Calculus Examples

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

Step 1

Find $\frac{\partial M}{\partial y}$ where $M (x, y) = 2 x + x y$ .

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Step 1.1

Differentiate $M$ with respect to $y$ .

$\frac{\partial M}{\partial y} = \frac{d}{d y} [2 x + x y]$

Step 1.2

Differentiate.

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Step 1.2.1

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

$\frac{\partial M}{\partial y} = \frac{d}{d y} [2 x] + \frac{d}{d y} [x y]$

Step 1.2.2

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

$\frac{\partial M}{\partial y} = 0 + \frac{d}{d y} [x y]$

Step 1.3

Evaluate $\frac{d}{d y} [x y]$ .

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Step 1.3.1

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

$\frac{\partial M}{\partial y} = 0 + x \frac{d}{d y} [y]$

Step 1.3.2

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

$\frac{\partial M}{\partial y} = 0 + x \cdot 1$

Step 1.3.3

Multiply $x$ by $1$ .

$\frac{\partial M}{\partial y} = 0 + x$

Step 1.4

Add $0$ and $x$ .

$\frac{\partial M}{\partial y} = x$

Step 2

Find $\frac{\partial N}{\partial x}$ where $N (x, y) = y$ .

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Step 2.1

Differentiate $N$ with respect to $x$ .

$\frac{\partial N}{\partial x} = \frac{d}{d x} [y]$

Step 2.2

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

$\frac{\partial N}{\partial x} = 0$

Step 3

Check that $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$ .

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Step 3.1

Substitute $x$ for $\frac{\partial M}{\partial y}$ and $0$ for $\frac{\partial N}{\partial x}$ .

$x = 0$

Step 3.2

Since the left side does not equal the right side, the equation is not an identity.

$x = 0$ is not an identity.

Step 4

Find the integration factor $μ (x, y) = e^{\int \frac{\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y}}{M} d y}$ .

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Step 4.1

Substitute $0$ for $\frac{\partial N}{\partial x}$ .

$\frac{0 - \frac{\partial M}{\partial y}}{M}$

Step 4.2

Substitute $x$ for $\frac{\partial M}{\partial y}$ .

$\frac{0 - x}{M}$

Step 4.3

Substitute $2 x + x y$ for $M$ .

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Step 4.3.1

Substitute $2 x + x y$ for $M$ .

$\frac{0 - x}{2 x + x y}$

Step 4.3.2

Subtract $x$ from $0$ .

$\frac{- x}{2 x + x y}$

Step 4.3.3

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

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Step 4.3.3.1

Factor $x$ out of $2 x$ .

$\frac{- x}{x \cdot 2 + x y}$

Step 4.3.3.2

Factor $x$ out of $x y$ .

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

Step 4.3.3.3

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

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

Step 4.3.4

Cancel the common factor of $x$ .

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Step 4.3.4.1

Cancel the common factor.

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

Step 4.3.4.2

Rewrite the expression.

$\frac{- 1}{2 + y}$

Step 4.3.5

Substitute $2 x + x y$ for $M$ .

$- \frac{1}{2 + y}$

Step 4.4

Find the integration factor $μ (x, y) = e^{\int \frac{\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y}}{M} d y}$ .

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

Step 5

Evaluate the integral $e^{\int - \frac{1}{2 + y} d y}$ .

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Step 5.1

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

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

Step 5.2

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

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Step 5.2.1

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

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Step 5.2.1.1

Differentiate $2 + y$ .

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

Step 5.2.1.2

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

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

Step 5.2.1.3

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

$0 + \frac{d}{d y} [y]$

Step 5.2.1.4

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

$0 + 1$

Step 5.2.1.5

Add $0$ and $1$ .

$1$

Step 5.2.2

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

$μ (x, y) = e^{- \int \frac{1}{u} d u}$

Step 5.3

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

$μ (x, y) = e^{- (\ln (| u |) + C)}$

Step 5.4

Simplify.

$μ (x, y) = e^{- \ln (| u |)} + C$

Step 5.5

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

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

Step 5.6

Simplify each term.

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Step 5.6.1

Simplify $- \ln (| 2 + y |)$ by moving $- 1$ inside the logarithm.

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

Step 5.6.2

Exponentiation and log are inverse functions.

$μ (x, y) = {| 2 + y |}^{- 1} + C$

Step 5.6.3

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$μ (x, y) = \frac{1}{| 2 + y |} + C$

Step 6

Multiply both sides of $(2 x + x y) d x + y d y = 0$ by the integration factor $\frac{1}{2 + y}$ .

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Step 6.1

Multiply $2 x + x y$ by $\frac{1}{2 + y}$ .

$(2 x + x y) \frac{1}{2 + y} d x + y d y = 0$

Step 6.2

Multiply $2 x + x y$ by $\frac{1}{2 + y}$ .

$\frac{2 x + x y}{2 + y} d x + y d y = 0$

Step 6.3

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

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Step 6.3.1

Factor $x$ out of $2 x$ .

$\frac{x \cdot 2 + x y}{2 + y} d x + y d y = 0$

Step 6.3.2

Factor $x$ out of $x y$ .

$\frac{x \cdot 2 + x (y)}{2 + y} d x + y d y = 0$

Step 6.3.3

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

$\frac{x (2 + y)}{2 + y} d x + y d y = 0$

Step 6.4

Cancel the common factor of $2 + y$ .

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Step 6.4.1

Cancel the common factor.

$\frac{x (2 + y)}{2 + y} d x + y d y = 0$

Step 6.4.2

Divide $x$ by $1$ .

$x d x + y d y = 0$

Step 6.5

Multiply $y$ by $\frac{1}{2 + y}$ .

$x d x + y \frac{1}{2 + y} d y = 0$

Step 6.6

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

$x d x + \frac{y}{2 + y} d y = 0$

Step 7

Set $f (x, y)$ equal to the integral of $M (x, y)$ .

$f (x, y) = \int x d x$

Step 8

Integrate $M (x, y) = x$ to find $f (x, y)$ .

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Step 8.1

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

$f (x, y) = \frac{1}{2} x^{2} + C$

Step 9

Since the integral of $g (y)$ will contain an integration constant, we can replace $C$ with $g (y)$ .

$f (x, y) = \frac{1}{2} x^{2} + g (y)$

Step 10

Set $\frac{\partial f}{\partial y} = N (x, y)$ .

$\frac{\partial f}{\partial y} = \frac{y}{2 + y}$

Step 11

Find $\frac{\partial f}{\partial y}$ .

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Step 11.1

Differentiate $f$ with respect to $y$ .

$\frac{d}{d y} [\frac{1}{2} x^{2} + g (y)] = \frac{y}{2 + y}$

Step 11.2

By the Sum Rule, the derivative of $\frac{1}{2} x^{2} + g (y)$ with respect to $y$ is $\frac{d}{d y} [\frac{1}{2} x^{2}] + \frac{d}{d y} [g (y)]$ .

$\frac{d}{d y} [\frac{1}{2} x^{2}] + \frac{d}{d y} [g (y)] = \frac{y}{2 + y}$

Step 11.3

Since $\frac{1}{2} x^{2}$ is constant with respect to $y$ , the derivative of $\frac{1}{2} x^{2}$ with respect to $y$ is $0$ .

$0 + \frac{d}{d y} [g (y)] = \frac{y}{2 + y}$

Step 11.4

Differentiate using the function rule which states that the derivative of $g (y)$ is $\frac{d g}{d y}$ .

$0 + \frac{d g}{d y} = \frac{y}{2 + y}$

Step 11.5

Add $0$ and $\frac{d g}{d y}$ .

$\frac{d g}{d y} = \frac{y}{2 + y}$

Step 12

Find the antiderivative of $\frac{y}{2 + y}$ to find $g (y)$ .

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Step 12.1

Integrate both sides of $\frac{d g}{d y} = \frac{y}{2 + y}$ .

$\int \frac{d g}{d y} d y = \int \frac{y}{2 + y} d y$

Step 12.2

Evaluate $\int \frac{d g}{d y} d y$ .

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

Step 12.3

Reorder $2$ and $y$ .

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

Step 12.4

Divide $y$ by $y + 2$ .

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Step 12.4.1

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

$y$

$2$

$y$

$0$

Step 12.4.2

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

					$1$
	$y$	+	$2$		$y$	+	$0$

Step 12.4.3

Multiply the new quotient term by the divisor.

				$1$
$y$	+	$2$		$y$	+	$0$
			+	$y$	+	$2$

Step 12.4.4

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

				$1$
$y$	+	$2$		$y$	+	$0$
			-	$y$	-	$2$

Step 12.4.5

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

				$1$
$y$	+	$2$		$y$	+	$0$
			-	$y$	-	$2$
					-	$2$

Step 12.4.6

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

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

Step 12.5

Split the single integral into multiple integrals.

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

Step 12.6

Apply the constant rule.

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

Step 12.7

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

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

Step 12.8

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

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

Step 12.9

Multiply $2$ by $- 1$ .

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

Step 12.10

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

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Step 12.10.1

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

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Step 12.10.1.1

Differentiate $y + 2$ .

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

Step 12.10.1.2

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

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

Step 12.10.1.3

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

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

Step 12.10.1.4

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

$1 + 0$

Step 12.10.1.5

Add $1$ and $0$ .

$1$

Step 12.10.2

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

$g (y) = y + C - 2 \int \frac{1}{u} d u$

Step 12.11

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

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

Step 12.12

Simplify.

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

Step 12.13

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

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

Step 13

Substitute for $g (y)$ in $f (x, y) = \frac{1}{2} x^{2} + g (y)$ .

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

Step 14

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

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Step 14.1

Simplify each term.

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Step 14.1.1

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

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

Step 14.1.2

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

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

Step 14.1.3

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

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

Step 14.2

To write $- \ln ({(y + 2)}^{2})$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

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

Step 14.3

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

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

Step 14.4

Combine the numerators over the common denominator.

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

Step 14.5

Simplify the numerator.

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Step 14.5.1

Multiply $- \ln ({(y + 2)}^{2}) \cdot 2$ .

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Step 14.5.1.1

Multiply $2$ by $- 1$ .

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

Step 14.5.1.2

Simplify $- 2 \ln ({(y + 2)}^{2})$ by moving $2$ inside the logarithm.

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

Step 14.5.2

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

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Step 14.5.2.1

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

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

Step 14.5.2.2

Multiply $2$ by $2$ .

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