Calculus Examples

$x (y^{2} - 4) d x + y d y = 0$

Step 1

Subtract $x (y^{2} - 4) d x$ from both sides of the equation.

$y d y = - x (y^{2} - 4) d x$

Step 2

Multiply both sides by $\frac{1}{y^{2} - 4}$ .

$\frac{1}{y^{2} - 4} y d y = \frac{1}{y^{2} - 4} (- x (y^{2} - 4)) d x$

Step 3

Simplify.

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Step 3.1

Simplify the denominator.

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Step 3.1.1

Rewrite $4$ as $2^{2}$ .

$\frac{1}{y^{2} - 2^{2}} y d y = \frac{1}{y^{2} - 4} (- x (y^{2} - 4)) d x$

Step 3.1.2

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

$\frac{1}{(y + 2) (y - 2)} y d y = \frac{1}{y^{2} - 4} (- x (y^{2} - 4)) d x$

Step 3.2

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

$\frac{y}{(y + 2) (y - 2)} d y = \frac{1}{y^{2} - 4} (- x (y^{2} - 4)) d x$

Step 3.3

Rewrite using the commutative property of multiplication.

$\frac{y}{(y + 2) (y - 2)} d y = - \frac{1}{y^{2} - 4} (x (y^{2} - 4)) d x$

Step 3.4

Cancel the common factor of $y^{2} - 4$ .

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Step 3.4.1

Move the leading negative in $- \frac{1}{y^{2} - 4}$ into the numerator.

$\frac{y}{(y + 2) (y - 2)} d y = \frac{- 1}{y^{2} - 4} (x (y^{2} - 4)) d x$

Step 3.4.2

Factor $y^{2} - 4$ out of $x (y^{2} - 4)$ .

$\frac{y}{(y + 2) (y - 2)} d y = \frac{- 1}{y^{2} - 4} ((y^{2} - 4) x) d x$

Step 3.4.3

Cancel the common factor.

$\frac{y}{(y + 2) (y - 2)} d y = \frac{- 1}{y^{2} - 4} ((y^{2} - 4) x) d x$

Step 3.4.4

Rewrite the expression.

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

Step 4

Integrate both sides.

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Step 4.1

Set up an integral on each side.

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

Step 4.2

Integrate the left side.

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Step 4.2.1

Let $u = (y + 2) (y - 2)$ . 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 4.2.1.1

Let $u = (y + 2) (y - 2)$ . Find $\frac{d u}{d y}$ .

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Step 4.2.1.1.1

Differentiate $(y + 2) (y - 2)$ .

$\frac{d}{d y} [(y + 2) (y - 2)]$

Step 4.2.1.1.2

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 $f (y) = y + 2$ and $g (y) = y - 2$ .

$(y + 2) \frac{d}{d y} [y - 2] + (y - 2) \frac{d}{d y} [y + 2]$

Step 4.2.1.1.3

Differentiate.

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Step 4.2.1.1.3.1

By the Sum Rule, the derivative of $y - 2$ with respect to $y$ is $\frac{d}{d y} [y] + \frac{d}{d y} [- 2]$ .

$(y + 2) (\frac{d}{d y} [y] + \frac{d}{d y} [- 2]) + (y - 2) \frac{d}{d y} [y + 2]$

Step 4.2.1.1.3.2

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

$(y + 2) (1 + \frac{d}{d y} [- 2]) + (y - 2) \frac{d}{d y} [y + 2]$

Step 4.2.1.1.3.3

Since $- 2$ is constant with respect to $y$ , the derivative of $- 2$ with respect to $y$ is $0$ .

$(y + 2) (1 + 0) + (y - 2) \frac{d}{d y} [y + 2]$

Step 4.2.1.1.3.4

Simplify the expression.

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Step 4.2.1.1.3.4.1

Add $1$ and $0$ .

$(y + 2) \cdot 1 + (y - 2) \frac{d}{d y} [y + 2]$

Step 4.2.1.1.3.4.2

Multiply $y + 2$ by $1$ .

$y + 2 + (y - 2) \frac{d}{d y} [y + 2]$

Step 4.2.1.1.3.5

By the Sum Rule, the derivative of $y + 2$ with respect to $y$ is $\frac{d}{d y} [y] + \frac{d}{d y} [2]$ .

$y + 2 + (y - 2) (\frac{d}{d y} [y] + \frac{d}{d y} [2])$

Step 4.2.1.1.3.6

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

$y + 2 + (y - 2) (1 + \frac{d}{d y} [2])$

Step 4.2.1.1.3.7

Since $2$ is constant with respect to $y$ , the derivative of $2$ with respect to $y$ is $0$ .

$y + 2 + (y - 2) (1 + 0)$

Step 4.2.1.1.3.8

Simplify by adding terms.

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Step 4.2.1.1.3.8.1

Add $1$ and $0$ .

$y + 2 + (y - 2) \cdot 1$

Step 4.2.1.1.3.8.2

Multiply $y - 2$ by $1$ .

$y + 2 + y - 2$

Step 4.2.1.1.3.8.3

Add $y$ and $y$ .

$2 y + 2 - 2$

Step 4.2.1.1.3.8.4

Simplify by subtracting numbers.

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Step 4.2.1.1.3.8.4.1

Subtract $2$ from $2$ .

$2 y + 0$

Step 4.2.1.1.3.8.4.2

Add $2 y$ and $0$ .

$2 y$

Step 4.2.1.2

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

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

Step 4.2.2

Simplify.

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Step 4.2.2.1

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

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

Step 4.2.2.2

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

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

Step 4.2.3

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 4.2.4

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 4.2.5

Simplify.

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

Step 4.2.6

Replace all occurrences of $u$ with $(y + 2) (y - 2)$ .

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

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.

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

Step 4.3.2

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

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

Step 4.3.3

Rewrite $- (\frac{1}{2} x^{2} + C_{2})$ as $- \frac{1}{2} x^{2} + C_{2}$ .

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

Step 4.4

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

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

Step 5

Solve for $y$ .

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Step 5.1

Multiply both sides of the equation by $2$ .

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

Step 5.2

Simplify both sides of the equation.

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Step 5.2.1

Simplify the left side.

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Step 5.2.1.1

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

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Step 5.2.1.1.1

Expand $(y + 2) (y - 2)$ using the FOIL Method.

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Step 5.2.1.1.1.1

Apply the distributive property.

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

Step 5.2.1.1.1.2

Apply the distributive property.

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

Step 5.2.1.1.1.3

Apply the distributive property.

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

Step 5.2.1.1.2

Simplify terms.

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Step 5.2.1.1.2.1

Combine the opposite terms in $y \cdot y + y \cdot - 2 + 2 y + 2 \cdot - 2$ .

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Step 5.2.1.1.2.1.1

Reorder the factors in the terms $y \cdot - 2$ and $2 y$ .

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

Step 5.2.1.1.2.1.2

Add $- 2 y$ and $2 y$ .

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

Step 5.2.1.1.2.1.3

Add $y \cdot y$ and $0$ .

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

Step 5.2.1.1.2.2

Simplify each term.

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Step 5.2.1.1.2.2.1

Multiply $y$ by $y$ .

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

Step 5.2.1.1.2.2.2

Multiply $2$ by $- 2$ .

$2 (\frac{1}{2} \ln (| y^{2} - 4 |)) = 2 (- \frac{1}{2} x^{2} + C)$

Step 5.2.1.1.2.3

Simplify terms.

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Step 5.2.1.1.2.3.1

Combine $\frac{1}{2}$ and $\ln (| y^{2} - 4 |)$ .

$2 \frac{\ln (| y^{2} - 4 |)}{2} = 2 (- \frac{1}{2} x^{2} + C)$

Step 5.2.1.1.2.3.2

Cancel the common factor of $2$ .

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Step 5.2.1.1.2.3.2.1

Cancel the common factor.

$2 \frac{\ln (| y^{2} - 4 |)}{2} = 2 (- \frac{1}{2} x^{2} + C)$

Step 5.2.1.1.2.3.2.2

Rewrite the expression.

$\ln (| y^{2} - 4 |) = 2 (- \frac{1}{2} x^{2} + C)$

Step 5.2.2

Simplify the right side.

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Step 5.2.2.1

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

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Step 5.2.2.1.1

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

$\ln (| y^{2} - 4 |) = 2 (- \frac{x^{2}}{2} + C)$

Step 5.2.2.1.2

Apply the distributive property.

$\ln (| y^{2} - 4 |) = 2 (- \frac{x^{2}}{2}) + 2 C$

Step 5.2.2.1.3

Cancel the common factor of $2$ .

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Step 5.2.2.1.3.1

Move the leading negative in $- \frac{x^{2}}{2}$ into the numerator.

$\ln (| y^{2} - 4 |) = 2 \frac{- x^{2}}{2} + 2 C$

Step 5.2.2.1.3.2

Cancel the common factor.

$\ln (| y^{2} - 4 |) = 2 \frac{- x^{2}}{2} + 2 C$

Step 5.2.2.1.3.3

Rewrite the expression.

$\ln (| y^{2} - 4 |) = - x^{2} + 2 C$

Step 5.3

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

$e^{\ln (| y^{2} - 4 |)} = e^{- x^{2} + 2 C}$

Step 5.4

Rewrite $\ln (| y^{2} - 4 |) = - x^{2} + 2 C$ 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 $b^{y} = x$ .

$e^{- x^{2} + 2 C} = | y^{2} - 4 |$

Step 5.5

Solve for $y$ .

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Step 5.5.1

Rewrite the equation as $| y^{2} - 4 | = e^{- x^{2} + 2 C}$ .

$| y^{2} - 4 | = e^{- x^{2} + 2 C}$

Step 5.5.2

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

$y^{2} - 4 = \pm e^{- x^{2} + 2 C}$

Step 5.5.3

Add $4$ to both sides of the equation.

$y^{2} = \pm e^{- x^{2} + 2 C} + 4$

Step 5.5.4

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

$y = \pm \sqrt{\pm e^{- x^{2} + 2 C} + 4}$

Step 6

Group the constant terms together.

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Step 6.1

Simplify the constant of integration.

$y = \pm \sqrt{\pm e^{- x^{2} + C} + 4}$

Step 6.2

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

$y = \pm \sqrt{\pm e^{- x^{2}} e^{C} + 4}$

Step 6.3

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

$y = \pm \sqrt{\pm e^{C} e^{- x^{2}} + 4}$

Step 6.4

Combine constants with the plus or minus.

$y = \pm \sqrt{C e^{- x^{2}} + 4}$