Calculus Examples

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

Step 1

Separate the variables.

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Step 1.1

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

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

Step 1.2

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

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Step 1.2.1

Cancel the common factor.

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

Step 1.2.2

Rewrite the expression.

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

Step 1.3

Rewrite the equation.

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

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

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

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 $1^{2}$ .

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

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 $a = y$ and $b = 1$ .

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

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 $A$ .

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

Step 2.2.1.1.3

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

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 $(y + 1) (y - 1)$ .

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

Step 2.2.1.1.5

Cancel the common factor of $y + 1$ .

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Step 2.2.1.1.5.1

Cancel the common factor.

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

Step 2.2.1.1.5.2

Rewrite the expression.

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

Step 2.2.1.1.6

Cancel the common factor of $y - 1$ .

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Step 2.2.1.1.6.1

Cancel the common factor.

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

Step 2.2.1.1.6.2

Rewrite the expression.

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

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 $y + 1$ .

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Step 2.2.1.1.7.1.1

Cancel the common factor.

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

Step 2.2.1.1.7.1.2

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

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

Step 2.2.1.1.7.2

Apply the distributive property.

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

Step 2.2.1.1.7.3

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

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

Step 2.2.1.1.7.4

Rewrite $- 1 A$ as $- A$ .

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

Step 2.2.1.1.7.5

Cancel the common factor of $y - 1$ .

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Step 2.2.1.1.7.5.1

Cancel the common factor.

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

Step 2.2.1.1.7.5.2

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

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

Step 2.2.1.1.7.6

Apply the distributive property.

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

Step 2.2.1.1.7.7

Multiply $B$ by $1$ .

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

Step 2.2.1.1.8

Move $- A$ .

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

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.

$0 = A + B$

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.

$1 = - 1 A + B$

Step 2.2.1.2.3

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

$0 = A + B$

$1 = - 1 A + B$

$0 = A + B$

$1 = - 1 A + B$

Step 2.2.1.3

Solve the system of equations.

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Step 2.2.1.3.1

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

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Step 2.2.1.3.1.1

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

$A + B = 0$

$1 = - 1 A + B$

Step 2.2.1.3.1.2

Subtract $B$ from both sides of the equation.

$A = - B$

$1 = - 1 A + B$

$A = - B$

$1 = - 1 A + B$

Step 2.2.1.3.2

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

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Step 2.2.1.3.2.1

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

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

$A = - B$

Step 2.2.1.3.2.2

Simplify the right side.

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Step 2.2.1.3.2.2.1

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

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Step 2.2.1.3.2.2.1.1

Multiply $- 1 (- B)$ .

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Step 2.2.1.3.2.2.1.1.1

Multiply $- 1$ by $- 1$ .

$1 = 1 B + B$

$A = - B$

Step 2.2.1.3.2.2.1.1.2

Multiply $B$ by $1$ .

$1 = B + B$

$A = - B$

$1 = B + B$

$A = - B$

Step 2.2.1.3.2.2.1.2

Add $B$ and $B$ .

$1 = 2 B$

$A = - B$

$1 = 2 B$

$A = - B$

$1 = 2 B$

$A = - B$

$1 = 2 B$

$A = - B$

Step 2.2.1.3.3

Solve for $B$ in $1 = 2 B$ .

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Step 2.2.1.3.3.1

Rewrite the equation as $2 B = 1$ .

$2 B = 1$

$A = - B$

Step 2.2.1.3.3.2

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

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Step 2.2.1.3.3.2.1

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

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

$A = - B$

Step 2.2.1.3.3.2.2

Simplify the left side.

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Step 2.2.1.3.3.2.2.1

Cancel the common factor of $2$ .

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Step 2.2.1.3.3.2.2.1.1

Cancel the common factor.

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

$A = - B$

Step 2.2.1.3.3.2.2.1.2

Divide $B$ by $1$ .

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

$A = - B$

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

$A = - B$

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

$A = - B$

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

$A = - B$

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

$A = - B$

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

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

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

Step 2.2.1.3.4.2

Simplify the right side.

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Step 2.2.1.3.4.2.1

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

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

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

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

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

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

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

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 $\frac{A}{y + 1} + \frac{B}{y - 1}$ with the values found for $A$ and $B$ .

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

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.

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

Step 2.2.1.5.2

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

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

Step 2.2.1.5.3

Move $2$ to the left of $y + 1$ .

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

Step 2.2.1.5.4

Multiply the numerator by the reciprocal of the denominator.

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

Step 2.2.1.5.5

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

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

Step 2.2.2

Split the single integral into multiple integrals.

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

Step 2.2.3

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

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

Step 2.2.4

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

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

Step 2.2.5

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

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Step 2.2.5.1

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

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Step 2.2.5.1.1

Differentiate $y + 1$ .

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

Step 2.2.5.1.2

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

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

Step 2.2.5.1.3

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

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

Step 2.2.5.1.4

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

$1 + 0$

Step 2.2.5.1.5

Add $1$ and $0$ .

$1$

Step 2.2.5.2

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

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

Step 2.2.6

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

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

Step 2.2.7

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

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

Step 2.2.8

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

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Step 2.2.8.1

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

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Step 2.2.8.1.1

Differentiate $y - 1$ .

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

Step 2.2.8.1.2

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

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

Step 2.2.8.1.3

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

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

Step 2.2.8.1.4

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

$1 + 0$

Step 2.2.8.1.5

Add $1$ and $0$ .

$1$

Step 2.2.8.2

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

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

Step 2.2.9

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

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

Step 2.2.10

Simplify.

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

Step 2.2.11

Substitute back in for each integration substitution variable.

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Step 2.2.11.1

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

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

Step 2.2.11.2

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

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

Step 2.3

Apply the constant rule.

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

Step 2.4

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

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

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

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

Step 3.1.1.2

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

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

Step 3.2

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

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Step 3.2.1

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

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

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

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

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

Step 3.2.2.1.1.2

Cancel the common factor.

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

Step 3.2.2.1.1.3

Rewrite the expression.

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

Step 3.2.2.1.2

Cancel the common factor of $2$ .

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Step 3.2.2.1.2.1

Cancel the common factor.

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

Step 3.2.2.1.2.2

Rewrite the expression.

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

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 $t$ .

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

Step 3.2.3.1.2

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

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

Step 3.3

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

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

Step 3.4

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

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

Step 3.5

Solve for $y$ .

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Step 3.5.1

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

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

Step 3.5.2

Expand the left side.

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Step 3.5.2.1

Expand $\ln (e^{2 t + 2 C + \ln (| y + 1 |)})$ by moving $2 t + 2 C + \ln (| y + 1 |)$ outside the logarithm.

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

Step 3.5.2.2

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

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

Step 3.5.2.3

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

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

Step 3.5.3

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

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

Step 3.5.4

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

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

Step 3.5.5

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

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

Step 3.5.6

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

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

Step 3.5.7

Solve for $y$ .

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Step 3.5.7.1

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

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

Step 3.5.7.2

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

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

Step 3.5.7.3

Simplify the left side.

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Step 3.5.7.3.1

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

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Step 3.5.7.3.1.1

Cancel the common factor.

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

Step 3.5.7.3.1.2

Rewrite the expression.

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

Step 3.5.7.4

Solve for $y$ .

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Step 3.5.7.4.1

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

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

Step 3.5.7.4.2

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

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Step 3.5.7.4.2.1

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

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

Step 3.5.7.4.2.2

Simplify the left side.

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Step 3.5.7.4.2.2.1

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

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Step 3.5.7.4.2.2.1.1

Cancel the common factor.

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

Step 3.5.7.4.2.2.1.2

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

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

Step 3.5.7.4.3

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

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

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

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

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

Step 3.5.7.4.4

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

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

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

Step 3.5.7.4.5

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

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Step 3.5.7.4.5.1

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

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

Step 3.5.7.4.5.2

Simplify.

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Step 3.5.7.4.5.2.1

Simplify the left side.

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Step 3.5.7.4.5.2.1.1

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

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Step 3.5.7.4.5.2.1.1.1

Apply the distributive property.

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

Step 3.5.7.4.5.2.1.1.2

Simplify the expression.

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Step 3.5.7.4.5.2.1.1.2.1

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

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

Step 3.5.7.4.5.2.1.1.2.2

Reorder $- 2 t$ and $- 2 C$ .

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

Step 3.5.7.4.5.2.1.1.2.3

Reorder $- 2 t$ and $- 2 C$ .

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

Step 3.5.7.4.5.2.2

Simplify the right side.

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Step 3.5.7.4.5.2.2.1

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

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Step 3.5.7.4.5.2.2.1.1

Cancel the common factor.

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

Step 3.5.7.4.5.2.2.1.2

Rewrite the expression.

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

Step 3.5.7.4.5.3

Solve for $y$ .

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Step 3.5.7.4.5.3.1

Subtract $y$ from both sides of the equation.

$y e^{- 2 C - 2 t} - e^{- 2 C - 2 t} - y = 1$

Step 3.5.7.4.5.3.2

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

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

Step 3.5.7.4.5.3.3

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

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Step 3.5.7.4.5.3.3.1

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

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

Step 3.5.7.4.5.3.3.2

Factor $y$ out of $- y$ .

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

Step 3.5.7.4.5.3.3.3

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

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

Step 3.5.7.4.5.3.4

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

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

Step 3.5.7.4.5.3.5

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

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

Step 3.5.7.4.5.3.6

Factor.

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Step 3.5.7.4.5.3.6.1

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = e^{- C - t}$ and $b = 1$ .

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

Step 3.5.7.4.5.3.6.2

Remove unnecessary parentheses.

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

Step 3.5.7.4.5.3.7

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

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Step 3.5.7.4.5.3.7.1

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

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

Step 3.5.7.4.5.3.7.2

Simplify the left side.

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Step 3.5.7.4.5.3.7.2.1

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

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Step 3.5.7.4.5.3.7.2.1.1

Cancel the common factor.

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

Step 3.5.7.4.5.3.7.2.1.2

Rewrite the expression.

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

Step 3.5.7.4.5.3.7.2.2

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

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Step 3.5.7.4.5.3.7.2.2.1

Cancel the common factor.

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

Step 3.5.7.4.5.3.7.2.2.2

Divide $y$ by $1$ .

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

Step 3.5.7.4.5.3.7.3

Simplify the right side.

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Step 3.5.7.4.5.3.7.3.1

Combine the numerators over the common denominator.

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

Step 3.5.7.4.6

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

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Step 3.5.7.4.6.1

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

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

Step 3.5.7.4.6.2

Simplify.

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Step 3.5.7.4.6.2.1

Simplify the left side.

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Step 3.5.7.4.6.2.1.1

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

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Step 3.5.7.4.6.2.1.1.1

Apply the distributive property.

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

Step 3.5.7.4.6.2.1.1.2

Simplify the expression.

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Step 3.5.7.4.6.2.1.1.2.1

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

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

Step 3.5.7.4.6.2.1.1.2.2

Reorder $- 2 t$ and $- 2 C$ .

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

Step 3.5.7.4.6.2.1.1.2.3

Reorder $- 2 t$ and $- 2 C$ .

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

Step 3.5.7.4.6.2.2

Simplify the right side.

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Step 3.5.7.4.6.2.2.1

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

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Step 3.5.7.4.6.2.2.1.1

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

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Step 3.5.7.4.6.2.2.1.1.1

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

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

Step 3.5.7.4.6.2.2.1.1.2

Cancel the common factor.

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

Step 3.5.7.4.6.2.2.1.1.3

Rewrite the expression.

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

Step 3.5.7.4.6.2.2.1.2

Apply the distributive property.

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

Step 3.5.7.4.6.2.2.1.3

Multiply $- 1$ by $1$ .

$y e^{- 2 C - 2 t} - e^{- 2 C - 2 t} = - y - 1$

Step 3.5.7.4.6.3

Solve for $y$ .

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Step 3.5.7.4.6.3.1

Add $y$ to both sides of the equation.

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

Step 3.5.7.4.6.3.2

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

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

Step 3.5.7.4.6.3.3

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

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Step 3.5.7.4.6.3.3.1

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

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

Step 3.5.7.4.6.3.3.2

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

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

Step 3.5.7.4.6.3.3.3

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

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

Step 3.5.7.4.6.3.3.4

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

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

Step 3.5.7.4.6.3.4

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

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Step 3.5.7.4.6.3.4.1

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

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

Step 3.5.7.4.6.3.4.2

Simplify the left side.

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Step 3.5.7.4.6.3.4.2.1

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

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Step 3.5.7.4.6.3.4.2.1.1

Cancel the common factor.

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

Step 3.5.7.4.6.3.4.2.1.2

Divide $y$ by $1$ .

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

Step 3.5.7.4.6.3.4.3

Simplify the right side.

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Step 3.5.7.4.6.3.4.3.1

Combine the numerators over the common denominator.

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

Step 3.5.7.4.6.3.4.3.2

Simplify the numerator.

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Step 3.5.7.4.6.3.4.3.2.1

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

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

Step 3.5.7.4.6.3.4.3.2.2

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

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

Step 3.5.7.4.6.3.4.3.2.3

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

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

Step 3.5.7.4.6.3.4.3.2.4

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = e^{- C - t}$ and $b = 1$ .

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

Step 3.5.7.4.7