Calculus Examples

$e^{x} (y - 1) d x + 2 (e^{x} + 4) d y = 0$

Step 1

Subtract $e^{x} (y - 1) d x$ from both sides of the equation.

$2 (e^{x} + 4) d y = - e^{x} (y - 1) d x$

Step 2

Multiply both sides by $\frac{1}{(e^{x} + 4) (y - 1)}$ .

$\frac{1}{(e^{x} + 4) (y - 1)} (2 (e^{x} + 4)) d y = \frac{1}{(e^{x} + 4) (y - 1)} (- e^{x} (y - 1)) d x$

Step 3

Simplify.

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Step 3.1

Rewrite using the commutative property of multiplication.

$2 \frac{1}{(e^{x} + 4) (y - 1)} (e^{x} + 4) d y = \frac{1}{(e^{x} + 4) (y - 1)} (- e^{x} (y - 1)) d x$

Step 3.2

Combine $2$ and $\frac{1}{(e^{x} + 4) (y - 1)}$ .

$\frac{2}{(e^{x} + 4) (y - 1)} (e^{x} + 4) d y = \frac{1}{(e^{x} + 4) (y - 1)} (- e^{x} (y - 1)) d x$

Step 3.3

Cancel the common factor of $e^{x} + 4$ .

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Step 3.3.1

Cancel the common factor.

$\frac{2}{(e^{x} + 4) (y - 1)} (e^{x} + 4) d y = \frac{1}{(e^{x} + 4) (y - 1)} (- e^{x} (y - 1)) d x$

Step 3.3.2

Rewrite the expression.

$\frac{2}{y - 1} d y = \frac{1}{(e^{x} + 4) (y - 1)} (- e^{x} (y - 1)) d x$

Step 3.4

Rewrite using the commutative property of multiplication.

$\frac{2}{y - 1} d y = - \frac{1}{(e^{x} + 4) (y - 1)} (e^{x} (y - 1)) d x$

Step 3.5

Cancel the common factor of $y - 1$ .

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Step 3.5.1

Move the leading negative in $- \frac{1}{(e^{x} + 4) (y - 1)}$ into the numerator.

$\frac{2}{y - 1} d y = \frac{- 1}{(e^{x} + 4) (y - 1)} (e^{x} (y - 1)) d x$

Step 3.5.2

Factor $y - 1$ out of $(e^{x} + 4) (y - 1)$ .

$\frac{2}{y - 1} d y = \frac{- 1}{(y - 1) (e^{x} + 4)} (e^{x} (y - 1)) d x$

Step 3.5.3

Factor $y - 1$ out of $e^{x} (y - 1)$ .

$\frac{2}{y - 1} d y = \frac{- 1}{(y - 1) (e^{x} + 4)} ((y - 1) e^{x}) d x$

Step 3.5.4

Cancel the common factor.

$\frac{2}{y - 1} d y = \frac{- 1}{(y - 1) (e^{x} + 4)} ((y - 1) e^{x}) d x$

Step 3.5.5

Rewrite the expression.

$\frac{2}{y - 1} d y = \frac{- 1}{e^{x} + 4} e^{x} d x$

Step 3.6

Combine $\frac{- 1}{e^{x} + 4}$ and $e^{x}$ .

$\frac{2}{y - 1} d y = \frac{- e^{x}}{e^{x} + 4} d x$

Step 3.7

Move the negative in front of the fraction.

$\frac{2}{y - 1} d y = - \frac{e^{x}}{e^{x} + 4} d x$

Step 4

Integrate both sides.

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Step 4.1

Set up an integral on each side.

$\int \frac{2}{y - 1} d y = \int - \frac{e^{x}}{e^{x} + 4} d x$

Step 4.2

Integrate the left side.

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Step 4.2.1

Since $2$ is constant with respect to $y$ , move $2$ out of the integral.

$2 \int \frac{1}{y - 1} d y = \int - \frac{e^{x}}{e^{x} + 4} d x$

Step 4.2.2

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

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Step 4.2.2.1

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

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Step 4.2.2.1.1

Differentiate $y - 1$ .

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

Step 4.2.2.1.2

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 4.2.2.1.3

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 4.2.2.1.4

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

$1 + 0$

Step 4.2.2.1.5

Add $1$ and $0$ .

$1$

Step 4.2.2.2

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

$2 \int \frac{1}{u_{1}} d u_{1} = \int - \frac{e^{x}}{e^{x} + 4} d x$

Step 4.2.3

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

$2 (\ln (| u_{1} |) + C_{1}) = \int - \frac{e^{x}}{e^{x} + 4} d x$

Step 4.2.4

Simplify.

$2 \ln (| u_{1} |) + C_{1} = \int - \frac{e^{x}}{e^{x} + 4} d x$

Step 4.2.5

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

$2 \ln (| y - 1 |) + C_{1} = \int - \frac{e^{x}}{e^{x} + 4} d x$

Step 4.3

Integrate the right side.

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Step 4.3.1

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

$2 \ln (| y - 1 |) + C_{1} = - \int \frac{e^{x}}{e^{x} + 4} d x$

Step 4.3.2

Let $u_{2} = e^{x} + 4$ . Then $d u_{2} = e^{x} d x$ , so $\frac{1}{e^{x}} d u_{2} = d x$ . Rewrite using $u_{2}$ and $d$ $u_{2}$ .

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Step 4.3.2.1

Let $u_{2} = e^{x} + 4$ . Find $\frac{d u_{2}}{d x}$ .

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Step 4.3.2.1.1

Differentiate $e^{x} + 4$ .

$\frac{d}{d x} [e^{x} + 4]$

Step 4.3.2.1.2

By the Sum Rule, the derivative of $e^{x} + 4$ with respect to $x$ is $\frac{d}{d x} [e^{x}] + \frac{d}{d x} [4]$ .

$\frac{d}{d x} [e^{x}] + \frac{d}{d x} [4]$

Step 4.3.2.1.3

Differentiate using the Exponential Rule which states that $\frac{d}{d x} [a^{x}]$ is $a^{x} \ln (a)$ where $a$ = $e$ .

$e^{x} + \frac{d}{d x} [4]$

Step 4.3.2.1.4

Differentiate using the Constant Rule.

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Step 4.3.2.1.4.1

Since $4$ is constant with respect to $x$ , the derivative of $4$ with respect to $x$ is $0$ .

$e^{x} + 0$

Step 4.3.2.1.4.2

Add $e^{x}$ and $0$ .

$e^{x}$

Step 4.3.2.2

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

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

Step 4.3.3

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

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

Step 4.3.4

Simplify.

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

Step 4.3.5

Replace all occurrences of $u_{2}$ with $e^{x} + 4$ .

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

Step 4.4

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

$2 \ln (| y - 1 |) = - \ln (| e^{x} + 4 |) + C$

Step 5

Solve for $y$ .

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Step 5.1

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

$2 \ln (| y - 1 |) + \ln (| e^{x} + 4 |) = C$

Step 5.2

Simplify the left side.

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Step 5.2.1

Simplify $2 \ln (| y - 1 |) + \ln (| e^{x} + 4 |)$ .

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Step 5.2.1.1

Simplify each term.

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Step 5.2.1.1.1

Simplify $2 \ln (| y - 1 |)$ by moving $2$ inside the logarithm.

$\ln ({| y - 1 |}^{2}) + \ln (| e^{x} + 4 |) = C$

Step 5.2.1.1.2

Remove the absolute value in ${| y - 1 |}^{2}$ because exponentiations with even powers are always positive.

$\ln ({(y - 1)}^{2}) + \ln (| e^{x} + 4 |) = C$

Step 5.2.1.2

Use the product property of logarithms, $\log_{b} (x) + \log_{b} (y) = \log_{b} (x y)$ .

$\ln ({(y - 1)}^{2} | e^{x} + 4 |) = C$

Step 5.2.1.3

Reorder factors in $\ln ({(y - 1)}^{2} | e^{x} + 4 |)$ .

$\ln (| e^{x} + 4 | {(y - 1)}^{2}) = C$

Step 5.3

Expand $\ln (| e^{x} + 4 | {(y - 1)}^{2})$ .

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Step 5.3.1

Rewrite $\ln (| e^{x} + 4 | {(y - 1)}^{2})$ as $\ln (| e^{x} + 4 |) + \ln ({(y - 1)}^{2})$ .

$\ln (| e^{x} + 4 |) + \ln ({(y - 1)}^{2})$

Step 5.3.2

Expand $\ln ({(y - 1)}^{2})$ by moving $2$ outside the logarithm.

$\ln (| e^{x} + 4 |) + 2 \ln (y - 1)$

Step 5.4

The expanded equation is $\ln (| e^{x} + 4 |) + 2 \ln (y - 1) = C$ .

$\ln (| e^{x} + 4 |) + 2 \ln (y - 1) = C$

Step 5.5

Subtract $\ln (| e^{x} + 4 |)$ from both sides of the equation.

$2 \ln (y - 1) = C - \ln (| e^{x} + 4 |)$

Step 5.6

Divide each term in $2 \ln (y - 1) = C - \ln (| e^{x} + 4 |)$ by $2$ and simplify.

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Step 5.6.1

Divide each term in $2 \ln (y - 1) = C - \ln (| e^{x} + 4 |)$ by $2$ .

$\frac{2 \ln (y - 1)}{2} = \frac{C}{2} + \frac{- \ln (| e^{x} + 4 |)}{2}$

Step 5.6.2

Simplify the left side.

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Step 5.6.2.1

Cancel the common factor of $2$ .

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Step 5.6.2.1.1

Cancel the common factor.

$\frac{2 \ln (y - 1)}{2} = \frac{C}{2} + \frac{- \ln (| e^{x} + 4 |)}{2}$

Step 5.6.2.1.2

Divide $\ln (y - 1)$ by $1$ .

$\ln (y - 1) = \frac{C}{2} + \frac{- \ln (| e^{x} + 4 |)}{2}$

Step 5.6.3

Simplify the right side.

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Step 5.6.3.1

Move the negative in front of the fraction.

$\ln (y - 1) = \frac{C}{2} - \frac{\ln (| e^{x} + 4 |)}{2}$

Step 5.7

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

$e^{\ln (y - 1)} = e^{\frac{C}{2} - \frac{\ln (| e^{x} + 4 |)}{2}}$

Step 5.8

Rewrite $\ln (y - 1) = \frac{C}{2} - \frac{\ln (| e^{x} + 4 |)}{2}$ 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 $b^{y} = x$ .

$e^{\frac{C}{2} - \frac{\ln (| e^{x} + 4 |)}{2}} = y - 1$

Step 5.9

Solve for $y$ .

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Step 5.9.1

Rewrite the equation as $y - 1 = e^{\frac{C}{2} - \frac{\ln (| e^{x} + 4 |)}{2}}$ .

$y - 1 = e^{\frac{C}{2} - \frac{\ln (| e^{x} + 4 |)}{2}}$

Step 5.9.2

Simplify $e^{\frac{C}{2} - \frac{\ln (| e^{x} + 4 |)}{2}}$ .

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Step 5.9.2.1

Rewrite.

$y - 1 = 0 + 0 + e^{\frac{C}{2} - \frac{\ln (| e^{x} + 4 |)}{2}}$

Step 5.9.2.2

Simplify by adding zeros.

$y - 1 = e^{\frac{C}{2} - \frac{\ln (| e^{x} + 4 |)}{2}}$

Step 5.9.2.3

Combine the numerators over the common denominator.

$y - 1 = e^{\frac{C - \ln (| e^{x} + 4 |)}{2}}$

Step 5.9.3

Move all terms not containing $y$ to the right side of the equation.

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Step 5.9.3.1

Add $1$ to both sides of the equation.

$y = e^{\frac{C - \ln (| e^{x} + 4 |)}{2}} + 1$

Step 5.9.3.2

Simplify each term.

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Step 5.9.3.2.1

Split the fraction $\frac{C - \ln (| e^{x} + 4 |)}{2}$ into two fractions.

$y = e^{\frac{C}{2} + \frac{- \ln (| e^{x} + 4 |)}{2}} + 1$

Step 5.9.3.2.2

Move the negative in front of the fraction.

$y = e^{\frac{C}{2} - \frac{\ln (| e^{x} + 4 |)}{2}} + 1$

Step 6

Group the constant terms together.

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Step 6.1

Simplify the constant of integration.

$y = e^{C - \frac{\ln (| e^{x} + 4 |)}{2}} + 1$

Step 6.2

Reorder terms.

$y = e^{- \frac{\ln (| e^{x} + 4 |)}{2} + C} + 1$

Step 6.3

Rewrite $e^{- \frac{\ln (| e^{x} + 4 |)}{2} + C}$ as $e^{- \frac{\ln (| e^{x} + 4 |)}{2}} e^{C}$ .

$y = e^{- \frac{\ln (| e^{x} + 4 |)}{2}} e^{C} + 1$

Step 6.4

Reorder $e^{- \frac{\ln (| e^{x} + 4 |)}{2}}$ and $e^{C}$ .

$y = e^{C} e^{- \frac{\ln (| e^{x} + 4 |)}{2}} + 1$