Calculus Examples

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

Step 1

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

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Step 1.1

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

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Step 1.1.1

Subtract $\frac{y^{2}}{x^{2}}$ from both sides of the equation.

$\frac{d y}{d x} + \frac{y}{x} - \frac{y^{2}}{x^{2}} = 0$

Step 1.1.2

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

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Step 1.1.2.1

Subtract $\frac{y}{x}$ from both sides of the equation.

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

Step 1.1.2.2

Add $\frac{y^{2}}{x^{2}}$ to both sides of the equation.

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

Step 1.2

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

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

Step 2

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

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

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

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

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.

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

Step 6.1.1.1.2

Subtract $V$ from $- V$ .

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

Step 6.1.1.2

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

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Step 6.1.1.2.1

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

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

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.

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

Step 6.1.1.2.2.1.2

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

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

Step 6.1.1.2.3

Simplify the right side.

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Step 6.1.1.2.3.1

Move the negative in front of the fraction.

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

Step 6.1.2

Factor.

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Step 6.1.2.1

Combine the numerators over the common denominator.

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

Step 6.1.2.2

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

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Step 6.1.2.2.1

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

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

Step 6.1.2.2.2

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

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

Step 6.1.2.2.3

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

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

Step 6.1.3

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

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

Step 6.1.4

Cancel the common factor of $V (V - 2)$ .

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Step 6.1.4.1

Cancel the common factor.

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

Step 6.1.4.2

Rewrite the expression.

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

Step 6.1.5

Rewrite the equation.

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

Step 6.2

Integrate both sides.

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Step 6.2.1

Set up an integral on each side.

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

Step 6.2.2

Integrate the left side.

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Step 6.2.2.1

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 $B$ .

$\frac{A}{V} + \frac{B}{V - 2}$

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 $V (V - 2)$ .

$\frac{1 (V (V - 2))}{V (V - 2)} = \frac{A (V (V - 2))}{V} + \frac{(B) (V (V - 2))}{V - 2}$

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.

$\frac{1 (V (V - 2))}{V (V - 2)} = \frac{A (V (V - 2))}{V} + \frac{(B) (V (V - 2))}{V - 2}$

Step 6.2.2.1.1.3.2

Rewrite the expression.

$\frac{1 (V - 2)}{V - 2} = \frac{A (V (V - 2))}{V} + \frac{(B) (V (V - 2))}{V - 2}$

Step 6.2.2.1.1.4

Cancel the common factor of $V - 2$ .

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Step 6.2.2.1.1.4.1

Cancel the common factor.

$\frac{1 (V - 2)}{V - 2} = \frac{A (V (V - 2))}{V} + \frac{(B) (V (V - 2))}{V - 2}$

Step 6.2.2.1.1.4.2

Rewrite the expression.

$1 = \frac{A (V (V - 2))}{V} + \frac{(B) (V (V - 2))}{V - 2}$

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.

$1 = \frac{A (V (V - 2))}{V} + \frac{(B) (V (V - 2))}{V - 2}$

Step 6.2.2.1.1.5.1.2

Divide $A (V - 2)$ by $1$ .

$1 = A (V - 2) + \frac{(B) (V (V - 2))}{V - 2}$

Step 6.2.2.1.1.5.2

Apply the distributive property.

$1 = A V + A \cdot - 2 + \frac{(B) (V (V - 2))}{V - 2}$

Step 6.2.2.1.1.5.3

Move $- 2$ to the left of $A$ .

$1 = A V - 2 \cdot A + \frac{(B) (V (V - 2))}{V - 2}$

Step 6.2.2.1.1.5.4

Cancel the common factor of $V - 2$ .

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Step 6.2.2.1.1.5.4.1

Cancel the common factor.

$1 = A V - 2 A + \frac{B (V (V - 2))}{V - 2}$

Step 6.2.2.1.1.5.4.2

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

$1 = A V - 2 A + (B) (V)$

$1 = A V - 2 A + B V$

Step 6.2.2.1.1.6

Move $- 2 A$ .

$1 = A V + B V - 2 A$

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.

$1 = - 2 A$

Step 6.2.2.1.2.3

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

$0 = A + B$

$1 = - 2 A$

$0 = A + B$

$1 = - 2 A$

Step 6.2.2.1.3

Solve the system of equations.

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Step 6.2.2.1.3.1

Solve for $A$ in $1 = - 2 A$ .

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Step 6.2.2.1.3.1.1

Rewrite the equation as $- 2 A = 1$ .

$- 2 A = 1$

$0 = A + B$

Step 6.2.2.1.3.1.2

Divide each term in $- 2 A = 1$ by $- 2$ and simplify.

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Step 6.2.2.1.3.1.2.1

Divide each term in $- 2 A = 1$ by $- 2$ .

$\frac{- 2 A}{- 2} = \frac{1}{- 2}$

$0 = A + B$

Step 6.2.2.1.3.1.2.2

Simplify the left side.

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Step 6.2.2.1.3.1.2.2.1

Cancel the common factor of $- 2$ .

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Step 6.2.2.1.3.1.2.2.1.1

Cancel the common factor.

$\frac{- 2 A}{- 2} = \frac{1}{- 2}$

$0 = A + B$

Step 6.2.2.1.3.1.2.2.1.2

Divide $A$ by $1$ .

$A = \frac{1}{- 2}$

$0 = A + B$

$A = \frac{1}{- 2}$

$0 = A + B$

$A = \frac{1}{- 2}$

$0 = A + B$

Step 6.2.2.1.3.1.2.3

Simplify the right side.

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Step 6.2.2.1.3.1.2.3.1

Move the negative in front of the fraction.

$A = - \frac{1}{2}$

$0 = A + B$

$A = - \frac{1}{2}$

$0 = A + B$

$A = - \frac{1}{2}$

$0 = A + B$

$A = - \frac{1}{2}$

$0 = A + B$

Step 6.2.2.1.3.2

Replace all occurrences of $A$ with $- \frac{1}{2}$ 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 $- \frac{1}{2}$ .

$0 = (- \frac{1}{2}) + B$

$A = - \frac{1}{2}$

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.

$0 = - \frac{1}{2} + B$

$A = - \frac{1}{2}$

$0 = - \frac{1}{2} + B$

$A = - \frac{1}{2}$

$0 = - \frac{1}{2} + B$

$A = - \frac{1}{2}$

Step 6.2.2.1.3.3

Solve for $B$ in $0 = - \frac{1}{2} + B$ .

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Step 6.2.2.1.3.3.1

Rewrite the equation as $- \frac{1}{2} + B = 0$ .

$- \frac{1}{2} + B = 0$

$A = - \frac{1}{2}$

Step 6.2.2.1.3.3.2

Add $\frac{1}{2}$ to both sides of the equation.

$B = \frac{1}{2}$

$A = - \frac{1}{2}$

$B = \frac{1}{2}$

$A = - \frac{1}{2}$

Step 6.2.2.1.3.4

Solve the system of equations.

$B = \frac{1}{2} A = - \frac{1}{2}$

Step 6.2.2.1.3.5

List all of the solutions.

$B = \frac{1}{2}, A = - \frac{1}{2}$

Step 6.2.2.1.4

Replace each of the partial fraction coefficients in $\frac{A}{V} + \frac{B}{V - 2}$ with the values found for $A$ and $B$ .

$- \frac{\frac{1}{2}}{V} + \frac{\frac{1}{2}}{V - 2}$

Step 6.2.2.1.5

Simplify.

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Step 6.2.2.1.5.1

Multiply the numerator by the reciprocal of the denominator.

$- \frac{1}{2} \cdot \frac{1}{V} + \frac{\frac{1}{2}}{V - 2}$

Step 6.2.2.1.5.2

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

$- \frac{1}{V \cdot 2} + \frac{\frac{1}{2}}{V - 2}$

Step 6.2.2.1.5.3

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

$- \frac{1}{2 V} + \frac{\frac{1}{2}}{V - 2}$

Step 6.2.2.1.5.4

Multiply the numerator by the reciprocal of the denominator.

$- \frac{1}{2 V} + \frac{1}{2} \cdot \frac{1}{V - 2}$

Step 6.2.2.1.5.5

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

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

Step 6.2.2.2

Split the single integral into multiple integrals.

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

Step 6.2.2.3

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

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

Step 6.2.2.4

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

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

Step 6.2.2.5

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

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

Step 6.2.2.6

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

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

Step 6.2.2.7

Let $u = V - 2$ . Then $d u = d V$ . Rewrite using $u$ and $d$ $u$ .

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Step 6.2.2.7.1

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

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Step 6.2.2.7.1.1

Differentiate $V - 2$ .

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

Step 6.2.2.7.1.2

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

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

Step 6.2.2.7.1.3

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

$1 + \frac{d}{d V} [- 2]$

Step 6.2.2.7.1.4

Since $- 2$ is constant with respect to $V$ , the derivative of $- 2$ with respect to $V$ is $0$ .

$1 + 0$

Step 6.2.2.7.1.5

Add $1$ and $0$ .

$1$

Step 6.2.2.7.2

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

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

Step 6.2.2.8

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

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

Step 6.2.2.9

Simplify.

$- \frac{1}{2} \ln (| V |) + \frac{1}{2} \ln (| u |) + C_{3} = \int \frac{1}{x} d x$

Step 6.2.2.10

Replace all occurrences of $u$ with $V - 2$ .

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

Step 6.2.3

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

$- \frac{1}{2} \ln (| V |) + \frac{1}{2} \ln (| V - 2 |) + C_{3} = \ln (| x |) + C_{4}$

Step 6.2.4

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

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

Step 6.3

Solve for $V$ .

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Step 6.3.1

Simplify the left side.

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Step 6.3.1.1

Simplify each term.

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Step 6.3.1.1.1

Combine $\ln (| | x | |)$ and $\frac{1}{2}$ .

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

Step 6.3.1.1.2

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

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

Step 6.3.2

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

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Step 6.3.2.1

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

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

Step 6.3.2.2

Simplify the left side.

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Step 6.3.2.2.1

Simplify each term.

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Step 6.3.2.2.1.1

Cancel the common factor of $2$ .

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Step 6.3.2.2.1.1.1

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

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

Step 6.3.2.2.1.1.2

Cancel the common factor.

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

Step 6.3.2.2.1.1.3

Rewrite the expression.

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

Step 6.3.2.2.1.2

Cancel the common factor of $2$ .

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Step 6.3.2.2.1.2.1

Cancel the common factor.

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

Step 6.3.2.2.1.2.2

Rewrite the expression.

$- \ln (| V |) + \ln (| V - 2 |) = \ln (| x |) \cdot 2 + C \cdot 2$

Step 6.3.2.3

Simplify the right side.

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Step 6.3.2.3.1

Simplify each term.

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Step 6.3.2.3.1.1

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

$- \ln (| V |) + \ln (| V - 2 |) = 2 \cdot \ln (| x |) + C \cdot 2$

Step 6.3.2.3.1.2

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

$- \ln (| V |) + \ln (| V - 2 |) = 2 \ln (| x |) + 2 C$

Step 6.3.3

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

$- \ln (| V |) + \ln (| V - 2 |) - 2 \ln (| x |) = 2 C$

Step 6.3.4

Simplify the left side.

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Step 6.3.4.1

Simplify $- \ln (| V |) + \ln (| V - 2 |) - 2 \ln (| x |)$ .

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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.

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

Step 6.3.4.1.1.2

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

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

Step 6.3.4.1.2

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

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

Step 6.3.5

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

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

Step 6.3.6

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

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

Step 6.3.7

Solve for $V$ .

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Step 6.3.7.1

Take the natural logarithm of both sides of the equation to remove the variable from the exponent.

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

Step 6.3.7.2

Expand the left side.

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Step 6.3.7.2.1

Expand $\ln (e^{2 C + \ln (| V |)})$ by moving $2 C + \ln (| V |)$ outside the logarithm.

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

Step 6.3.7.2.2

The natural logarithm of $e$ is $1$ .

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

Step 6.3.7.2.3

Multiply $2 C + \ln (| V |)$ by $1$ .

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

Step 6.3.7.3

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

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

Step 6.3.7.4

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

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

Step 6.3.7.5

Multiply the numerator by the reciprocal of the denominator.

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

Step 6.3.7.6

Combine $| V |$ and $\frac{x^{2}}{| V - 2 |}$ .

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

Step 6.3.7.7

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

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

Step 6.3.7.8

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

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

Step 6.3.7.9

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

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

Step 6.3.7.10

Solve for $V$ .

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Step 6.3.7.10.1

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

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

Step 6.3.7.10.2

Multiply both sides by $| V - 2 |$ .

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

Step 6.3.7.10.3

Simplify the left side.

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Step 6.3.7.10.3.1

Cancel the common factor of $| V - 2 |$ .

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Step 6.3.7.10.3.1.1

Cancel the common factor.

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

Step 6.3.7.10.3.1.2

Rewrite the expression.

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

Step 6.3.7.10.4

Solve for $V$ .

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Step 6.3.7.10.4.1

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

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

Step 6.3.7.10.4.2

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

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Step 6.3.7.10.4.2.1

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

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

Step 6.3.7.10.4.2.2

Simplify the left side.

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Step 6.3.7.10.4.2.2.1

Cancel the common factor of $e^{- 2 C}$ .

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Step 6.3.7.10.4.2.2.1.1

Cancel the common factor.

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

Step 6.3.7.10.4.2.2.1.2

Divide $| V - 2 |$ by $1$ .

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

Step 6.3.7.10.4.3

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

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

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

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

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

Step 6.3.7.10.4.4

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

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

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

Step 6.3.7.10.4.5

Solve $V - 2 = \frac{x^{2} (V)}{e^{- 2 C}}$ for $V$ .

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Step 6.3.7.10.4.5.1

Multiply both sides by $e^{- 2 C}$ .

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

Step 6.3.7.10.4.5.2

Simplify.

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Step 6.3.7.10.4.5.2.1

Simplify the left side.

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Step 6.3.7.10.4.5.2.1.1

Apply the distributive property.

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

Step 6.3.7.10.4.5.2.2

Simplify the right side.

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Step 6.3.7.10.4.5.2.2.1

Cancel the common factor of $e^{- 2 C}$ .

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Step 6.3.7.10.4.5.2.2.1.1

Cancel the common factor.

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

Step 6.3.7.10.4.5.2.2.1.2

Rewrite the expression.

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

Step 6.3.7.10.4.5.3

Solve for $V$ .

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Step 6.3.7.10.4.5.3.1

Subtract $x^{2} V$ from both sides of the equation.

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

Step 6.3.7.10.4.5.3.2

Add $2 e^{- 2 C}$ to both sides of the equation.

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

Step 6.3.7.10.4.5.3.3

Factor $V$ out of $V e^{- 2 C} - x^{2} V$ .

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Step 6.3.7.10.4.5.3.3.1

Factor $V$ out of $V e^{- 2 C}$ .

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

Step 6.3.7.10.4.5.3.3.2

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

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

Step 6.3.7.10.4.5.3.3.3

Factor $V$ out of $V e^{- 2 C} + V (- x^{2})$ .

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

Step 6.3.7.10.4.5.3.4

Rewrite $e^{- 2 C}$ as ${(e^{- C})}^{2}$ .

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

Step 6.3.7.10.4.5.3.5

Factor.

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Step 6.3.7.10.4.5.3.5.1

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

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

Step 6.3.7.10.4.5.3.5.2

Remove unnecessary parentheses.

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

Step 6.3.7.10.4.5.3.6

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

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Step 6.3.7.10.4.5.3.6.1

Divide each term in $V (e^{- C} + x) (e^{- C} - x) = 2 e^{- 2 C}$ by $(e^{- C} + x) (e^{- C} - x)$ .

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

Step 6.3.7.10.4.5.3.6.2

Simplify the left side.

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Step 6.3.7.10.4.5.3.6.2.1

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

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Step 6.3.7.10.4.5.3.6.2.1.1

Cancel the common factor.

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

Step 6.3.7.10.4.5.3.6.2.1.2

Rewrite the expression.

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

Step 6.3.7.10.4.5.3.6.2.2

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

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Step 6.3.7.10.4.5.3.6.2.2.1

Cancel the common factor.

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

Step 6.3.7.10.4.5.3.6.2.2.2

Divide $V$ by $1$ .

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

Step 6.3.7.10.4.6

Solve $V - 2 = - \frac{x^{2} V}{e^{- 2 C}}$ for $V$ .

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Step 6.3.7.10.4.6.1

Multiply both sides by $e^{- 2 C}$ .

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

Step 6.3.7.10.4.6.2

Simplify.

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Step 6.3.7.10.4.6.2.1

Simplify the left side.

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Step 6.3.7.10.4.6.2.1.1

Apply the distributive property.

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

Step 6.3.7.10.4.6.2.2

Simplify the right side.

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Step 6.3.7.10.4.6.2.2.1

Cancel the common factor of $e^{- 2 C}$ .

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Step 6.3.7.10.4.6.2.2.1.1

Move the leading negative in $- \frac{x^{2} V}{e^{- 2 C}}$ into the numerator.

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

Step 6.3.7.10.4.6.2.2.1.2

Cancel the common factor.

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

Step 6.3.7.10.4.6.2.2.1.3

Rewrite the expression.

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

Step 6.3.7.10.4.6.3

Solve for $V$ .

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Step 6.3.7.10.4.6.3.1

Add $x^{2} V$ to both sides of the equation.

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

Step 6.3.7.10.4.6.3.2

Add $2 e^{- 2 C}$ to both sides of the equation.

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

Step 6.3.7.10.4.6.3.3

Factor $V$ out of $V e^{- 2 C} + x^{2} V$ .

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Step 6.3.7.10.4.6.3.3.1

Factor $V$ out of $V e^{- 2 C}$ .

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

Step 6.3.7.10.4.6.3.3.2

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

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

Step 6.3.7.10.4.6.3.3.3

Factor $V$ out of $V e^{- 2 C} + V x^{2}$ .

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

Step 6.3.7.10.4.6.3.4

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

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Step 6.3.7.10.4.6.3.4.1

Divide each term in $V (e^{- 2 C} + x^{2}) = 2 e^{- 2 C}$ by $e^{- 2 C} + x^{2}$ .

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

Step 6.3.7.10.4.6.3.4.2

Simplify the left side.

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Step 6.3.7.10.4.6.3.4.2.1

Cancel the common factor of $e^{- 2 C} + x^{2}$ .

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Step 6.3.7.10.4.6.3.4.2.1.1

Cancel the common factor.

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

Step 6.3.7.10.4.6.3.4.2.1.2

Divide $V$ by $1$ .

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

Step 6.3.7.10.4.7

List all of the solutions.

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

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

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

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

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

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

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

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

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

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

Step 6.4

Simplify the constant of integration.

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

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

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

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

Step 7

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

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

Step 8

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

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Step 8.1

Rewrite.

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

Step 8.2

Multiply the numerator of the first fraction by the denominator of the second fraction. Set this equal to the product of the denominator of the first fraction and the numerator of the second fraction.

$y ((C + x) (C - x)) = x C$

Step 8.3

Divide each term in $y (C + x) (C - x) = x C$ by $(C + x) (C - x)$ and simplify.

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Step 8.3.1

Divide each term in $y (C + x) (C - x) = x C$ by $(C + x) (C - x)$ .

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

Step 8.3.2

Simplify the left side.

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Step 8.3.2.1

Cancel the common factor of $C + x$ .

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Step 8.3.2.1.1

Cancel the common factor.

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

Step 8.3.2.1.2

Rewrite the expression.

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

Step 8.3.2.2

Divide $y$ by $1$ .

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

Step 9

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

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Step 9.1

Rewrite.

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

Step 9.2

$y (C + x^{2}) = x C$

Step 9.3

Divide each term in $y (C + x^{2}) = x C$ by $C + x^{2}$ and simplify.

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Step 9.3.1

Divide each term in $y (C + x^{2}) = x C$ by $C + x^{2}$ .

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

Step 9.3.2

Simplify the left side.

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Step 9.3.2.1

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

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Step 9.3.2.1.1

Cancel the common factor.

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

Step 9.3.2.1.2

Divide $y$ by $1$ .

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

Step 10

List the solutions.

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