Calculus Examples

$(1 - x) y' = y^{2}$

Step 1

Rewrite the differential equation.

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

Step 2

Separate the variables.

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Step 2.1

Divide each term in $(1 - x) \frac{d y}{d x} = y^{2}$ by $1 - x$ and simplify.

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Step 2.1.1

Divide each term in $(1 - x) \frac{d y}{d x} = y^{2}$ by $1 - x$ .

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

Step 2.1.2

Simplify the left side.

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Step 2.1.2.1

Cancel the common factor of $1 - x$ .

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Step 2.1.2.1.1

Cancel the common factor.

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

Step 2.1.2.1.2

Divide $\frac{d y}{d x}$ by $1$ .

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

Step 2.2

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

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

Step 2.3

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

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Step 2.3.1

Cancel the common factor.

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

Step 2.3.2

Rewrite the expression.

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

Step 2.4

Rewrite the equation.

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

Step 3

Integrate both sides.

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Step 3.1

Set up an integral on each side.

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

Step 3.2

Integrate the left side.

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Step 3.2.1

Apply basic rules of exponents.

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Step 3.2.1.1

Move $y^{2}$ out of the denominator by raising it to the $- 1$ power.

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

Step 3.2.1.2

Multiply the exponents in ${(y^{2})}^{- 1}$ .

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Step 3.2.1.2.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

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

Step 3.2.1.2.2

Multiply $2$ by $- 1$ .

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

Step 3.2.2

By the Power Rule, the integral of $y^{- 2}$ with respect to $y$ is $- y^{- 1}$ .

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

Step 3.2.3

Rewrite $- y^{- 1} + C_{1}$ as $- \frac{1}{y} + C_{1}$ .

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

Step 3.3

Integrate the right side.

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Step 3.3.1

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

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Step 3.3.1.1

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

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Step 3.3.1.1.1

Rewrite.

$\frac{- 1}{1}$

Step 3.3.1.1.2

Divide $- 1$ by $1$ .

$- 1$

Step 3.3.1.2

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

$- \frac{1}{y} + C_{1} = \int \frac{- 1}{u} d u$

Step 3.3.2

Split the fraction into multiple fractions.

$- \frac{1}{y} + C_{1} = \int - \frac{1}{u} d u$

Step 3.3.3

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

$- \frac{1}{y} + C_{1} = - \int \frac{1}{u} d u$

Step 3.3.4

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

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

Step 3.3.5

Simplify.

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

Step 3.3.6

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

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

Step 3.4

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

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

Step 4

Solve for $y$ .

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Step 4.1

Find the LCD of the terms in the equation.

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Step 4.1.1

Finding the LCD of a list of values is the same as finding the LCM of the denominators of those values.

$y, 1, 1$

Step 4.1.2

The LCM of one and any expression is the expression.

$y$

Step 4.2

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

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Step 4.2.1

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

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

Step 4.2.2

Simplify the left side.

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Step 4.2.2.1

Cancel the common factor of $y$ .

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Step 4.2.2.1.1

Move the leading negative in $- \frac{1}{y}$ into the numerator.

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

Step 4.2.2.1.2

Cancel the common factor.

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

Step 4.2.2.1.3

Rewrite the expression.

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

Step 4.2.3

Simplify the right side.

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Step 4.2.3.1

Reorder factors in $- \ln (| 1 - x |) y + C y$ .

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

Step 4.3

Solve the equation.

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Step 4.3.1

Rewrite the equation as $- y \ln (| 1 - x |) + C y = - 1$ .

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

Step 4.3.2

Factor $y$ out of $- y \ln (| 1 - x |) + C y$ .

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Step 4.3.2.1

Factor $y$ out of $- y \ln (| 1 - x |)$ .

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

Step 4.3.2.2

Factor $y$ out of $C y$ .

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

Step 4.3.2.3

Factor $y$ out of $y (- 1 \ln (| 1 - x |)) + y C$ .

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

Step 4.3.3

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

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

Step 4.3.4

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

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Step 4.3.4.1

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

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

Step 4.3.4.2

Simplify the left side.

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Step 4.3.4.2.1

Cancel the common factor of $- \ln (| 1 - x |) + C$ .

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Step 4.3.4.2.1.1

Cancel the common factor.

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

Step 4.3.4.2.1.2

Divide $y$ by $1$ .

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

Step 4.3.4.3

Simplify the right side.

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Step 4.3.4.3.1

Move the negative in front of the fraction.

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

Step 4.3.4.3.2

Factor $- 1$ out of $- \ln (| 1 - x |)$ .

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

Step 4.3.4.3.3

Factor $- 1$ out of $C$ .

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

Step 4.3.4.3.4

Factor $- 1$ out of $- (\ln (| 1 - x |)) - 1 (- C)$ .

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

Step 4.3.4.3.5

Simplify the expression.

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Step 4.3.4.3.5.1

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

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

Step 4.3.4.3.5.2

Move the negative in front of the fraction.

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

Step 4.3.4.3.5.3

Multiply $- 1$ by $- 1$ .

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

Step 4.3.4.3.5.4

Multiply $\frac{1}{\ln (| 1 - x |) - C}$ by $1$ .

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

Step 5

Simplify the constant of integration.

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