Calculus Examples

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

Step 1

Separate the variables.

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Step 1.1

Multiply both sides by $\frac{1}{1 - y^{2}}$ .

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

Step 1.2

Cancel the common factor of $1 - y^{2}$ .

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Step 1.2.1

Cancel the common factor.

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

Step 1.2.2

Rewrite the expression.

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

Step 1.3

Rewrite the equation.

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

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

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

Step 2.2

Integrate the left side.

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Step 2.2.1

Write the fraction using partial fraction decomposition.

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Step 2.2.1.1

Decompose the fraction and multiply through by the common denominator.

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Step 2.2.1.1.1

Factor the fraction.

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Step 2.2.1.1.1.1

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

$\frac{1}{1^{2} - y^{2}}$

Step 2.2.1.1.1.2

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 $b = y$ .

$\frac{1}{(1 + y) (1 - y)}$

Step 2.2.1.1.2

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 $A$ .

$\frac{A}{1 + y}$

Step 2.2.1.1.3

$\frac{A}{1 + y} + \frac{B}{1 - y}$

Step 2.2.1.1.4

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

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

Step 2.2.1.1.5

Cancel the common factor of $1 + y$ .

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Step 2.2.1.1.5.1

Cancel the common factor.

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

Step 2.2.1.1.5.2

Rewrite the expression.

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

Step 2.2.1.1.6

Cancel the common factor of $1 - y$ .

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Step 2.2.1.1.6.1

Cancel the common factor.

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

Step 2.2.1.1.6.2

Rewrite the expression.

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

Step 2.2.1.1.7

Simplify each term.

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Step 2.2.1.1.7.1

Cancel the common factor of $1 + y$ .

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Step 2.2.1.1.7.1.1

Cancel the common factor.

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

Step 2.2.1.1.7.1.2

Divide $(A) (1 - y)$ by $1$ .

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

Step 2.2.1.1.7.2

Apply the distributive property.

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

Step 2.2.1.1.7.3

Multiply $A$ by $1$ .

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

Step 2.2.1.1.7.4

Rewrite using the commutative property of multiplication.

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

Step 2.2.1.1.7.5

Cancel the common factor of $1 - y$ .

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Step 2.2.1.1.7.5.1

Cancel the common factor.

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

Step 2.2.1.1.7.5.2

Divide $(B) (1 + y)$ by $1$ .

$1 = A - A y + (B) (1 + y)$

Step 2.2.1.1.7.6

Apply the distributive property.

$1 = A - A y + B \cdot 1 + B y$

Step 2.2.1.1.7.7

Multiply $B$ by $1$ .

$1 = A - A y + B + B y$

Step 2.2.1.1.8

Simplify the expression.

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Step 2.2.1.1.8.1

Move $A$ .

$1 = A - y A + B + B y$

Step 2.2.1.1.8.2

Reorder $B$ and $y$ .

$1 = A - y A + B + B y$

Step 2.2.1.1.8.3

Move $B$ .

$1 = A - y A + B y + B$

Step 2.2.1.1.8.4

Move $A$ .

$1 = - y A + B y + A + B$

Step 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 2.2.1.2.1

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

$0 = - 1 A + B$

Step 2.2.1.2.2

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

$1 = A + B$

Step 2.2.1.2.3

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

$0 = - 1 A + B$

$1 = A + B$

$0 = - 1 A + B$

$1 = A + B$

Step 2.2.1.3

Solve the system of equations.

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Step 2.2.1.3.1

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

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Step 2.2.1.3.1.1

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

$A + B = 1$

$0 = - 1 A + B$

Step 2.2.1.3.1.2

Subtract $B$ from both sides of the equation.

$A = 1 - B$

$0 = - 1 A + B$

$A = 1 - B$

$0 = - 1 A + B$

Step 2.2.1.3.2

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

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Step 2.2.1.3.2.1

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

$0 = - 1 (1 - B) + B$

$A = 1 - B$

Step 2.2.1.3.2.2

Simplify the right side.

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Step 2.2.1.3.2.2.1

Simplify $- 1 (1 - B) + B$ .

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Step 2.2.1.3.2.2.1.1

Simplify each term.

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Step 2.2.1.3.2.2.1.1.1

Apply the distributive property.

$0 = - 1 \cdot 1 - 1 (- B) + B$

$A = 1 - B$

Step 2.2.1.3.2.2.1.1.2

Multiply $- 1$ by $1$ .

$0 = - 1 - 1 (- B) + B$

$A = 1 - B$

Step 2.2.1.3.2.2.1.1.3

Multiply $- 1 (- B)$ .

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Step 2.2.1.3.2.2.1.1.3.1

Multiply $- 1$ by $- 1$ .

$0 = - 1 + 1 B + B$

$A = 1 - B$

Step 2.2.1.3.2.2.1.1.3.2

Multiply $B$ by $1$ .

$0 = - 1 + B + B$

$A = 1 - B$

$0 = - 1 + B + B$

$A = 1 - B$

$0 = - 1 + B + B$

$A = 1 - B$

Step 2.2.1.3.2.2.1.2

Add $B$ and $B$ .

$0 = - 1 + 2 B$

$A = 1 - B$

$0 = - 1 + 2 B$

$A = 1 - B$

$0 = - 1 + 2 B$

$A = 1 - B$

$0 = - 1 + 2 B$

$A = 1 - B$

Step 2.2.1.3.3

Solve for $B$ in $0 = - 1 + 2 B$ .

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Step 2.2.1.3.3.1

Rewrite the equation as $- 1 + 2 B = 0$ .

$- 1 + 2 B = 0$

$A = 1 - B$

Step 2.2.1.3.3.2

Add $1$ to both sides of the equation.

$2 B = 1$

$A = 1 - B$

Step 2.2.1.3.3.3

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

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Step 2.2.1.3.3.3.1

Divide each term in $2 B = 1$ by $2$ .

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

$A = 1 - B$

Step 2.2.1.3.3.3.2

Simplify the left side.

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Step 2.2.1.3.3.3.2.1

Cancel the common factor of $2$ .

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Step 2.2.1.3.3.3.2.1.1

Cancel the common factor.

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

$A = 1 - B$

Step 2.2.1.3.3.3.2.1.2

Divide $B$ by $1$ .

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

$A = 1 - B$

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

$A = 1 - B$

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

$A = 1 - B$

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

$A = 1 - B$

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

$A = 1 - B$

Step 2.2.1.3.4

Replace all occurrences of $B$ with $\frac{1}{2}$ in each equation.

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Step 2.2.1.3.4.1

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

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

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

Step 2.2.1.3.4.2

Simplify the right side.

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Step 2.2.1.3.4.2.1

Simplify $1 - (\frac{1}{2})$ .

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Step 2.2.1.3.4.2.1.1

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

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

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

Step 2.2.1.3.4.2.1.2

Combine the numerators over the common denominator.

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

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

Step 2.2.1.3.4.2.1.3

Subtract $1$ from $2$ .

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

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

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

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

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

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

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

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

Step 2.2.1.3.5

List all of the solutions.

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

Step 2.2.1.4

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

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

Step 2.2.1.5

Simplify.

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Step 2.2.1.5.1

Multiply the numerator by the reciprocal of the denominator.

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

Step 2.2.1.5.2

Multiply $\frac{1}{2}$ by $\frac{1}{1 + y}$ .

$\frac{1}{2 (1 + y)} + \frac{\frac{1}{2}}{1 - y}$

Step 2.2.1.5.3

Multiply the numerator by the reciprocal of the denominator.

$\frac{1}{2 (1 + y)} + \frac{1}{2} \cdot \frac{1}{1 - y}$

Step 2.2.1.5.4

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

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

Step 2.2.2

Split the single integral into multiple integrals.

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

Step 2.2.3

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

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

Step 2.2.4

Let $u_{1} = 1 + y$ . Then $d u_{1} = d y$ . Rewrite using $u_{1}$ and $d$ $u_{1}$ .

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Step 2.2.4.1

Let $u_{1} = 1 + y$ . Find $\frac{d u_{1}}{d y}$ .

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Step 2.2.4.1.1

Differentiate $1 + y$ .

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

Step 2.2.4.1.2

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

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

Step 2.2.4.1.3

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

$0 + \frac{d}{d y} [y]$

Step 2.2.4.1.4

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

$0 + 1$

Step 2.2.4.1.5

Add $0$ and $1$ .

$1$

Step 2.2.4.2

Rewrite the problem using $u_{1}$ and $d u_{1}$ .

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

Step 2.2.5

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

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

Step 2.2.6

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

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

Step 2.2.7

Let $u_{2} = 1 - y$ . Then $d u_{2} = - d y$ , so $- d u_{2} = d y$ . Rewrite using $u_{2}$ and $d$ $u_{2}$ .

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Step 2.2.7.1

Let $u_{2} = 1 - y$ . Find $\frac{d u_{2}}{d y}$ .

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Step 2.2.7.1.1

Rewrite.

$\frac{- 1}{1}$

Step 2.2.7.1.2

Divide $- 1$ by $1$ .

$- 1$

Step 2.2.7.2

Rewrite the problem using $u_{2}$ and $d u_{2}$ .

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

Step 2.2.8

Move the negative in front of the fraction.

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

Step 2.2.9

Since $- 1$ is constant with respect to $u_{2}$ , move $- 1$ out of the integral.

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

Step 2.2.10

The integral of $\frac{1}{u_{2}}$ with respect to $u_{2}$ is $\ln (| u_{2} |)$ .

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

Step 2.2.11

Simplify.

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Step 2.2.11.1

Simplify.

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

Step 2.2.11.2

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

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

Step 2.2.12

Substitute back in for each integration substitution variable.

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Step 2.2.12.1

Replace all occurrences of $u_{1}$ with $1 + y$ .

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

Step 2.2.12.2

Replace all occurrences of $u_{2}$ with $1 - y$ .

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

Step 2.2.13

Reorder terms.

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

Step 2.3

Apply the constant rule.

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

Step 2.4

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

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

Step 3

Solve for $y$ .

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Step 3.1

Simplify the left side.

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Step 3.1.1

Simplify each term.

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Step 3.1.1.1

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

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

Step 3.1.1.2

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

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

Step 3.2

Multiply each term in $\frac{\ln (| 1 + y |)}{2} - \frac{\ln (| 1 - y |)}{2} = x + C$ by $2$ to eliminate the fractions.

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Step 3.2.1

Multiply each term in $\frac{\ln (| 1 + y |)}{2} - \frac{\ln (| 1 - y |)}{2} = x + C$ by $2$ .

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

Step 3.2.2

Simplify the left side.

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Step 3.2.2.1

Simplify each term.

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Step 3.2.2.1.1

Cancel the common factor of $2$ .

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Step 3.2.2.1.1.1

Cancel the common factor.

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

Step 3.2.2.1.1.2

Rewrite the expression.

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

Step 3.2.2.1.2

Cancel the common factor of $2$ .

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Step 3.2.2.1.2.1

Move the leading negative in $- \frac{\ln (| 1 - y |)}{2}$ into the numerator.

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

Step 3.2.2.1.2.2

Cancel the common factor.

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

Step 3.2.2.1.2.3

Rewrite the expression.

$\ln (| 1 + y |) - \ln (| 1 - y |) = x \cdot 2 + C \cdot 2$

Step 3.2.3

Simplify the right side.

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Step 3.2.3.1

Simplify each term.

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Step 3.2.3.1.1

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

$\ln (| 1 + y |) - \ln (| 1 - y |) = 2 \cdot x + C \cdot 2$

Step 3.2.3.1.2

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

$\ln (| 1 + y |) - \ln (| 1 - y |) = 2 x + 2 C$

Step 3.3

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

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

Step 3.4

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

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

Step 3.5

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

$e^{2 x + 2 C} = \frac{| 1 + y |}{| 1 - y |}$

Step 3.6

Solve for $y$ .

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Step 3.6.1

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

$\frac{| 1 + y |}{| 1 - y |} = e^{2 x + 2 C}$

Step 3.6.2

Multiply both sides by $| 1 - y |$ .

$\frac{| 1 + y |}{| 1 - y |} | 1 - y | = e^{2 x + 2 C} | 1 - y |$

Step 3.6.3

Simplify the left side.

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Step 3.6.3.1

Cancel the common factor of $| 1 - y |$ .

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Step 3.6.3.1.1

Cancel the common factor.

$\frac{| 1 + y |}{| 1 - y |} | 1 - y | = e^{2 x + 2 C} | 1 - y |$

Step 3.6.3.1.2

Rewrite the expression.

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

Step 3.6.4

Solve for $y$ .

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Step 3.6.4.1

Rewrite the equation as $e^{2 x + 2 C} | 1 - y | = | 1 + y |$ .

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

Step 3.6.4.2

Divide each term in $e^{2 x + 2 C} | 1 - y | = | 1 + y |$ by $e^{2 x + 2 C}$ and simplify.

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Step 3.6.4.2.1

Divide each term in $e^{2 x + 2 C} | 1 - y | = | 1 + y |$ by $e^{2 x + 2 C}$ .

$\frac{e^{2 x + 2 C} | 1 - y |}{e^{2 x + 2 C}} = \frac{| 1 + y |}{e^{2 x + 2 C}}$

Step 3.6.4.2.2

Simplify the left side.

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Step 3.6.4.2.2.1

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

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Step 3.6.4.2.2.1.1

Cancel the common factor.

$\frac{e^{2 x + 2 C} | 1 - y |}{e^{2 x + 2 C}} = \frac{| 1 + y |}{e^{2 x + 2 C}}$

Step 3.6.4.2.2.1.2

Divide $| 1 - y |$ by $1$ .

$| 1 - y | = \frac{| 1 + y |}{e^{2 x + 2 C}}$

Step 3.6.4.3

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

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

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

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

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

Step 3.6.4.4

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

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

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

Step 3.6.4.5

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

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Step 3.6.4.5.1

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

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

Step 3.6.4.5.2

Simplify.

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Step 3.6.4.5.2.1

Simplify the left side.

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Step 3.6.4.5.2.1.1

Simplify $(1 - y) e^{2 x + 2 C}$ .

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Step 3.6.4.5.2.1.1.1

Apply the distributive property.

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

Step 3.6.4.5.2.1.1.2

Simplify the expression.

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Step 3.6.4.5.2.1.1.2.1

Multiply $e^{2 x + 2 C}$ by $1$ .

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

Step 3.6.4.5.2.1.1.2.2

Reorder $2 x$ and $2 C$ .

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

Step 3.6.4.5.2.1.1.2.3

Reorder $2 x$ and $2 C$ .

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

Step 3.6.4.5.2.1.1.2.4

Reorder $e^{2 C + 2 x}$ and $- y e^{2 C + 2 x}$ .

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

Step 3.6.4.5.2.2

Simplify the right side.

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Step 3.6.4.5.2.2.1

Simplify $\frac{1 + y}{e^{2 x + 2 C}} e^{2 x + 2 C}$ .

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Step 3.6.4.5.2.2.1.1

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

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Step 3.6.4.5.2.2.1.1.1

Cancel the common factor.

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

Step 3.6.4.5.2.2.1.1.2

Rewrite the expression.

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

Step 3.6.4.5.2.2.1.2

Reorder $1$ and $y$ .

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

Step 3.6.4.5.3

Solve for $y$ .

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Step 3.6.4.5.3.1

Subtract $y$ from both sides of the equation.

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

Step 3.6.4.5.3.2

Subtract $e^{2 C + 2 x}$ from both sides of the equation.

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

Step 3.6.4.5.3.3

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

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Step 3.6.4.5.3.3.1

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

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

Step 3.6.4.5.3.3.2

Factor $y$ out of $- y$ .

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

Step 3.6.4.5.3.3.3

Factor $y$ out of $y (- 1 e^{2 C + 2 x}) + y \cdot - 1$ .

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

Step 3.6.4.5.3.4

Rewrite $- 1 e^{2 C + 2 x}$ as $- e^{2 C + 2 x}$ .

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

Step 3.6.4.5.3.5

Divide each term in $y (- e^{2 C + 2 x} - 1) = 1 - e^{2 C + 2 x}$ by $- e^{2 C + 2 x} - 1$ and simplify.

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Step 3.6.4.5.3.5.1

Divide each term in $y (- e^{2 C + 2 x} - 1) = 1 - e^{2 C + 2 x}$ by $- e^{2 C + 2 x} - 1$ .

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

Step 3.6.4.5.3.5.2

Simplify the left side.

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Step 3.6.4.5.3.5.2.1

Cancel the common factor of $- e^{2 C + 2 x} - 1$ .

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Step 3.6.4.5.3.5.2.1.1

Cancel the common factor.

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

Step 3.6.4.5.3.5.2.1.2

Divide $y$ by $1$ .

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

Step 3.6.4.5.3.5.3

Simplify the right side.

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Step 3.6.4.5.3.5.3.1

Combine the numerators over the common denominator.

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

Step 3.6.4.5.3.5.3.2

Simplify the numerator.

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Step 3.6.4.5.3.5.3.2.1

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

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

Step 3.6.4.5.3.5.3.2.2

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

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

Step 3.6.4.5.3.5.3.2.3

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 $b = e^{C + x}$ .

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

Step 3.6.4.5.3.5.3.3

Simplify with factoring out.

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Step 3.6.4.5.3.5.3.3.1

Factor $- 1$ out of $- e^{2 C + 2 x}$ .

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

Step 3.6.4.5.3.5.3.3.2

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

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

Step 3.6.4.5.3.5.3.3.3

Factor $- 1$ out of $- e^{2 C + 2 x} - 1 (1)$ .

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

Step 3.6.4.5.3.5.3.3.4

Rewrite negatives.

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Step 3.6.4.5.3.5.3.3.4.1

Rewrite $- (e^{2 C + 2 x} + 1)$ as $- 1 (e^{2 C + 2 x} + 1)$ .

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

Step 3.6.4.5.3.5.3.3.4.2

Move the negative in front of the fraction.

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

Step 3.6.4.6

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

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Step 3.6.4.6.1

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

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

Step 3.6.4.6.2

Simplify.

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Step 3.6.4.6.2.1

Simplify the left side.

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Step 3.6.4.6.2.1.1

Simplify $(1 - y) e^{2 x + 2 C}$ .

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Step 3.6.4.6.2.1.1.1

Apply the distributive property.

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

Step 3.6.4.6.2.1.1.2

Simplify the expression.

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Step 3.6.4.6.2.1.1.2.1

Multiply $e^{2 x + 2 C}$ by $1$ .

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

Step 3.6.4.6.2.1.1.2.2

Reorder $2 x$ and $2 C$ .

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

Step 3.6.4.6.2.1.1.2.3

Reorder $2 x$ and $2 C$ .

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

Step 3.6.4.6.2.1.1.2.4

Reorder $e^{2 C + 2 x}$ and $- y e^{2 C + 2 x}$ .

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

Step 3.6.4.6.2.2

Simplify the right side.

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Step 3.6.4.6.2.2.1

Simplify $- \frac{1 + y}{e^{2 x + 2 C}} e^{2 x + 2 C}$ .

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Step 3.6.4.6.2.2.1.1

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

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Step 3.6.4.6.2.2.1.1.1

Move the leading negative in $- \frac{1 + y}{e^{2 x + 2 C}}$ into the numerator.

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

Step 3.6.4.6.2.2.1.1.2

Cancel the common factor.

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

Step 3.6.4.6.2.2.1.1.3

Rewrite the expression.

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

Step 3.6.4.6.2.2.1.2

Apply the distributive property.

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

Step 3.6.4.6.2.2.1.3

Simplify the expression.

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Step 3.6.4.6.2.2.1.3.1

Multiply $- 1$ by $1$ .

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

Step 3.6.4.6.2.2.1.3.2

Reorder $- 1$ and $- y$ .

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

Step 3.6.4.6.3

Solve for $y$ .

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Step 3.6.4.6.3.1

Add $y$ to both sides of the equation.

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

Step 3.6.4.6.3.2

Subtract $e^{2 C + 2 x}$ from both sides of the equation.

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

Step 3.6.4.6.3.3

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

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Step 3.6.4.6.3.3.1

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

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

Step 3.6.4.6.3.3.2

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

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

Step 3.6.4.6.3.3.3

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

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

Step 3.6.4.6.3.3.4

Factor $y$ out of $y (- 1 e^{2 C + 2 x}) + y \cdot 1$ .

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

Step 3.6.4.6.3.4

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

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

Step 3.6.4.6.3.5

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

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

Step 3.6.4.6.3.6

Reorder $- {(e^{C + x})}^{2}$ and $1^{2}$ .

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

Step 3.6.4.6.3.7

Factor.

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Step 3.6.4.6.3.7.1

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 $b = e^{C + x}$ .

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

Step 3.6.4.6.3.7.2

Remove unnecessary parentheses.

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

Step 3.6.4.6.3.8

Divide each term in $y (1 + e^{C + x}) (1 - e^{C + x}) = - 1 - e^{2 C + 2 x}$ by $(1 + e^{C + x}) (1 - e^{C + x})$ and simplify.

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Step 3.6.4.6.3.8.1

Divide each term in $y (1 + e^{C + x}) (1 - e^{C + x}) = - 1 - e^{2 C + 2 x}$ by $(1 + e^{C + x}) (1 - e^{C + x})$ .

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

Step 3.6.4.6.3.8.2

Simplify the left side.

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Step 3.6.4.6.3.8.2.1

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

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Step 3.6.4.6.3.8.2.1.1

Cancel the common factor.

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

Step 3.6.4.6.3.8.2.1.2

Rewrite the expression.

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

Step 3.6.4.6.3.8.2.2

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

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Step 3.6.4.6.3.8.2.2.1

Cancel the common factor.

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

Step 3.6.4.6.3.8.2.2.2

Divide $y$ by $1$ .

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

Step 3.6.4.6.3.8.3

Simplify the right side.

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Step 3.6.4.6.3.8.3.1

Combine the numerators over the common denominator.

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

Step 3.6.4.6.3.8.3.2

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

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

Step 3.6.4.6.3.8.3.3

Factor $- 1$ out of $- e^{2 C + 2 x}$ .

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

Step 3.6.4.6.3.8.3.4

Factor $- 1$ out of $- 1 (1) - e^{2 C + 2 x}$ .

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

Step 3.6.4.6.3.8.3.5

Move the negative in front of the fraction.

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

Step 3.6.4.7