from each side of the equation. For the equation to be equal, the equivalent coefficients on each side of the equation must be equal.

$0 = A + B$

Step 6.2.2.1.2.2

Create an equation for the partial fraction variables by equating the coefficients of the terms not containing $V$ . For the equation to be equal, the equivalent coefficients on each side of the equation must be equal.

$1 = A$

Step 6.2.2.1.2.3

Set up the system of equations to find the coefficients of the partial fractions.

$0 = A + B$

$1 = A$

$0 = A + B$

$1 = A$

Step 6.2.2.1.3

Solve the system of equations.

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Step 6.2.2.1.3.1

Rewrite the equation as $A = 1$ .

$A = 1$

$0 = A + B$

Step 6.2.2.1.3.2

Replace all occurrences of $A$ with $1$ in each equation.

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Step 6.2.2.1.3.2.1

Replace all occurrences of $A$ in $0 = A + B$ with $1$ .

$0 = (1) + B$

$A = 1$

Step 6.2.2.1.3.2.2

Simplify the right side.

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Step 6.2.2.1.3.2.2.1

Remove parentheses.

$0 = 1 + B$

$A = 1$

$0 = 1 + B$

$A = 1$

$0 = 1 + B$

$A = 1$

Step 6.2.2.1.3.3

Solve for $B$ in $0 = 1 + B$ .

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Step 6.2.2.1.3.3.1

Rewrite the equation as $1 + B = 0$ .

$1 + B = 0$

$A = 1$

Step 6.2.2.1.3.3.2

Subtract $1$ from both sides of the equation.

$B = - 1$

$A = 1$

$B = - 1$

$A = 1$

Step 6.2.2.1.3.4

Solve the system of equations.

$B = - 1 A = 1$

Step 6.2.2.1.3.5

List all of the solutions.

$B = - 1, A = 1$

Step 6.2.2.1.4

Replace each of the partial fraction coefficients in $\frac{A}{V} + \frac{B}{V + 1}$ with the values found for $A$ and $B$ .

$\frac{1}{V} + \frac{- 1}{V + 1}$

Step 6.2.2.1.5

Move the negative in front of the fraction.

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

Step 6.2.2.2

Split the single integral into multiple integrals.

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

Step 6.2.2.3

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

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

Step 6.2.2.4

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

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

Step 6.2.2.5

Let $u = V + 1$ . Then $d u = d V$ . Rewrite using $u$ and $d$ $u$ .

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Step 6.2.2.5.1

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

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Step 6.2.2.5.1.1

Differentiate $V + 1$ .

$\frac{d}{d V} [V + 1]$

Step 6.2.2.5.1.2

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

$\frac{d}{d V} [V] + \frac{d}{d V} [1]$

Step 6.2.2.5.1.3

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

$1 + \frac{d}{d V} [1]$

Step 6.2.2.5.1.4

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

$1 + 0$

Step 6.2.2.5.1.5

Add $1$ and $0$ .

$1$

Step 6.2.2.5.2

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

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

Step 6.2.2.6

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

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

Step 6.2.2.7

Simplify.

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

Step 6.2.2.8

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

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

Step 6.2.2.9

Replace all occurrences of $u$ with $V + 1$ .

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

Step 6.2.3

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

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

Step 6.2.4

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

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

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.

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

Step 6.3.2

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

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

Step 6.3.3

Multiply the numerator by the reciprocal of the denominator.

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

Step 6.3.4

Multiply $\frac{| V |}{| V + 1 |} \cdot \frac{1}{| x |}$ .

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Step 6.3.4.1

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

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

Step 6.3.4.2

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

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

Step 6.3.5

Simplify the denominator.

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Step 6.3.5.1

Apply the distributive property.

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

Step 6.3.5.2

Multiply $x$ by $1$ .

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

Step 6.3.5.3

Factor $x$ out of $V x + x$ .

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Step 6.3.5.3.1

Factor $x$ out of $V x$ .

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

Step 6.3.5.3.2

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

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

Step 6.3.5.3.3

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

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

Step 6.3.5.3.4

Factor $x$ out of $x V + x \cdot 1$ .

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

Step 6.3.6

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

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

Step 6.3.7

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

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

Step 6.3.8

Solve for $V$ .

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Step 6.3.8.1

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

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

Step 6.3.8.2

Multiply both sides by $| x (V + 1) |$ .

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

Step 6.3.8.3

Simplify.

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Step 6.3.8.3.1

Simplify the left side.

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Step 6.3.8.3.1.1

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

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Step 6.3.8.3.1.1.1

Cancel the common factor.

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

Step 6.3.8.3.1.1.2

Rewrite the expression.

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

Step 6.3.8.3.2

Simplify the right side.

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Step 6.3.8.3.2.1

Simplify $e^{C} | x (V + 1) |$ .

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Step 6.3.8.3.2.1.1

Apply the distributive property.

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

Step 6.3.8.3.2.1.2

Multiply $x$ by $1$ .

$| V | = e^{C} | x V + x |$

Step 6.3.8.4

Solve for $V$ .

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Step 6.3.8.4.1

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

$e^{C} | x V + x | = | V |$

Step 6.3.8.4.2

Divide each term in $e^{C} | x V + x | = | V |$ by $e^{C}$ and simplify.

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Step 6.3.8.4.2.1

Divide each term in $e^{C} | x V + x | = | V |$ by $e^{C}$ .

$\frac{e^{C} | x V + x |}{e^{C}} = \frac{| V |}{e^{C}}$

Step 6.3.8.4.2.2

Simplify the left side.

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Step 6.3.8.4.2.2.1

Cancel the common factor of $e^{C}$ .

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Step 6.3.8.4.2.2.1.1

Cancel the common factor.

$\frac{e^{C} | x V + x |}{e^{C}} = \frac{| V |}{e^{C}}$

Step 6.3.8.4.2.2.1.2

Divide $| x V + x |$ by $1$ .

$| x V + x | = \frac{| V |}{e^{C}}$

Step 6.3.8.4.3

Rewrite the absolute value equation as four equations without absolute value bars.

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

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

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

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

Step 6.3.8.4.4

After simplifying, there are only two unique equations to be solved.

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

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

Step 6.3.8.4.5

Solve $x V + x = \frac{V}{e^{C}}$ for $V$ .

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Step 6.3.8.4.5.1

Multiply both sides by $e^{C}$ .

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

Step 6.3.8.4.5.2

Simplify.

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Step 6.3.8.4.5.2.1

Simplify the left side.

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Step 6.3.8.4.5.2.1.1

Apply the distributive property.

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

Step 6.3.8.4.5.2.2

Simplify the right side.

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Step 6.3.8.4.5.2.2.1

Cancel the common factor of $e^{C}$ .

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Step 6.3.8.4.5.2.2.1.1

Cancel the common factor.

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

Step 6.3.8.4.5.2.2.1.2

Rewrite the expression.

$x V e^{C} + x e^{C} = V$

Step 6.3.8.4.5.3

Solve for $V$ .

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Step 6.3.8.4.5.3.1

Subtract $V$ from both sides of the equation.

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

Step 6.3.8.4.5.3.2

Subtract $x e^{C}$ from both sides of the equation.

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

Step 6.3.8.4.5.3.3

Factor $V$ out of $x V e^{C} - V$ .

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Step 6.3.8.4.5.3.3.1

Factor $V$ out of $x V e^{C}$ .

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

Step 6.3.8.4.5.3.3.2

Factor $V$ out of $- V$ .

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

Step 6.3.8.4.5.3.3.3

Factor $V$ out of $V (x e^{C}) + V \cdot - 1$ .

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

Step 6.3.8.4.5.3.4

Divide each term in $V (x e^{C} - 1) = - x e^{C}$ by $x e^{C} - 1$ and simplify.

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Step 6.3.8.4.5.3.4.1

Divide each term in $V (x e^{C} - 1) = - x e^{C}$ by $x e^{C} - 1$ .

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

Step 6.3.8.4.5.3.4.2

Simplify the left side.

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Step 6.3.8.4.5.3.4.2.1

Cancel the common factor of $x e^{C} - 1$ .

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Step 6.3.8.4.5.3.4.2.1.1

Cancel the common factor.

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

Step 6.3.8.4.5.3.4.2.1.2

Divide $V$ by $1$ .

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

Step 6.3.8.4.5.3.4.3

Simplify the right side.

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Step 6.3.8.4.5.3.4.3.1

Move the negative in front of the fraction.

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

Step 6.3.8.4.6

Solve $x V + x = - \frac{V}{e^{C}}$ for $V$ .

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Step 6.3.8.4.6.1

Multiply both sides by $e^{C}$ .

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

Step 6.3.8.4.6.2

Simplify.

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Step 6.3.8.4.6.2.1

Simplify the left side.

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Step 6.3.8.4.6.2.1.1

Apply the distributive property.

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

Step 6.3.8.4.6.2.2

Simplify the right side.

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Step 6.3.8.4.6.2.2.1

Cancel the common factor of $e^{C}$ .

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Step 6.3.8.4.6.2.2.1.1

Move the leading negative in $- \frac{V}{e^{C}}$ into the numerator.

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

Step 6.3.8.4.6.2.2.1.2

Cancel the common factor.

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

Step 6.3.8.4.6.2.2.1.3

Rewrite the expression.

$x V e^{C} + x e^{C} = - V$

Step 6.3.8.4.6.3

Solve for $V$ .

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Step 6.3.8.4.6.3.1

Add $V$ to both sides of the equation.

$x V e^{C} + x e^{C} + V = 0$

Step 6.3.8.4.6.3.2

Subtract $x e^{C}$ from both sides of the equation.

$x V e^{C} + V = - x e^{C}$

Step 6.3.8.4.6.3.3

Factor $V$ out of $x V e^{C} + V$ .

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Step 6.3.8.4.6.3.3.1

Factor $V$ out of $x V e^{C}$ .

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

Step 6.3.8.4.6.3.3.2

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

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

Step 6.3.8.4.6.3.3.3

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

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

Step 6.3.8.4.6.3.3.4

Factor $V$ out of $V (x e^{C}) + V \cdot 1$ .

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

Step 6.3.8.4.6.3.4

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

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Step 6.3.8.4.6.3.4.1

Divide each term in $V (x e^{C} + 1) = - x e^{C}$ by $x e^{C} + 1$ .

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

Step 6.3.8.4.6.3.4.2

Simplify the left side.

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Step 6.3.8.4.6.3.4.2.1

Cancel the common factor of $x e^{C} + 1$ .

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Step 6.3.8.4.6.3.4.2.1.1

Cancel the common factor.

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

Step 6.3.8.4.6.3.4.2.1.2

Divide $V$ by $1$ .

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

Step 6.3.8.4.6.3.4.3

Simplify the right side.

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Step 6.3.8.4.6.3.4.3.1

Move the negative in front of the fraction.