Calculus Examples

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

Step 1

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

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Step 1.1

Differentiate $M$ with respect to $y$ .

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

Step 1.2

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

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

Step 2

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

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Step 2.1

Differentiate $N$ with respect to $x$ .

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

Step 2.2

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

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

Step 2.3

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

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Step 2.3.1

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

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

Step 2.3.2

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

$\frac{\partial N}{\partial x} = y \cdot 1 + \frac{d}{d x} [- y]$

Step 2.3.3

Multiply $y$ by $1$ .

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

Step 2.4

Differentiate using the Constant Rule.

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Step 2.4.1

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

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

Step 2.4.2

Add $y$ and $0$ .

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

Step 3

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

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Step 3.1

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

$0 = y$

Step 3.2

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

$0 = y$ is not an identity.

Step 4

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

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Step 4.1

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

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

Step 4.2

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

$\frac{0 - y}{N}$

Step 4.3

Substitute $x y - y$ for $N$ .

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Step 4.3.1

Substitute $x y - y$ for $N$ .

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

Step 4.3.2

Subtract $y$ from $0$ .

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

Step 4.3.3

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

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Step 4.3.3.1

Factor $y$ out of $x y$ .

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

Step 4.3.3.2

Factor $y$ out of $- y$ .

$\frac{- y}{y x + y \cdot - 1}$

Step 4.3.3.3

Factor $y$ out of $y x + y \cdot - 1$ .

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

Step 4.3.4

Cancel the common factor of $y$ .

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Step 4.3.4.1

Cancel the common factor.

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

Step 4.3.4.2

Rewrite the expression.

$\frac{- 1}{x - 1}$

Step 4.3.5

Move the negative in front of the fraction.

$- \frac{1}{x - 1}$

Step 4.4

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

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

Step 5

Evaluate the integral $e^{\int - \frac{1}{x - 1} d x}$ .

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Step 5.1

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

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

Step 5.2

Let $u = x - 1$ . Then $d u = d x$ . Rewrite using $u$ and $d$ $u$ .

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Step 5.2.1

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

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Step 5.2.1.1

Differentiate $x - 1$ .

$\frac{d}{d x} [x - 1]$

Step 5.2.1.2

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

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

Step 5.2.1.3

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

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

Step 5.2.1.4

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

$1 + 0$

Step 5.2.1.5

Add $1$ and $0$ .

$1$

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 $x - 1$ .

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

Step 5.6

Simplify each term.

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Step 5.6.1

Simplify $- \ln (| x - 1 |)$ by moving $- 1$ inside the logarithm.

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

Step 5.6.2

Exponentiation and log are inverse functions.

$μ (x, y) = {| x - 1 |}^{- 1} + C$

Step 5.6.3

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

$μ (x, y) = \frac{1}{| x - 1 |} + C$

Step 6

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

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Step 6.1

Multiply $x^{2}$ by $\frac{1}{x - 1}$ .

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

Step 6.2

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

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

Step 6.3

Multiply $x y - y$ by $\frac{1}{x - 1}$ .

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

Step 6.4

Multiply $x y - y$ by $\frac{1}{x - 1}$ .

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

Step 6.5

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

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Step 6.5.1

Factor $y$ out of $x y$ .

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

Step 6.5.2

Factor $y$ out of $- y$ .

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

Step 6.5.3

Factor $y$ out of $y x + y \cdot - 1$ .

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

Step 6.6

Cancel the common factor of $x - 1$ .

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Step 6.6.1

Cancel the common factor.

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

Step 6.6.2

Divide $y$ by $1$ .

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

Step 7

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

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

Step 8

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

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Step 8.1

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

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

Step 9

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

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

Step 10

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

$\frac{\partial f}{\partial x} = \frac{x^{2}}{x - 1}$

Step 11

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

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Step 11.1

Differentiate $f$ with respect to $x$ .

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

Step 11.2

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

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

Step 11.3

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

$0 + \frac{d}{d x} [g (x)] = \frac{x^{2}}{x - 1}$

Step 11.4

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

$0 + \frac{d g}{d x} = \frac{x^{2}}{x - 1}$

Step 11.5

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

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

Step 12

Find the antiderivative of $\frac{x^{2}}{x - 1}$ to find $g (x)$ .

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Step 12.1

Integrate both sides of $\frac{d g}{d x} = \frac{x^{2}}{x - 1}$ .

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

Step 12.2

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

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

Step 12.3

Divide $x^{2}$ by $x - 1$ .

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Step 12.3.1

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

$x$

$1$

$x^{2}$

$0 x$

$0$

Step 12.3.2

Divide the highest order term in the dividend $x^{2}$ by the highest order term in divisor $x$ .

					$x$
	$x$	-	$1$		$x^{2}$	+	$0 x$	+	$0$

Step 12.3.3

Multiply the new quotient term by the divisor.

				$x$
$x$	-	$1$		$x^{2}$	+	$0 x$	+	$0$
			+	$x^{2}$	-	$x$

Step 12.3.4

The expression needs to be subtracted from the dividend, so change all the signs in $x^{2} - x$

				$x$
$x$	-	$1$		$x^{2}$	+	$0 x$	+	$0$
			-	$x^{2}$	+	$x$

Step 12.3.5

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

				$x$
$x$	-	$1$		$x^{2}$	+	$0 x$	+	$0$
			-	$x^{2}$	+	$x$
					+	$x$

Step 12.3.6

Pull the next terms from the original dividend down into the current dividend.

				$x$
$x$	-	$1$		$x^{2}$	+	$0 x$	+	$0$
			-	$x^{2}$	+	$x$
					+	$x$	+	$0$

Step 12.3.7

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

				$x$	+	$1$
$x$	-	$1$		$x^{2}$	+	$0 x$	+	$0$
			-	$x^{2}$	+	$x$
					+	$x$	+	$0$

Step 12.3.8

Multiply the new quotient term by the divisor.

				$x$	+	$1$
$x$	-	$1$		$x^{2}$	+	$0 x$	+	$0$
			-	$x^{2}$	+	$x$
					+	$x$	+	$0$
					+	$x$	-	$1$

Step 12.3.9

The expression needs to be subtracted from the dividend, so change all the signs in $x - 1$

				$x$	+	$1$
$x$	-	$1$		$x^{2}$	+	$0 x$	+	$0$
			-	$x^{2}$	+	$x$
					+	$x$	+	$0$
					-	$x$	+	$1$

Step 12.3.10

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

				$x$	+	$1$
$x$	-	$1$		$x^{2}$	+	$0 x$	+	$0$
			-	$x^{2}$	+	$x$
					+	$x$	+	$0$
					-	$x$	+	$1$
							+	$1$

Step 12.3.11

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

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

Step 12.4

Split the single integral into multiple integrals.

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

Step 12.5

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

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

Step 12.6

Apply the constant rule.

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

Step 12.7

Let $u = x - 1$ . Then $d u = d x$ . Rewrite using $u$ and $d$ $u$ .

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Step 12.7.1

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

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Step 12.7.1.1

Differentiate $x - 1$ .

$\frac{d}{d x} [x - 1]$

Step 12.7.1.2

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

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

Step 12.7.1.3

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

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

Step 12.7.1.4

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

$1 + 0$

Step 12.7.1.5

Add $1$ and $0$ .

$1$

Step 12.7.2

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

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

Step 12.8

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

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

Step 12.9

Simplify.

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

Step 12.10

Replace all occurrences of $u$ with $x - 1$ .

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

Step 13

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

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

Step 14

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

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Step 14.1

Simplify each term.

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Step 14.1.1

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

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

Step 14.1.2

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

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

Step 14.2

To write $\ln (| x - 1 |)$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

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

Step 14.3

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

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

Step 14.4

Combine the numerators over the common denominator.

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

Step 14.5

Simplify the numerator.

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Step 14.5.1

Multiply $\ln (| x - 1 |) \cdot 2$ .

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Step 14.5.1.1

Reorder $\ln (| x - 1 |)$ and $2$ .

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

Step 14.5.1.2

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

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

Step 14.5.2

Remove the absolute value in ${| x - 1 |}^{2}$ because exponentiations with even powers are always positive.

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