Calculus Examples

$(x y) d x - (x + 2) d y = 0$

Step 1

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

$- (x + 2) d y = - x y d x$

Step 2

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

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

Step 3

Simplify.

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Step 3.1

Rewrite using the commutative property of multiplication.

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

Step 3.2

Cancel the common factor of $x + 2$ .

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Step 3.2.1

Move the leading negative in $- \frac{1}{(x + 2) y}$ into the numerator.

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

Step 3.2.2

Factor $x + 2$ out of $(x + 2) y$ .

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

Step 3.2.3

Cancel the common factor.

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

Step 3.2.4

Rewrite the expression.

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

Step 3.3

Move the negative in front of the fraction.

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

Step 3.4

Rewrite using the commutative property of multiplication.

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

Step 3.5

Cancel the common factor of $y$ .

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Step 3.5.1

Move the leading negative in $- \frac{1}{(x + 2) y}$ into the numerator.

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

Step 3.5.2

Factor $y$ out of $(x + 2) y$ .

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

Step 3.5.3

Factor $y$ out of $x y$ .

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

Step 3.5.4

Cancel the common factor.

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

Step 3.5.5

Rewrite the expression.

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

Step 3.6

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

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

Step 3.7

Move the negative in front of the fraction.

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

Step 4

Integrate both sides.

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Step 4.1

Set up an integral on each side.

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

Step 4.2

Integrate the left side.

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Step 4.2.1

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

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

Step 4.2.2

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

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

Step 4.2.3

Simplify.

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

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.

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

Step 4.3.2

Divide $x$ by $x + 2$ .

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Step 4.3.2.1

Set up the polynomials to be divided. If there is not a term for every exponent, insert one with a value of $0$ .

$x$

$2$

$x$

$0$

Step 4.3.2.2

Divide the highest order term in the dividend $x$ by the highest order term in divisor $x$ .

					$1$
	$x$	+	$2$		$x$	+	$0$

Step 4.3.2.3

Multiply the new quotient term by the divisor.

				$1$
$x$	+	$2$		$x$	+	$0$
			+	$x$	+	$2$

Step 4.3.2.4

The expression needs to be subtracted from the dividend, so change all the signs in $x + 2$

				$1$
$x$	+	$2$		$x$	+	$0$
			-	$x$	-	$2$

Step 4.3.2.5

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

				$1$
$x$	+	$2$		$x$	+	$0$
			-	$x$	-	$2$
					-	$2$

Step 4.3.2.6

The final answer is the quotient plus the remainder over the divisor.

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

Step 4.3.3

Split the single integral into multiple integrals.

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

Step 4.3.4

Apply the constant rule.

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

Step 4.3.5

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

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

Step 4.3.6

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

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

Step 4.3.7

Multiply $2$ by $- 1$ .

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

Step 4.3.8

Let $u = x + 2$ . Then $d u = d x$ . Rewrite using $u$ and $d$ $u$ .

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Step 4.3.8.1

Let $u = x + 2$ . Find $\frac{d u}{d x}$ .

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Step 4.3.8.1.1

Differentiate $x + 2$ .

$\frac{d}{d x} [x + 2]$

Step 4.3.8.1.2

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

$\frac{d}{d x} [x] + \frac{d}{d x} [2]$

Step 4.3.8.1.3

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

$1 + \frac{d}{d x} [2]$

Step 4.3.8.1.4

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

$1 + 0$

Step 4.3.8.1.5

Add $1$ and $0$ .

$1$

Step 4.3.8.2

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

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

Step 4.3.9

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

$- \ln (| y |) + C_{1} = - (x + C_{2} - 2 (\ln (| u |) + C_{3}))$

Step 4.3.10

Simplify.

$- \ln (| y |) + C_{1} = - (x - 2 \ln (| u |)) + C_{4}$

Step 4.3.11

Replace all occurrences of $u$ with $x + 2$ .

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

Step 4.3.12

Simplify.

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Step 4.3.12.1

Apply the distributive property.

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

Step 4.3.12.2

Multiply $- 2$ by $- 1$ .

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

Step 4.4

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

$- \ln (| y |) = - x + 2 \ln (| x + 2 |) + C$

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.

$- \ln (| y |) - 2 \ln (| x + 2 |) = - x + C$

Step 5.2

Simplify the left side.

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Step 5.2.1

Simplify each term.

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Step 5.2.1.1

Simplify $- 2 \ln (| x + 2 |)$ by moving $2$ inside the logarithm.

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

Step 5.2.1.2

Remove the absolute value in ${| x + 2 |}^{2}$ because exponentiations with even powers are always positive.

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

Step 5.3

Add $\ln ({(x + 2)}^{2})$ to both sides of the equation.

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

Step 5.4

Divide each term in $- \ln (| y |) = - x + C + \ln ({(x + 2)}^{2})$ by $- 1$ and simplify.

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Step 5.4.1

Divide each term in $- \ln (| y |) = - x + C + \ln ({(x + 2)}^{2})$ by $- 1$ .

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

Step 5.4.2

Simplify the left side.

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Step 5.4.2.1

Dividing two negative values results in a positive value.

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

Step 5.4.2.2

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

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

Step 5.4.3

Simplify the right side.

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Step 5.4.3.1

Simplify each term.

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Step 5.4.3.1.1

Dividing two negative values results in a positive value.

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

Step 5.4.3.1.2

Divide $x$ by $1$ .

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

Step 5.4.3.1.3

Move the negative one from the denominator of $\frac{C}{- 1}$ .

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

Step 5.4.3.1.4

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

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

Step 5.4.3.1.5

Move the negative one from the denominator of $\frac{\ln ({(x + 2)}^{2})}{- 1}$ .

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

Step 5.4.3.1.6

Rewrite $- 1 \cdot \ln ({(x + 2)}^{2})$ as $- \ln ({(x + 2)}^{2})$ .

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

Step 5.5

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

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

Step 5.6

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

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

Step 5.7

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

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

Step 5.8

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

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

Step 5.9

Solve for $y$ .

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Step 5.9.1

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

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

Step 5.9.2

Divide each term in $| y | {(x + 2)}^{2} = e^{x - C}$ by ${(x + 2)}^{2}$ and simplify.

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Step 5.9.2.1

Divide each term in $| y | {(x + 2)}^{2} = e^{x - C}$ by ${(x + 2)}^{2}$ .

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

Step 5.9.2.2

Simplify the left side.

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Step 5.9.2.2.1

Cancel the common factor of ${(x + 2)}^{2}$ .

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Step 5.9.2.2.1.1

Cancel the common factor.

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

Step 5.9.2.2.1.2

Divide $| y |$ by $1$ .

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

Step 5.9.3

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

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

Step 6

Group the constant terms together.

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Step 6.1

Simplify the constant of integration.

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

Step 6.2

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

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

Step 6.3

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

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

Step 6.4

Combine constants with the plus or minus.

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