Calculus Examples

$\frac{d y}{d x} + x y = \frac{x}{y}$

Step 1

Separate the variables.

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Step 1.1

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

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

Step 1.2

Factor.

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Step 1.2.1

Factor $x$ out of $\frac{x}{y} - x y$ .

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Step 1.2.1.1

Factor $x$ out of $\frac{x}{y}$ .

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

Step 1.2.1.2

Factor $x$ out of $- x y$ .

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

Step 1.2.1.3

Factor $x$ out of $x \frac{1}{y} + x (- 1 y)$ .

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

Step 1.2.2

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

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

Step 1.2.3

To write $- y$ as a fraction with a common denominator, multiply by $\frac{y}{y}$ .

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

Step 1.2.4

Combine $- y$ and $\frac{y}{y}$ .

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

Step 1.2.5

Combine the numerators over the common denominator.

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

Step 1.2.6

Simplify the numerator.

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Step 1.2.6.1

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

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

Step 1.2.6.2

Rewrite $y \cdot y$ as $y^{2}$ .

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

Step 1.2.6.3

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

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

Step 1.3

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

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

Step 1.4

Simplify.

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Step 1.4.1

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

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

Step 1.4.2

Cancel the common factor of $y$ .

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Step 1.4.2.1

Cancel the common factor.

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

Step 1.4.2.2

Rewrite the expression.

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

Step 1.4.3

Cancel the common factor of $(1 + y) (1 - y)$ .

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Step 1.4.3.1

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

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

Step 1.4.3.2

Cancel the common factor.

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

Step 1.4.3.3

Rewrite the expression.

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

Step 1.5

Rewrite the equation.

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

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

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

Step 2.2

Integrate the left side.

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Step 2.2.1

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

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Step 2.2.1.1

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

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Step 2.2.1.1.1

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

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

Step 2.2.1.1.2

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

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

Step 2.2.1.1.3

Differentiate.

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Step 2.2.1.1.3.1

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

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

Step 2.2.1.1.3.2

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

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

Step 2.2.1.1.3.3

Add $0$ and $\frac{d}{d y} [- y]$ .

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

Step 2.2.1.1.3.4

Since $- 1$ is constant with respect to $y$ , the derivative of $- y$ with respect to $y$ is $- \frac{d}{d y} [y]$ .

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

Step 2.2.1.1.3.5

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

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

Step 2.2.1.1.3.6

Simplify the expression.

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Step 2.2.1.1.3.6.1

Multiply $- 1$ by $1$ .

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

Step 2.2.1.1.3.6.2

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

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

Step 2.2.1.1.3.6.3

Rewrite $- 1 (1 + y)$ as $- (1 + y)$ .

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

Step 2.2.1.1.3.7

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

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

Step 2.2.1.1.3.8

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

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

Step 2.2.1.1.3.9

Add $0$ and $\frac{d}{d y} [y]$ .

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

Step 2.2.1.1.3.10

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

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

Step 2.2.1.1.3.11

Multiply $1 - y$ by $1$ .

$- (1 + y) + 1 - y$

Step 2.2.1.1.4

Simplify.

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Step 2.2.1.1.4.1

Apply the distributive property.

$- 1 \cdot 1 - y + 1 - y$

Step 2.2.1.1.4.2

Combine terms.

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Step 2.2.1.1.4.2.1

Multiply $- 1$ by $1$ .

$- 1 - y + 1 - y$

Step 2.2.1.1.4.2.2

Add $- 1$ and $1$ .

$- y + 0 - y$

Step 2.2.1.1.4.2.3

Add $- y$ and $0$ .

$- y - y$

Step 2.2.1.1.4.2.4

Subtract $y$ from $- y$ .

$- 2 y$

Step 2.2.1.2

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

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

Step 2.2.2

Simplify.

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Step 2.2.2.1

Move the negative in front of the fraction.

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

Step 2.2.2.2

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

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

Step 2.2.2.3

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

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

Step 2.2.3

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

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

Step 2.2.4

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

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

Step 2.2.5

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

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

Step 2.2.6

Simplify.

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

Step 2.2.7

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

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

Step 2.3

By the Power Rule, the integral of $x$ with respect to $x$ is $\frac{1}{2} x^{2}$ .

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

Step 2.4

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

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

Step 3

Solve for $y$ .

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Step 3.1

Multiply both sides of the equation by $- 2$ .

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

Step 3.2

Simplify both sides of the equation.

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Step 3.2.1

Simplify the left side.

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Step 3.2.1.1

Simplify $- 2 (- \frac{1}{2} \ln (| (1 + y) (1 - y) |))$ .

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Step 3.2.1.1.1

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

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Step 3.2.1.1.1.1

Apply the distributive property.

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

Step 3.2.1.1.1.2

Apply the distributive property.

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

Step 3.2.1.1.1.3

Apply the distributive property.

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

Step 3.2.1.1.2

Simplify and combine like terms.

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Step 3.2.1.1.2.1

Simplify each term.

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Step 3.2.1.1.2.1.1

Multiply $1$ by $1$ .

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

Step 3.2.1.1.2.1.2

Multiply $- y$ by $1$ .

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

Step 3.2.1.1.2.1.3

Multiply $y$ by $1$ .

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

Step 3.2.1.1.2.1.4

Rewrite using the commutative property of multiplication.

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

Step 3.2.1.1.2.1.5

Multiply $y$ by $y$ by adding the exponents.

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Step 3.2.1.1.2.1.5.1

Move $y$ .

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

Step 3.2.1.1.2.1.5.2

Multiply $y$ by $y$ .

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

Step 3.2.1.1.2.2

Add $- y$ and $y$ .

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

Step 3.2.1.1.2.3

Add $1$ and $0$ .

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

Step 3.2.1.1.3

Combine $\ln (| 1 - y^{2} |)$ and $\frac{1}{2}$ .

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

Step 3.2.1.1.4

Cancel the common factor of $2$ .

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Step 3.2.1.1.4.1

Move the leading negative in $- \frac{\ln (| 1 - y^{2} |)}{2}$ into the numerator.

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

Step 3.2.1.1.4.2

Factor $2$ out of $- 2$ .

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

Step 3.2.1.1.4.3

Cancel the common factor.

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

Step 3.2.1.1.4.4

Rewrite the expression.

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

Step 3.2.1.1.5

Multiply.

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Step 3.2.1.1.5.1

Multiply $- 1$ by $- 1$ .

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

Step 3.2.1.1.5.2

Multiply $\ln (| 1 - y^{2} |)$ by $1$ .

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

Step 3.2.2

Simplify the right side.

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Step 3.2.2.1

Simplify $- 2 (\frac{1}{2} x^{2} + C)$ .

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Step 3.2.2.1.1

Combine $\frac{1}{2}$ and $x^{2}$ .

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

Step 3.2.2.1.2

Apply the distributive property.

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

Step 3.2.2.1.3

Cancel the common factor of $2$ .

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Step 3.2.2.1.3.1

Factor $2$ out of $- 2$ .

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

Step 3.2.2.1.3.2

Cancel the common factor.

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

Step 3.2.2.1.3.3

Rewrite the expression.

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

Step 3.3

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

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

Step 3.4

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

$e^{- x^{2} - 2 C} = | 1 - y^{2} |$

Step 3.5

Solve for $y$ .

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Step 3.5.1

Rewrite the equation as $| 1 - y^{2} | = e^{- x^{2} - 2 C}$ .

$| 1 - y^{2} | = e^{- x^{2} - 2 C}$

Step 3.5.2

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

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

Step 3.5.3

Subtract $1$ from both sides of the equation.

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

Step 3.5.4

Divide each term in $- y^{2} = \pm e^{- x^{2} - 2 C} - 1$ by $- 1$ and simplify.

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Step 3.5.4.1

Divide each term in $- y^{2} = \pm e^{- x^{2} - 2 C} - 1$ by $- 1$ .

$\frac{- y^{2}}{- 1} = \frac{\pm e^{- x^{2} - 2 C}}{- 1} + \frac{- 1}{- 1}$

Step 3.5.4.2

Simplify the left side.

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Step 3.5.4.2.1

Dividing two negative values results in a positive value.

$\frac{y^{2}}{1} = \frac{\pm e^{- x^{2} - 2 C}}{- 1} + \frac{- 1}{- 1}$

Step 3.5.4.2.2

Divide $y^{2}$ by $1$ .

$y^{2} = \frac{\pm e^{- x^{2} - 2 C}}{- 1} + \frac{- 1}{- 1}$

Step 3.5.4.3

Simplify the right side.

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Step 3.5.4.3.1

Simplify each term.

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Step 3.5.4.3.1.1

Move the negative one from the denominator of $\frac{\pm e^{- x^{2} - 2 C}}{- 1}$ .

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

Step 3.5.4.3.1.2

Rewrite $- 1 \cdot (\pm e^{- x^{2} - 2 C})$ as $- (\pm e^{- x^{2} - 2 C})$ .

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

Step 3.5.4.3.1.3

Divide $- 1$ by $- 1$ .

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

Step 3.5.5

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

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

Step 4

Group the constant terms together.

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Step 4.1

Simplify the constant of integration.

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

Step 4.2

Rewrite $e^{- x^{2} + C}$ as $e^{- x^{2}} e^{C}$ .

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

Step 4.3

Reorder $e^{- x^{2}}$ and $e^{C}$ .

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

Step 4.4

Combine constants with the plus or minus.

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