Calculus Examples

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

Step 1

Separate the variables.

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Step 1.1

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

$\frac{1}{1 - y} \frac{d y}{d x} = \frac{1}{1 - y} \cdot \frac{1 - y}{1 - x}$

Step 1.2

Cancel the common factor of $1 - y$ .

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Step 1.2.1

Cancel the common factor.

$\frac{1}{1 - y} \frac{d y}{d x} = \frac{1}{1 - y} \cdot \frac{1 - y}{1 - x}$

Step 1.2.2

Rewrite the expression.

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

Step 1.3

Rewrite the equation.

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

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

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

Step 2.2

Integrate the left side.

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Step 2.2.1

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

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Step 2.2.1.1

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

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Step 2.2.1.1.1

Rewrite.

$\frac{- 1}{1}$

Step 2.2.1.1.2

Divide $- 1$ by $1$ .

$- 1$

Step 2.2.1.2

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

$\int \frac{- 1}{u_{1}} d u_{1} = \int \frac{1}{1 - x} d x$

Step 2.2.2

Split the fraction into multiple fractions.

$\int - \frac{1}{u_{1}} d u_{1} = \int \frac{1}{1 - x} d x$

Step 2.2.3

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

$- \int \frac{1}{u_{1}} d u_{1} = \int \frac{1}{1 - x} d x$

Step 2.2.4

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

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

Step 2.2.5

Simplify.

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

Step 2.2.6

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

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

Step 2.3

Integrate the right side.

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Step 2.3.1

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

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Step 2.3.1.1

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

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Step 2.3.1.1.1

Rewrite.

$\frac{- 1}{1}$

Step 2.3.1.1.2

Divide $- 1$ by $1$ .

$- 1$

Step 2.3.1.2

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

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

Step 2.3.2

Split the fraction into multiple fractions.

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

Step 2.3.3

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

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

Step 2.3.4

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

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

Step 2.3.5

Simplify.

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

Step 2.3.6

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

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

Step 2.4

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

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

Step 3

Solve for $y$ .

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Step 3.1

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

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

Step 3.2

Subtract $\ln (| 1 - x |)$ from both sides of the equation.

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

Step 3.3

Divide each term in $- \ln (| 1 - y |) = C - \ln (| 1 - x |)$ by $- 1$ and simplify.

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Step 3.3.1

Divide each term in $- \ln (| 1 - y |) = C - \ln (| 1 - x |)$ by $- 1$ .

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

Step 3.3.2

Simplify the left side.

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Step 3.3.2.1

Dividing two negative values results in a positive value.

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

Step 3.3.2.2

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

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

Step 3.3.3

Simplify the right side.

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Step 3.3.3.1

Simplify each term.

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Step 3.3.3.1.1

Move the negative one from the denominator of $\frac{C}{- 1}$ .

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

Step 3.3.3.1.2

Rewrite $- 1 \cdot C$ as $- C$ .

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

Step 3.3.3.1.3

Dividing two negative values results in a positive value.

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

Step 3.3.3.1.4

Divide $\ln (| 1 - x |)$ by $1$ .

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

Step 3.4

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

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

Step 3.5

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

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

Step 3.6

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

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

Step 3.7

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

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

Step 3.8

Solve for $y$ .

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Step 3.8.1

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

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

Step 3.8.2

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

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

Step 3.8.3

Simplify the left side.

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Step 3.8.3.1

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

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Step 3.8.3.1.1

Cancel the common factor.

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

Step 3.8.3.1.2

Rewrite the expression.

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

Step 3.8.4

Solve for $y$ .

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Step 3.8.4.1

Reorder factors in $e^{- C} | 1 - x |$ .

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

Step 3.8.4.2

Remove the absolute value term. This creates a $\pm$ on the right side of the equation because $| x | = \pm x$ .

$1 - y = \pm | 1 - x | e^{- C}$

Step 3.8.4.3

Reorder factors in $\pm | 1 - x | e^{- C}$ .

$1 - y = \pm e^{- C} | 1 - x |$

Step 3.8.4.4

Subtract $1$ from both sides of the equation.

$- y = \pm e^{- C} | 1 - x | - 1$

Step 3.8.4.5

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

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Step 3.8.4.5.1

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

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

Step 3.8.4.5.2

Simplify the left side.

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Step 3.8.4.5.2.1

Dividing two negative values results in a positive value.

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

Step 3.8.4.5.2.2

Divide $y$ by $1$ .

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

Step 3.8.4.5.3

Simplify the right side.

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Step 3.8.4.5.3.1

Simplify each term.

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Step 3.8.4.5.3.1.1

Move the negative one from the denominator of $\frac{\pm e^{- C} | 1 - x |}{- 1}$ .

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

Step 3.8.4.5.3.1.2

Rewrite $- 1 \cdot (\pm e^{- C} | 1 - x |)$ as $- (\pm e^{- C} | 1 - x |)$ .

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

Step 3.8.4.5.3.1.3

Divide $- 1$ by $- 1$ .

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

Step 4

Group the constant terms together.

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Step 4.1

Simplify the constant of integration.

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

Step 4.2

Combine constants with the plus or minus.

$y = - (C | 1 - x |) + 1$