Calculus Examples

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

Step 1

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

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

Step 2

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

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

Step 3

Simplify.

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Step 3.1

Rewrite using the commutative property of multiplication.

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

Step 3.2

Cancel the common factor of $2 + x$ .

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Step 3.2.1

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

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

Step 3.2.2

Cancel the common factor.

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

Step 3.2.3

Rewrite the expression.

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

Step 3.3

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

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

Step 3.4

Move the negative in front of the fraction.

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

Step 4

Integrate both sides.

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Step 4.1

Set up an integral on each side.

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

Step 4.2

Apply the constant rule.

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

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.

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

Step 4.3.2

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

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Step 4.3.2.1

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

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Step 4.3.2.1.1

Differentiate $2 + x$ .

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

Step 4.3.2.1.2

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

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

Step 4.3.2.1.3

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

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

Step 4.3.2.1.4

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

$0 + 1$

Step 4.3.2.1.5

Add $0$ and $1$ .

$1$

Step 4.3.2.2

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

$- y + C_{1} = - \int \frac{1}{u} d u$

Step 4.3.3

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

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

Step 4.3.4

Simplify.

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

Step 4.3.5

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

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

Step 4.4

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

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

Step 5

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

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Step 5.1

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

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

Step 5.2

Simplify the left side.

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Step 5.2.1

Dividing two negative values results in a positive value.

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

Step 5.2.2

Divide $y$ by $1$ .

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

Step 5.3

Simplify the right side.

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Step 5.3.1

Simplify each term.

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Step 5.3.1.1

Dividing two negative values results in a positive value.

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

Step 5.3.1.2

Divide $\ln (| 2 + x |)$ by $1$ .

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

Step 5.3.1.3

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

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

Step 5.3.1.4

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

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

Step 6

Simplify the constant of integration.

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