Calculus Examples

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

Step 1

Subtract $2 x y d x$ from both sides of the equation.

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

Step 2

Multiply both sides by $\frac{1}{(x^{2} - 1) y}$ .

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

Step 3

Simplify.

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Step 3.1

Cancel the common factor of $x^{2} - 1$ .

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Step 3.1.1

Factor $x^{2} - 1$ out of $(x^{2} - 1) y$ .

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

Step 3.1.2

Cancel the common factor.

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

Step 3.1.3

Rewrite the expression.

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

Step 3.2

Rewrite using the commutative property of multiplication.

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

Step 3.3

Simplify the denominator.

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Step 3.3.1

Rewrite $1$ as $1^{2}$ .

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

Step 3.3.2

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 1$ .

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

Step 3.4

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

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

Step 3.5

Cancel the common factor of $y$ .

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Step 3.5.1

Factor $y$ out of $(x + 1) (x - 1) y$ .

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

Step 3.5.2

Factor $y$ out of $x y$ .

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

Step 3.5.3

Cancel the common factor.

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

Step 3.5.4

Rewrite the expression.

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

Step 3.6

Combine $\frac{- 2}{(x + 1) (x - 1)}$ and $x$ .

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

Step 3.7

Move the negative in front of the fraction.

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

Step 4

Integrate both sides.

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Step 4.1

Set up an integral on each side.

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

Step 4.2

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

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

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.

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

Step 4.3.2

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

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

Step 4.3.3

Multiply $2$ by $- 1$ .

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

Step 4.3.4

Let $u = (x + 1) (x - 1)$ . Then $d u = 2 x d x$ , so $\frac{1}{2} d u = x d x$ . Rewrite using $u$ and $d$ $u$ .

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Step 4.3.4.1

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

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Step 4.3.4.1.1

Differentiate $(x + 1) (x - 1)$ .

$\frac{d}{d x} [(x + 1) (x - 1)]$

Step 4.3.4.1.2

Differentiate using the Product Rule which states that $\frac{d}{d x} [f (x) g (x)]$ is $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ where $f (x) = x + 1$ and $g (x) = x - 1$ .

$(x + 1) \frac{d}{d x} [x - 1] + (x - 1) \frac{d}{d x} [x + 1]$

Step 4.3.4.1.3

Differentiate.

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Step 4.3.4.1.3.1

By the Sum Rule, the derivative of $x - 1$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [- 1]$ .

$(x + 1) (\frac{d}{d x} [x] + \frac{d}{d x} [- 1]) + (x - 1) \frac{d}{d x} [x + 1]$

Step 4.3.4.1.3.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

$(x + 1) (1 + \frac{d}{d x} [- 1]) + (x - 1) \frac{d}{d x} [x + 1]$

Step 4.3.4.1.3.3

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

$(x + 1) (1 + 0) + (x - 1) \frac{d}{d x} [x + 1]$

Step 4.3.4.1.3.4

Simplify the expression.

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Step 4.3.4.1.3.4.1

Add $1$ and $0$ .

$(x + 1) \cdot 1 + (x - 1) \frac{d}{d x} [x + 1]$

Step 4.3.4.1.3.4.2

Multiply $x + 1$ by $1$ .

$x + 1 + (x - 1) \frac{d}{d x} [x + 1]$

Step 4.3.4.1.3.5

By the Sum Rule, the derivative of $x + 1$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [1]$ .

$x + 1 + (x - 1) (\frac{d}{d x} [x] + \frac{d}{d x} [1])$

Step 4.3.4.1.3.6

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

$x + 1 + (x - 1) (1 + \frac{d}{d x} [1])$

Step 4.3.4.1.3.7

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

$x + 1 + (x - 1) (1 + 0)$

Step 4.3.4.1.3.8

Simplify by adding terms.

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Step 4.3.4.1.3.8.1

Add $1$ and $0$ .

$x + 1 + (x - 1) \cdot 1$

Step 4.3.4.1.3.8.2

Multiply $x - 1$ by $1$ .

$x + 1 + x - 1$

Step 4.3.4.1.3.8.3

Add $x$ and $x$ .

$2 x + 1 - 1$

Step 4.3.4.1.3.8.4

Simplify by subtracting numbers.

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Step 4.3.4.1.3.8.4.1

Subtract $1$ from $1$ .

$2 x + 0$

Step 4.3.4.1.3.8.4.2

Add $2 x$ and $0$ .

$2 x$

Step 4.3.4.2

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

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

Step 4.3.5

Simplify.

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Step 4.3.5.1

Multiply $\frac{1}{u}$ by $\frac{1}{2}$ .

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

Step 4.3.5.2

Move $2$ to the left of $u$ .

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

Step 4.3.6

Since $\frac{1}{2}$ is constant with respect to $u$ , move $\frac{1}{2}$ out of the integral.

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

Step 4.3.7

Simplify.

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Step 4.3.7.1

Combine $\frac{1}{2}$ and $- 2$ .

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

Step 4.3.7.2

Cancel the common factor of $- 2$ and $2$ .

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Step 4.3.7.2.1

Factor $2$ out of $- 2$ .

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

Step 4.3.7.2.2

Cancel the common factors.

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Step 4.3.7.2.2.1

Factor $2$ out of $2$ .

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

Step 4.3.7.2.2.2

Cancel the common factor.

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

Step 4.3.7.2.2.3

Rewrite the expression.

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

Step 4.3.7.2.2.4

Divide $- 1$ by $1$ .

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

Step 4.3.8

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

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

Step 4.3.9

Simplify.

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

Step 4.3.10

Replace all occurrences of $u$ with $(x + 1) (x - 1)$ .

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

Step 4.4

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

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

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.

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

Step 5.2

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

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

Step 5.3

Expand $(x + 1) (x - 1)$ using the FOIL Method.

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Step 5.3.1

Apply the distributive property.

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

Step 5.3.2

Apply the distributive property.

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

Step 5.3.3

Apply the distributive property.

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

Step 5.4

Simplify and combine like terms.

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Step 5.4.1

Simplify each term.

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Step 5.4.1.1

Multiply $x$ by $x$ .

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

Step 5.4.1.2

Move $- 1$ to the left of $x$ .

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

Step 5.4.1.3

Rewrite $- 1 x$ as $- x$ .

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

Step 5.4.1.4

Multiply $x$ by $1$ .

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

Step 5.4.1.5

Multiply $- 1$ by $1$ .

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

Step 5.4.2

Add $- x$ and $x$ .

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

Step 5.4.3

Add $x^{2}$ and $0$ .

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

Step 5.5

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

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

Step 5.6

Simplify by multiplying through.

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Step 5.6.1

Apply the distributive property.

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

Step 5.6.2

Move $- 1$ to the left of $y$ .

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

Step 5.7

Rewrite $- 1 y$ as $- y$ .

$\ln (| y x^{2} - y |) = C$

Step 5.8

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

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

Step 5.9

Rewrite $\ln (| y x^{2} - y |) = C$ 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 $b^{y} = x$ .

$e^{C} = | y x^{2} - y |$

Step 5.10

Solve for $y$ .

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Step 5.10.1

Rewrite the equation as $| y x^{2} - y | = e^{C}$ .

$| y x^{2} - y | = e^{C}$

Step 5.10.2

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

$y x^{2} - y = \pm e^{C}$

Step 5.10.3

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

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Step 5.10.3.1

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

$y (x^{2}) - y = \pm e^{C}$

Step 5.10.3.2

Factor $y$ out of $- y$ .

$y (x^{2}) + y \cdot - 1 = \pm e^{C}$

Step 5.10.3.3

Factor $y$ out of $y (x^{2}) + y \cdot - 1$ .

$y (x^{2} - 1) = \pm e^{C}$

Step 5.10.4

Rewrite $1$ as $1^{2}$ .

$y (x^{2} - 1^{2}) = \pm e^{C}$

Step 5.10.5

Factor.

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Step 5.10.5.1

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 1$ .

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

Step 5.10.5.2

Remove unnecessary parentheses.

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

Step 5.10.6

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

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Step 5.10.6.1

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

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

Step 5.10.6.2

Simplify the left side.

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Step 5.10.6.2.1

Cancel the common factor of $x + 1$ .

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Step 5.10.6.2.1.1

Cancel the common factor.

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

Step 5.10.6.2.1.2

Rewrite the expression.

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

Step 5.10.6.2.2

Cancel the common factor of $x - 1$ .

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Step 5.10.6.2.2.1

Cancel the common factor.

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

Step 5.10.6.2.2.2

Divide $y$ by $1$ .

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

Step 6

Simplify the constant of integration.

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