Calculus Examples

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

Step 1

Separate the variables.

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Step 1.1

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

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

Step 1.2

Cancel the common factor of $1 - y$ .

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Step 1.2.1

Cancel the common factor.

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

Step 1.2.2

Rewrite the expression.

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

Step 1.3

Rewrite the equation.

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

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

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

Step 2.2

Integrate the left side.

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Step 2.2.1

Let $u = 1 - y$ . Then $d u = - d y$ , so $- d u = d y$ . Rewrite using $u$ and $d$ $u$ .

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Step 2.2.1.1

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

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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 $u$ and $d u$ .

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

Step 2.2.2

Split the fraction into multiple fractions.

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

Step 2.2.3

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

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

Step 2.2.4

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

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

Step 2.2.5

Simplify.

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

Step 2.2.6

Replace all occurrences of $u$ with $1 - y$ .

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

Step 2.3

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

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

Step 2.4

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

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

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.

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

Step 3.2

Add $\ln (| x |)$ to both sides of the equation.

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

Step 3.3

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

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Step 3.3.1

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

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

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.

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

Step 3.3.2.2

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

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

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 $\frac{C}{- 1}$ .

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

Step 3.3.3.1.2

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

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

Step 3.3.3.1.3

Move the negative one from the denominator of $\frac{\ln (| x |)}{- 1}$ .

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

Step 3.3.3.1.4

Rewrite $- 1 \cdot \ln (| x |)$ as $- \ln (| x |)$ .

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

Step 3.4

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

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

Step 3.5

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

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

Step 3.6

To multiply absolute values, multiply the terms inside each absolute value.

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

Step 3.7

Apply the distributive property.

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

Step 3.8

Multiply $x$ by $1$ .

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

Step 3.9

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

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

Step 3.10

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

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

Step 3.11

Solve for $y$ .

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Step 3.11.1

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

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

Step 3.11.2

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

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

Step 3.11.3

Subtract $x$ from both sides of the equation.

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

Step 3.11.4

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

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Step 3.11.4.1

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

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

Step 3.11.4.2

Simplify the left side.

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Step 3.11.4.2.1

Dividing two negative values results in a positive value.

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

Step 3.11.4.2.2

Cancel the common factor of $x$ .

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Step 3.11.4.2.2.1

Cancel the common factor.

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

Step 3.11.4.2.2.2

Divide $y$ by $1$ .

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

Step 3.11.4.3

Simplify the right side.

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Step 3.11.4.3.1

Simplify each term.

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Step 3.11.4.3.1.1

Simplify $\frac{\pm e^{- C}}{- x}$ .

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

Step 3.11.4.3.1.2

Dividing two negative values results in a positive value.

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

Step 3.11.4.3.1.3

Cancel the common factor of $x$ .

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Step 3.11.4.3.1.3.1

Cancel the common factor.

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

Step 3.11.4.3.1.3.2

Rewrite the expression.

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

Step 4

Simplify the constant of integration.

$y = \frac{C}{x} + 1$