Calculus Examples

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

Step 1

Separate the variables.

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Step 1.1

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

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

Step 1.2

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

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Step 1.2.1

Cancel the common factor.

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

Step 1.2.2

Rewrite the expression.

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

Step 1.3

Rewrite the equation.

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

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

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

Step 2.2

Integrate the left side.

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Step 2.2.1

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

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Step 2.2.1.1

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

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Step 2.2.1.1.1

Differentiate $2 + y$ .

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

Step 2.2.1.1.2

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 2.2.1.1.3

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 2.2.1.1.4

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 2.2.1.1.5

Add $0$ and $1$ .

$1$

Step 2.2.1.2

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

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

Step 2.2.2

Apply basic rules of exponents.

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Step 2.2.2.1

Move $u^{2}$ out of the denominator by raising it to the $- 1$ power.

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

Step 2.2.2.2

Multiply the exponents in ${(u^{2})}^{- 1}$ .

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Step 2.2.2.2.1

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

$\int u^{2 \cdot - 1} d u = \int d x$

Step 2.2.2.2.2

Multiply $2$ by $- 1$ .

$\int u^{- 2} d u = \int d x$

Step 2.2.3

By the Power Rule, the integral of $u^{- 2}$ with respect to $u$ is $- u^{- 1}$ .

$- u^{- 1} + C_{1} = \int d x$

Step 2.2.4

Rewrite $- u^{- 1} + C_{1}$ as $- \frac{1}{u} + C_{1}$ .

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

Step 2.2.5

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

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

Step 2.3

Apply the constant rule.

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

Step 2.4

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

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

Step 3

Solve for $y$ .

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Step 3.1

Find the LCD of the terms in the equation.

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Step 3.1.1

Finding the LCD of a list of values is the same as finding the LCM of the denominators of those values.

$2 + y, 1, 1$

Step 3.1.2

Remove parentheses.

$2 + y, 1, 1$

Step 3.1.3

The LCM of one and any expression is the expression.

$2 + y$

Step 3.2

Multiply each term in $- \frac{1}{2 + y} = x + K$ by $2 + y$ to eliminate the fractions.

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Step 3.2.1

Multiply each term in $- \frac{1}{2 + y} = x + K$ by $2 + y$ .

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

Step 3.2.2

Simplify the left side.

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Step 3.2.2.1

Cancel the common factor of $2 + y$ .

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Step 3.2.2.1.1

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

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

Step 3.2.2.1.2

Cancel the common factor.

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

Step 3.2.2.1.3

Rewrite the expression.

$- 1 = x (2 + y) + K (2 + y)$

Step 3.2.3

Simplify the right side.

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Step 3.2.3.1

Simplify each term.

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Step 3.2.3.1.1

Apply the distributive property.

$- 1 = x \cdot 2 + x y + K (2 + y)$

Step 3.2.3.1.2

Move $2$ to the left of $x$ .

$- 1 = 2 \cdot x + x y + K (2 + y)$

Step 3.2.3.1.3

Apply the distributive property.

$- 1 = 2 x + x y + K \cdot 2 + K y$

Step 3.2.3.1.4

Move $2$ to the left of $K$ .

$- 1 = 2 x + x y + 2 K + K y$

Step 3.3

Solve the equation.

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Step 3.3.1

Rewrite the equation as $2 x + x y + 2 K + K y = - 1$ .

$2 x + x y + 2 K + K y = - 1$

Step 3.3.2

Move all terms not containing $y$ to the right side of the equation.

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Step 3.3.2.1

Subtract $2 x$ from both sides of the equation.

$x y + 2 K + K y = - 1 - 2 x$

Step 3.3.2.2

Subtract $2 K$ from both sides of the equation.

$x y + K y = - 1 - 2 x - 2 K$

Step 3.3.3

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

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Step 3.3.3.1

Factor $y$ out of $x y$ .

$y x + K y = - 1 - 2 x - 2 K$

Step 3.3.3.2

Factor $y$ out of $K y$ .

$y x + y K = - 1 - 2 x - 2 K$

Step 3.3.3.3

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

$y (x + K) = - 1 - 2 x - 2 K$

Step 3.3.4

Divide each term in $y (x + K) = - 1 - 2 x - 2 K$ by $x + K$ and simplify.

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Step 3.3.4.1

Divide each term in $y (x + K) = - 1 - 2 x - 2 K$ by $x + K$ .

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

Step 3.3.4.2

Simplify the left side.

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Step 3.3.4.2.1

Cancel the common factor of $x + K$ .

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Step 3.3.4.2.1.1

Cancel the common factor.

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

Step 3.3.4.2.1.2

Divide $y$ by $1$ .

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

Step 3.3.4.3

Simplify the right side.

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Step 3.3.4.3.1

Simplify each term.

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Step 3.3.4.3.1.1

Move the negative in front of the fraction.

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

Step 3.3.4.3.1.2

Move the negative in front of the fraction.

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

Step 3.3.4.3.1.3

Move the negative in front of the fraction.

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

Step 3.3.4.3.2

Simplify terms.

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Step 3.3.4.3.2.1

Combine the numerators over the common denominator.

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

Step 3.3.4.3.2.2

Combine the numerators over the common denominator.

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

Step 3.3.4.3.2.3

Rewrite $- 1$ as $- 1 (1)$ .

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

Step 3.3.4.3.2.4

Factor $- 1$ out of $- 2 x$ .

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

Step 3.3.4.3.2.5

Factor $- 1$ out of $- 1 (1) - (2 x)$ .

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

Step 3.3.4.3.2.6

Factor $- 1$ out of $- 2 K$ .

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

Step 3.3.4.3.2.7

Factor $- 1$ out of $- 1 (1 + 2 x) - (2 K)$ .

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

Step 3.3.4.3.2.8

Move the negative in front of the fraction.

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

Step 4

Simplify the constant of integration.

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