Calculus Examples

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

Step 1

Rewrite the differential equation as $\frac{d y}{d x} + M (x) y = Q (x)$ .

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Step 1.1

Divide each term in $(x - 1) \frac{d y}{d x} + x y = (x - 1) e^{- x}$ by $x - 1$ .

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

Step 1.2

Cancel the common factor of $x - 1$ .

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Step 1.2.1

Cancel the common factor.

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

Step 1.2.2

Divide $\frac{d y}{d x}$ by $1$ .

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

Step 1.3

Cancel the common factor of $x - 1$ .

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Step 1.3.1

Cancel the common factor.

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

Step 1.3.2

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

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

Step 1.4

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

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

Step 1.5

Reorder $y$ and $\frac{x}{x - 1}$ .

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

Step 2

The integrating factor is defined by the formula $e^{\int P (x) d x}$ , where $P (x) = \frac{x}{x - 1}$ .

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Step 2.1

Set up the integration.

$e^{\int \frac{x}{x - 1} d x}$

Step 2.2

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

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Step 2.2.1

Divide $x$ by $x - 1$ .

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Step 2.2.1.1

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

$x$

$1$

$x$

$0$

Step 2.2.1.2

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

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

Step 2.2.1.3

Multiply the new quotient term by the divisor.

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

Step 2.2.1.4

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

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

Step 2.2.1.5

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

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

Step 2.2.1.6

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

$e^{\int 1 + \frac{1}{x - 1} d x}$

Step 2.2.2

Split the single integral into multiple integrals.

$e^{\int d x + \int \frac{1}{x - 1} d x}$

Step 2.2.3

Apply the constant rule.

$e^{x + C + \int \frac{1}{x - 1} d x}$

Step 2.2.4

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

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Step 2.2.4.1

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

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Step 2.2.4.1.1

Differentiate $x - 1$ .

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

Step 2.2.4.1.2

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 2.2.4.1.3

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 2.2.4.1.4

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

$1 + 0$

Step 2.2.4.1.5

Add $1$ and $0$ .

$1$

Step 2.2.4.2

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

$e^{x + C + \int \frac{1}{u} d u}$

Step 2.2.5

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

$e^{x + C + \ln (| u |) + C}$

Step 2.2.6

Simplify.

$e^{x + \ln (| u |) + C}$

Step 2.2.7

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

$e^{x + \ln (| x - 1 |) + C}$

Step 2.3

Remove the constant of integration.

$e^{x + \ln (x - 1)}$

Step 3

Multiply each term by the integrating factor $e^{x + \ln (x - 1)}$ .

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Step 3.1

Multiply each term by $e^{x + \ln (x - 1)}$ .

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

Step 3.2

Simplify each term.

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Step 3.2.1

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

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

Step 3.2.2

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

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

Step 3.3

To write $e^{x + \ln (x - 1)} \frac{d y}{d x}$ as a fraction with a common denominator, multiply by $\frac{x - 1}{x - 1}$ .

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

Step 3.4

Combine the numerators over the common denominator.

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

Step 3.5

Simplify the numerator.

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Step 3.5.1

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

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Step 3.5.1.1

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

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

Step 3.5.1.2

Factor $e^{x + \ln (x - 1)}$ out of $e^{x + \ln (x - 1)} x y$ .

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

Step 3.5.1.3

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

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

Step 3.5.2

Apply the distributive property.

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

Step 3.5.3

Move $- 1$ to the left of $\frac{d y}{d x}$ .

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

Step 3.5.4

Rewrite $- 1 \frac{d y}{d x}$ as $- \frac{d y}{d x}$ .

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

Step 3.6

Multiply $e^{x + \ln (x - 1)}$ by $e^{- x}$ by adding the exponents.

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Step 3.6.1

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

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

Step 3.6.2

Combine the opposite terms in $x + \ln (x - 1) - x$ .

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Step 3.6.2.1

Subtract $x$ from $x$ .

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

Step 3.6.2.2

Add $0$ and $\ln (x - 1)$ .

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

Step 3.7

Exponentiation and log are inverse functions.

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

Step 3.8

Reorder factors in $\frac{e^{x + \ln (x - 1)} (\frac{d y}{d x} x - \frac{d y}{d x} + x y)}{x - 1} = x - 1$ .

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

Step 4

Rewrite the left side as a result of differentiating a product.

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

Step 5

Set up an integral on each side.

$\int \frac{d}{d x} [e^{x + \ln (x - 1)} y] d x = \int x - 1 d x$

Step 6

Integrate the left side.

$e^{x + \ln (x - 1)} y = \int x - 1 d x$

Step 7

Integrate the right side.

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Step 7.1

Split the single integral into multiple integrals.

$e^{x + \ln (x - 1)} y = \int x d x + \int - 1 d x$

Step 7.2

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

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

Step 7.3

Apply the constant rule.

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

Step 7.4

Simplify.

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

Step 8

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

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Step 8.1

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

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

Step 8.2

Simplify the left side.

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Step 8.2.1

Cancel the common factor of $e^{x + \ln (x - 1)}$ .

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Step 8.2.1.1

Cancel the common factor.

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

Step 8.2.1.2

Divide $y$ by $1$ .

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

Step 8.3

Simplify the right side.

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Step 8.3.1

Simplify each term.

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Step 8.3.1.1

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

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

Step 8.3.1.2

Multiply the numerator by the reciprocal of the denominator.

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

Step 8.3.1.3

Combine.

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

Step 8.3.1.4

Multiply $x^{2}$ by $1$ .

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

Step 8.3.1.5

Move the negative in front of the fraction.

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