Calculus Examples

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

Step 1

Separate the variables.

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Step 1.1

Multiply both sides by $\frac{1}{5 y + y^{2}}$ .

$\frac{1}{5 y + y^{2}} \frac{d y}{d x} = \frac{1}{5 y + y^{2}} (5 y + y^{2})$

Step 1.2

Cancel the common factor of $5 y + y^{2}$ .

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Step 1.2.1

Cancel the common factor.

$\frac{1}{5 y + y^{2}} \frac{d y}{d x} = \frac{1}{5 y + y^{2}} (5 y + y^{2})$

Step 1.2.2

Rewrite the expression.

$\frac{1}{5 y + y^{2}} \frac{d y}{d x} = 1$

Step 1.3

Rewrite the equation.

$\frac{1}{5 y + y^{2}} d y = 1 d x$

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

$\int \frac{1}{5 y + y^{2}} d y = \int d x$

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 $y$ out of $5 y + y^{2}$ .

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Step 2.2.1.1.1.1

Factor $y$ out of $5 y$ .

$\frac{1}{y \cdot 5 + y^{2}}$

Step 2.2.1.1.1.2

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

$\frac{1}{y \cdot 5 + y \cdot y}$

Step 2.2.1.1.1.3

Factor $y$ out of $y \cdot 5 + y \cdot y$ .

$\frac{1}{y \cdot (5 + y)}$

Step 2.2.1.1.1.4

Multiply $y$ by $5 + y$ .

$\frac{1}{y (5 + y)}$

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 $B$ .

$\frac{A}{y} + \frac{B}{5 + y}$

Step 2.2.1.1.3

Multiply each fraction in the equation by the denominator of the original expression. In this case, the denominator is $y (5 + y)$ .

$\frac{1 (y (5 + y))}{y (5 + y)} = \frac{A (y (5 + y))}{y} + \frac{(B) (y (5 + y))}{5 + y}$

Step 2.2.1.1.4

Cancel the common factor of $y$ .

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Step 2.2.1.1.4.1

Cancel the common factor.

$\frac{1 (y (5 + y))}{y (5 + y)} = \frac{A (y (5 + y))}{y} + \frac{(B) (y (5 + y))}{5 + y}$

Step 2.2.1.1.4.2

Rewrite the expression.

$\frac{1 (5 + y)}{5 + y} = \frac{A (y (5 + y))}{y} + \frac{(B) (y (5 + y))}{5 + y}$

Step 2.2.1.1.5

Cancel the common factor of $5 + y$ .

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Step 2.2.1.1.5.1

Cancel the common factor.

$\frac{1 (5 + y)}{5 + y} = \frac{A (y (5 + y))}{y} + \frac{(B) (y (5 + y))}{5 + y}$

Step 2.2.1.1.5.2

Rewrite the expression.

$1 = \frac{A (y (5 + y))}{y} + \frac{(B) (y (5 + y))}{5 + y}$

Step 2.2.1.1.6

Simplify each term.

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Step 2.2.1.1.6.1

Cancel the common factor of $y$ .

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Step 2.2.1.1.6.1.1

Cancel the common factor.

$1 = \frac{A (y (5 + y))}{y} + \frac{(B) (y (5 + y))}{5 + y}$

Step 2.2.1.1.6.1.2

Divide $A (5 + y)$ by $1$ .

$1 = A (5 + y) + \frac{(B) (y (5 + y))}{5 + y}$

Step 2.2.1.1.6.2

Apply the distributive property.

$1 = A \cdot 5 + A y + \frac{(B) (y (5 + y))}{5 + y}$

Step 2.2.1.1.6.3

Move $5$ to the left of $A$ .

$1 = 5 \cdot A + A y + \frac{(B) (y (5 + y))}{5 + y}$

Step 2.2.1.1.6.4

Cancel the common factor of $5 + y$ .

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Step 2.2.1.1.6.4.1

Cancel the common factor.

$1 = 5 A + A y + \frac{B (y (5 + y))}{5 + y}$

Step 2.2.1.1.6.4.2

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

$1 = 5 A + A y + (B) (y)$

$1 = 5 A + A y + B y$

Step 2.2.1.1.7

Move $5 A$ .

$1 = A y + B y + 5 A$

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.

$1 = 5 A$

Step 2.2.1.2.3

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

$0 = A + B$

$1 = 5 A$

$0 = A + B$

$1 = 5 A$

Step 2.2.1.3

Solve the system of equations.

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Step 2.2.1.3.1

Solve for $A$ in $1 = 5 A$ .

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Step 2.2.1.3.1.1

Rewrite the equation as $5 A = 1$ .

$5 A = 1$

$0 = A + B$

Step 2.2.1.3.1.2

Divide each term in $5 A = 1$ by $5$ and simplify.

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Step 2.2.1.3.1.2.1

Divide each term in $5 A = 1$ by $5$ .

$\frac{5 A}{5} = \frac{1}{5}$

$0 = A + B$

Step 2.2.1.3.1.2.2

Simplify the left side.

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Step 2.2.1.3.1.2.2.1

Cancel the common factor of $5$ .

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Step 2.2.1.3.1.2.2.1.1

Cancel the common factor.

$\frac{5 A}{5} = \frac{1}{5}$

$0 = A + B$

Step 2.2.1.3.1.2.2.1.2

Divide $A$ by $1$ .

$A = \frac{1}{5}$

$0 = A + B$

$A = \frac{1}{5}$

$0 = A + B$

$A = \frac{1}{5}$

$0 = A + B$

$A = \frac{1}{5}$

$0 = A + B$

$A = \frac{1}{5}$

$0 = A + B$

Step 2.2.1.3.2

Replace all occurrences of $A$ with $\frac{1}{5}$ in each equation.

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Step 2.2.1.3.2.1

Replace all occurrences of $A$ in $0 = A + B$ with $\frac{1}{5}$ .

$0 = (\frac{1}{5}) + B$

$A = \frac{1}{5}$

Step 2.2.1.3.2.2

Simplify the right side.

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Step 2.2.1.3.2.2.1

Remove parentheses.

$0 = \frac{1}{5} + B$

$A = \frac{1}{5}$

$0 = \frac{1}{5} + B$

$A = \frac{1}{5}$

$0 = \frac{1}{5} + B$

$A = \frac{1}{5}$

Step 2.2.1.3.3

Solve for $B$ in $0 = \frac{1}{5} + B$ .

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Step 2.2.1.3.3.1

Rewrite the equation as $\frac{1}{5} + B = 0$ .

$\frac{1}{5} + B = 0$

$A = \frac{1}{5}$

Step 2.2.1.3.3.2

Subtract $\frac{1}{5}$ from both sides of the equation.

$B = - \frac{1}{5}$

$A = \frac{1}{5}$

$B = - \frac{1}{5}$

$A = \frac{1}{5}$

Step 2.2.1.3.4

Solve the system of equations.

$B = - \frac{1}{5} A = \frac{1}{5}$

Step 2.2.1.3.5

List all of the solutions.

$B = - \frac{1}{5}, A = \frac{1}{5}$

Step 2.2.1.4

Replace each of the partial fraction coefficients in $\frac{A}{y} + \frac{B}{5 + y}$ with the values found for $A$ and $B$ .

$\frac{\frac{1}{5}}{y} + \frac{- \frac{1}{5}}{5 + y}$

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.

$\frac{1}{5} \cdot \frac{1}{y} + \frac{- \frac{1}{5}}{5 + y}$

Step 2.2.1.5.2

Multiply $\frac{1}{5}$ by $\frac{1}{y}$ .

$\frac{1}{5 y} + \frac{- \frac{1}{5}}{5 + y}$

Step 2.2.1.5.3

Multiply the numerator by the reciprocal of the denominator.

$\frac{1}{5 y} - \frac{1}{5} \cdot \frac{1}{5 + y}$

Step 2.2.1.5.4

Multiply $\frac{1}{5 + y}$ by $\frac{1}{5}$ .

$\frac{1}{5 y} - \frac{1}{(5 + y) \cdot 5}$

Step 2.2.1.5.5

Move $5$ to the left of $5 + y$ .

$\int \frac{1}{5 y} - \frac{1}{5 (5 + y)} d y = \int d x$

Step 2.2.2

Split the single integral into multiple integrals.

$\int \frac{1}{5 y} d y + \int - \frac{1}{5 (5 + y)} d y = \int d x$

Step 2.2.3

Since $\frac{1}{5}$ is constant with respect to $y$ , move $\frac{1}{5}$ out of the integral.

$\frac{1}{5} \int \frac{1}{y} d y + \int - \frac{1}{5 (5 + y)} d y = \int d x$

Step 2.2.4

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

$\frac{1}{5} (\ln (| y |) + C_{1}) + \int - \frac{1}{5 (5 + y)} d y = \int d x$

Step 2.2.5

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

$\frac{1}{5} (\ln (| y |) + C_{1}) - \int \frac{1}{5 (5 + y)} d y = \int d x$

Step 2.2.6

Since $\frac{1}{5}$ is constant with respect to $y$ , move $\frac{1}{5}$ out of the integral.

$\frac{1}{5} (\ln (| y |) + C_{1}) - (\frac{1}{5} \int \frac{1}{5 + y} d y) = \int d x$

Step 2.2.7

Let $u = 5 + y$ . Then $d u = d y$ . Rewrite using $u$ and $d$ $u$ .

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Step 2.2.7.1

Let $u = 5 + y$ . Find $\frac{d u}{d y}$ .

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Step 2.2.7.1.1

Differentiate $5 + y$ .

$\frac{d}{d y} [5 + y]$

Step 2.2.7.1.2

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

$\frac{d}{d y} [5] + \frac{d}{d y} [y]$

Step 2.2.7.1.3

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

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

Step 2.2.7.1.4

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 2.2.7.1.5

Add $0$ and $1$ .

$1$

Step 2.2.7.2

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

$\frac{1}{5} (\ln (| y |) + C_{1}) - \frac{1}{5} \int \frac{1}{u} d u = \int d x$

Step 2.2.8

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

$\frac{1}{5} (\ln (| y |) + C_{1}) - \frac{1}{5} (\ln (| u |) + C_{2}) = \int d x$

Step 2.2.9

Simplify.

$\frac{1}{5} \ln (| y |) - \frac{1}{5} \ln (| u |) + C_{3} = \int d x$

Step 2.2.10

Replace all occurrences of $u$ with $5 + y$ .

$\frac{1}{5} \ln (| y |) - \frac{1}{5} \ln (| 5 + y |) + C_{3} = \int d x$

Step 2.3

Apply the constant rule.

$\frac{1}{5} \ln (| y |) - \frac{1}{5} \ln (| 5 + y |) + C_{3} = x + C_{4}$

Step 2.4

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

$\frac{1}{5} \ln (| y |) - \frac{1}{5} \ln (| 5 + y |) = x + C$

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 $\frac{1}{5}$ and $\ln (| y |)$ .

$\frac{\ln (| y |)}{5} - \frac{1}{5} \ln (| 5 + y |) = x + C$

Step 3.1.1.2

Combine $\ln (| 5 + y |)$ and $\frac{1}{5}$ .

$\frac{\ln (| y |)}{5} - \frac{\ln (| 5 + y |)}{5} = x + C$

Step 3.2

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

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Step 3.2.1

Multiply each term in $\frac{\ln (| y |)}{5} - \frac{\ln (| 5 + y |)}{5} = x + C$ by $5$ .

$\frac{\ln (| y |)}{5} \cdot 5 - \frac{\ln (| 5 + y |)}{5} \cdot 5 = x \cdot 5 + C \cdot 5$

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 $5$ .

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Step 3.2.2.1.1.1

Cancel the common factor.

$\frac{\ln (| y |)}{5} \cdot 5 - \frac{\ln (| 5 + y |)}{5} \cdot 5 = x \cdot 5 + C \cdot 5$

Step 3.2.2.1.1.2

Rewrite the expression.

$\ln (| y |) - \frac{\ln (| 5 + y |)}{5} \cdot 5 = x \cdot 5 + C \cdot 5$

Step 3.2.2.1.2

Cancel the common factor of $5$ .

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Step 3.2.2.1.2.1

Move the leading negative in $- \frac{\ln (| 5 + y |)}{5}$ into the numerator.

$\ln (| y |) + \frac{- \ln (| 5 + y |)}{5} \cdot 5 = x \cdot 5 + C \cdot 5$

Step 3.2.2.1.2.2

Cancel the common factor.

$\ln (| y |) + \frac{- \ln (| 5 + y |)}{5} \cdot 5 = x \cdot 5 + C \cdot 5$

Step 3.2.2.1.2.3

Rewrite the expression.

$\ln (| y |) - \ln (| 5 + y |) = x \cdot 5 + C \cdot 5$

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 $5$ to the left of $x$ .

$\ln (| y |) - \ln (| 5 + y |) = 5 \cdot x + C \cdot 5$

Step 3.2.3.1.2

Move $5$ to the left of $C$ .

$\ln (| y |) - \ln (| 5 + y |) = 5 x + 5 C$

Step 3.3

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

$\ln (\frac{| y |}{| 5 + y |}) = 5 x + 5 C$

Step 3.4

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

$e^{\ln (\frac{| y |}{| 5 + y |})} = e^{5 x + 5 C}$

Step 3.5

Rewrite $\ln (\frac{| y |}{| 5 + y |}) = 5 x + 5 C$ 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 $b^{y} = x$ .

$e^{5 x + 5 C} = \frac{| y |}{| 5 + y |}$

Step 3.6

Solve for $y$ .

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Step 3.6.1

Rewrite the equation as $\frac{| y |}{| 5 + y |} = e^{5 x + 5 C}$ .

$\frac{| y |}{| 5 + y |} = e^{5 x + 5 C}$

Step 3.6.2

Multiply both sides by $| 5 + y |$ .

$\frac{| y |}{| 5 + y |} | 5 + y | = e^{5 x + 5 C} | 5 + y |$

Step 3.6.3

Simplify the left side.

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Step 3.6.3.1

Cancel the common factor of $| 5 + y |$ .

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Step 3.6.3.1.1

Cancel the common factor.

$\frac{| y |}{| 5 + y |} | 5 + y | = e^{5 x + 5 C} | 5 + y |$

Step 3.6.3.1.2

Rewrite the expression.

$| y | = e^{5 x + 5 C} | 5 + y |$

Step 3.6.4

Solve for $y$ .

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Step 3.6.4.1

Rewrite the equation as $e^{5 x + 5 C} | 5 + y | = | y |$ .

$e^{5 x + 5 C} | 5 + y | = | y |$

Step 3.6.4.2

Divide each term in $e^{5 x + 5 C} | 5 + y | = | y |$ by $e^{5 x + 5 C}$ and simplify.

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Step 3.6.4.2.1

Divide each term in $e^{5 x + 5 C} | 5 + y | = | y |$ by $e^{5 x + 5 C}$ .

$\frac{e^{5 x + 5 C} | 5 + y |}{e^{5 x + 5 C}} = \frac{| y |}{e^{5 x + 5 C}}$

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 $e^{5 x + 5 C}$ .

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Step 3.6.4.2.2.1.1

Cancel the common factor.

$\frac{e^{5 x + 5 C} | 5 + y |}{e^{5 x + 5 C}} = \frac{| y |}{e^{5 x + 5 C}}$

Step 3.6.4.2.2.1.2

Divide $| 5 + y |$ by $1$ .

$| 5 + y | = \frac{| y |}{e^{5 x + 5 C}}$

Step 3.6.4.3

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

$5 + y = \frac{y}{e^{5 x + 5 C}}$

$5 + y = - \frac{y}{e^{5 x + 5 C}}$

$- (5 + y) = \frac{y}{e^{5 x + 5 C}}$

$- (5 + y) = - \frac{y}{e^{5 x + 5 C}}$

Step 3.6.4.4

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

$5 + y = \frac{y}{e^{5 x + 5 C}}$

$5 + y = - \frac{y}{e^{5 x + 5 C}}$

Step 3.6.4.5

Solve $5 + y = \frac{y}{e^{5 x + 5 C}}$ for $y$ .

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Step 3.6.4.5.1

Multiply both sides by $e^{5 x + 5 C}$ .

$(5 + y) e^{5 x + 5 C} = \frac{y}{e^{5 x + 5 C}} e^{5 x + 5 C}$

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 $(5 + y) e^{5 x + 5 C}$ .

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Step 3.6.4.5.2.1.1.1

Apply the distributive property.

$5 e^{5 x + 5 C} + y e^{5 x + 5 C} = \frac{y}{e^{5 x + 5 C}} e^{5 x + 5 C}$

Step 3.6.4.5.2.1.1.2

Simplify with commuting.

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Step 3.6.4.5.2.1.1.2.1

Reorder $5 x$ and $5 C$ .

$5 e^{5 C + 5 x} + y e^{5 x + 5 C} = \frac{y}{e^{5 x + 5 C}} e^{5 x + 5 C}$

Step 3.6.4.5.2.1.1.2.2

Reorder $5 x$ and $5 C$ .

$5 e^{5 C + 5 x} + y e^{5 C + 5 x} = \frac{y}{e^{5 x + 5 C}} e^{5 x + 5 C}$

Step 3.6.4.5.2.1.1.2.3

Reorder $5 e^{5 C + 5 x}$ and $y e^{5 C + 5 x}$ .

$y e^{5 C + 5 x} + 5 e^{5 C + 5 x} = \frac{y}{e^{5 x + 5 C}} e^{5 x + 5 C}$

Step 3.6.4.5.2.2

Simplify the right side.

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Step 3.6.4.5.2.2.1

Cancel the common factor of $e^{5 x + 5 C}$ .

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Step 3.6.4.5.2.2.1.1

Cancel the common factor.

$y e^{5 C + 5 x} + 5 e^{5 C + 5 x} = \frac{y}{e^{5 x + 5 C}} e^{5 x + 5 C}$

Step 3.6.4.5.2.2.1.2

Rewrite the expression.

$y e^{5 C + 5 x} + 5 e^{5 C + 5 x} = y$

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.

$y e^{5 C + 5 x} + 5 e^{5 C + 5 x} - y = 0$

Step 3.6.4.5.3.2

Subtract $5 e^{5 C + 5 x}$ from both sides of the equation.

$y e^{5 C + 5 x} - y = - 5 e^{5 C + 5 x}$

Step 3.6.4.5.3.3

Factor $y$ out of $y e^{5 C + 5 x} - y$ .

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Step 3.6.4.5.3.3.1

Factor $y$ out of $y e^{5 C + 5 x}$ .

$y e^{5 C + 5 x} - y = - 5 e^{5 C + 5 x}$

Step 3.6.4.5.3.3.2

Factor $y$ out of $- y$ .

$y e^{5 C + 5 x} + y \cdot - 1 = - 5 e^{5 C + 5 x}$

Step 3.6.4.5.3.3.3

Factor $y$ out of $y e^{5 C + 5 x} + y \cdot - 1$ .

$y (e^{5 C + 5 x} - 1) = - 5 e^{5 C + 5 x}$

Step 3.6.4.5.3.4

Divide each term in $y (e^{5 C + 5 x} - 1) = - 5 e^{5 C + 5 x}$ by $e^{5 C + 5 x} - 1$ and simplify.

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Step 3.6.4.5.3.4.1

Divide each term in $y (e^{5 C + 5 x} - 1) = - 5 e^{5 C + 5 x}$ by $e^{5 C + 5 x} - 1$ .

$\frac{y (e^{5 C + 5 x} - 1)}{e^{5 C + 5 x} - 1} = \frac{- 5 e^{5 C + 5 x}}{e^{5 C + 5 x} - 1}$

Step 3.6.4.5.3.4.2

Simplify the left side.

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Step 3.6.4.5.3.4.2.1

Cancel the common factor of $e^{5 C + 5 x} - 1$ .

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Step 3.6.4.5.3.4.2.1.1

Cancel the common factor.

$\frac{y (e^{5 C + 5 x} - 1)}{e^{5 C + 5 x} - 1} = \frac{- 5 e^{5 C + 5 x}}{e^{5 C + 5 x} - 1}$

Step 3.6.4.5.3.4.2.1.2

Divide $y$ by $1$ .

$y = \frac{- 5 e^{5 C + 5 x}}{e^{5 C + 5 x} - 1}$

Step 3.6.4.5.3.4.3

Simplify the right side.

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Step 3.6.4.5.3.4.3.1

Move the negative in front of the fraction.

$y = - \frac{5 e^{5 C + 5 x}}{e^{5 C + 5 x} - 1}$

Step 3.6.4.6

Solve $5 + y = - \frac{y}{e^{5 x + 5 C}}$ for $y$ .

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Step 3.6.4.6.1

Multiply both sides by $e^{5 x + 5 C}$ .

$(5 + y) e^{5 x + 5 C} = - \frac{y}{e^{5 x + 5 C}} e^{5 x + 5 C}$

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 $(5 + y) e^{5 x + 5 C}$ .

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Step 3.6.4.6.2.1.1.1

Apply the distributive property.

$5 e^{5 x + 5 C} + y e^{5 x + 5 C} = - \frac{y}{e^{5 x + 5 C}} e^{5 x + 5 C}$

Step 3.6.4.6.2.1.1.2

Simplify with commuting.

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Step 3.6.4.6.2.1.1.2.1

Reorder $5 x$ and $5 C$ .

$5 e^{5 C + 5 x} + y e^{5 x + 5 C} = - \frac{y}{e^{5 x + 5 C}} e^{5 x + 5 C}$

Step 3.6.4.6.2.1.1.2.2

Reorder $5 x$ and $5 C$ .

$5 e^{5 C + 5 x} + y e^{5 C + 5 x} = - \frac{y}{e^{5 x + 5 C}} e^{5 x + 5 C}$

Step 3.6.4.6.2.1.1.2.3

Reorder $5 e^{5 C + 5 x}$ and $y e^{5 C + 5 x}$ .

$y e^{5 C + 5 x} + 5 e^{5 C + 5 x} = - \frac{y}{e^{5 x + 5 C}} e^{5 x + 5 C}$

Step 3.6.4.6.2.2

Simplify the right side.

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Step 3.6.4.6.2.2.1

Cancel the common factor of $e^{5 x + 5 C}$ .

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Step 3.6.4.6.2.2.1.1

Move the leading negative in $- \frac{y}{e^{5 x + 5 C}}$ into the numerator.

$y e^{5 C + 5 x} + 5 e^{5 C + 5 x} = \frac{- y}{e^{5 x + 5 C}} e^{5 x + 5 C}$

Step 3.6.4.6.2.2.1.2

Cancel the common factor.

$y e^{5 C + 5 x} + 5 e^{5 C + 5 x} = \frac{- y}{e^{5 x + 5 C}} e^{5 x + 5 C}$

Step 3.6.4.6.2.2.1.3

Rewrite the expression.

$y e^{5 C + 5 x} + 5 e^{5 C + 5 x} = - y$

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.

$y e^{5 C + 5 x} + 5 e^{5 C + 5 x} + y = 0$

Step 3.6.4.6.3.2

Subtract $5 e^{5 C + 5 x}$ from both sides of the equation.

$y e^{5 C + 5 x} + y = - 5 e^{5 C + 5 x}$

Step 3.6.4.6.3.3

Factor $y$ out of $y e^{5 C + 5 x} + y$ .

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Step 3.6.4.6.3.3.1

Factor $y$ out of $y e^{5 C + 5 x}$ .

$y e^{5 C + 5 x} + y = - 5 e^{5 C + 5 x}$

Step 3.6.4.6.3.3.2

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

$y e^{5 C + 5 x} + y = - 5 e^{5 C + 5 x}$

Step 3.6.4.6.3.3.3

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

$y e^{5 C + 5 x} + y \cdot 1 = - 5 e^{5 C + 5 x}$

Step 3.6.4.6.3.3.4

Factor $y$ out of $y e^{5 C + 5 x} + y \cdot 1$ .

$y (e^{5 C + 5 x} + 1) = - 5 e^{5 C + 5 x}$

Step 3.6.4.6.3.4

Divide each term in $y (e^{5 C + 5 x} + 1) = - 5 e^{5 C + 5 x}$ by $e^{5 C + 5 x} + 1$ and simplify.

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Step 3.6.4.6.3.4.1

Divide each term in $y (e^{5 C + 5 x} + 1) = - 5 e^{5 C + 5 x}$ by $e^{5 C + 5 x} + 1$ .

$\frac{y (e^{5 C + 5 x} + 1)}{e^{5 C + 5 x} + 1} = \frac{- 5 e^{5 C + 5 x}}{e^{5 C + 5 x} + 1}$

Step 3.6.4.6.3.4.2

Simplify the left side.

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Step 3.6.4.6.3.4.2.1

Cancel the common factor of $e^{5 C + 5 x} + 1$ .

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Step 3.6.4.6.3.4.2.1.1

Cancel the common factor.

$\frac{y (e^{5 C + 5 x} + 1)}{e^{5 C + 5 x} + 1} = \frac{- 5 e^{5 C + 5 x}}{e^{5 C + 5 x} + 1}$

Step 3.6.4.6.3.4.2.1.2

Divide $y$ by $1$ .

$y = \frac{- 5 e^{5 C + 5 x}}{e^{5 C + 5 x} + 1}$

Step 3.6.4.6.3.4.3

Simplify the right side.

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Step 3.6.4.6.3.4.3.1

Move the negative in front of the fraction.

$y = - \frac{5 e^{5 C + 5 x}}{e^{5 C + 5 x} + 1}$

Step 3.6.4.7

List all of the solutions.

$y = - \frac{5 e^{5 C + 5 x}}{e^{5 C + 5 x} - 1}$

$y = - \frac{5 e^{5 C + 5 x}}{e^{5 C + 5 x} + 1}$

$y = - \frac{5 e^{5 C + 5 x}}{e^{5 C + 5 x} - 1}$

$y = - \frac{5 e^{5 C + 5 x}}{e^{5 C + 5 x} + 1}$

$y = - \frac{5 e^{5 C + 5 x}}{e^{5 C + 5 x} - 1}$

$y = - \frac{5 e^{5 C + 5 x}}{e^{5 C + 5 x} + 1}$

$y = - \frac{5 e^{5 C + 5 x}}{e^{5 C + 5 x} - 1}$

$y = - \frac{5 e^{5 C + 5 x}}{e^{5 C + 5 x} + 1}$

Step 4

Group the constant terms together.

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Step 4.1

Simplify the constant of integration.

$y = - \frac{5 e^{C + 5 x}}{e^{C + 5 x} - 1}$

$y = - \frac{5 e^{C + 5 x}}{e^{C + 5 x} + 1}$

Step 4.2

Reorder terms.

$y = - \frac{5 e^{5 x + C}}{e^{C + 5 x} - 1}$

$y = - \frac{5 e^{C + 5 x}}{e^{C + 5 x} + 1}$

Step 4.3

Rewrite $e^{5 x + C}$ as $e^{5 x} e^{C}$ .

$y = - \frac{5 (e^{5 x} e^{C})}{e^{C + 5 x} - 1}$

$y = - \frac{5 e^{C + 5 x}}{e^{C + 5 x} + 1}$

Step 4.4

Reorder $e^{5 x}$ and $e^{C}$ .

$y = - \frac{5 (e^{C} e^{5 x})}{e^{C + 5 x} - 1}$

$y = - \frac{5 e^{C + 5 x}}{e^{C + 5 x} + 1}$

Step 4.5

Reorder terms.

$y = - \frac{5 (e^{C} e^{5 x})}{e^{5 x + C} - 1}$

$y = - \frac{5 e^{C + 5 x}}{e^{C + 5 x} + 1}$

Step 4.6

Rewrite $e^{5 x + C}$ as $e^{5 x} e^{C}$ .

$y = - \frac{5 (e^{C} e^{5 x})}{e^{5 x} e^{C} - 1}$

$y = - \frac{5 e^{C + 5 x}}{e^{C + 5 x} + 1}$

Step 4.7

Reorder $e^{5 x}$ and $e^{C}$ .