Calculus Examples

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

Step 1

Separate the variables.

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Step 1.1

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

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

Step 1.2

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

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Step 1.2.1

Factor ${(2 - y)}^{2}$ out of $e^{x + 2} {(2 - y)}^{2}$ .

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

Step 1.2.2

Cancel the common factor.

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

Step 1.2.3

Rewrite the expression.

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

Step 1.3

Rewrite the equation.

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

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

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

Step 2.2

Integrate the left side.

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Step 2.2.1

Let $u_{1} = 2 - y$ . Then $d u_{1} = - d y$ , so $- d u_{1} = d y$ . Rewrite using $u_{1}$ and $d$ $u_{1}$ .

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Step 2.2.1.1

Let $u_{1} = 2 - y$ . Find $\frac{d u_{1}}{d y}$ .

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Step 2.2.1.1.1

Rewrite.

$\frac{- 1}{1}$

Step 2.2.1.1.2

Divide $- 1$ by $1$ .

$- 1$

Step 2.2.1.2

Rewrite the problem using $u_{1}$ and $d u_{1}$ .

$\int \frac{- 1}{{u_{1}}^{2}} d u_{1} = \int e^{x + 2} d x$

Step 2.2.2

Move the negative in front of the fraction.

$\int - \frac{1}{{u_{1}}^{2}} d u_{1} = \int e^{x + 2} d x$

Step 2.2.3

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

$- \int \frac{1}{{u_{1}}^{2}} d u_{1} = \int e^{x + 2} d x$

Step 2.2.4

Apply basic rules of exponents.

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Step 2.2.4.1

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

$- \int {({u_{1}}^{2})}^{- 1} d u_{1} = \int e^{x + 2} d x$

Step 2.2.4.2

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

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Step 2.2.4.2.1

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

$- \int {u_{1}}^{2 \cdot - 1} d u_{1} = \int e^{x + 2} d x$

Step 2.2.4.2.2

Multiply $2$ by $- 1$ .

$- \int {u_{1}}^{- 2} d u_{1} = \int e^{x + 2} d x$

Step 2.2.5

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

$- (- {u_{1}}^{- 1} + C_{1}) = \int e^{x + 2} d x$

Step 2.2.6

Simplify.

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Step 2.2.6.1

Rewrite $- (- {u_{1}}^{- 1} + C_{1})$ as $- - \frac{1}{u_{1}} + C_{1}$ .

$- - \frac{1}{u_{1}} + C_{1} = \int e^{x + 2} d x$

Step 2.2.6.2

Simplify.

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Step 2.2.6.2.1

Multiply $- 1$ by $- 1$ .

$1 \frac{1}{u_{1}} + C_{1} = \int e^{x + 2} d x$

Step 2.2.6.2.2

Multiply $\frac{1}{u_{1}}$ by $1$ .

$\frac{1}{u_{1}} + C_{1} = \int e^{x + 2} d x$

Step 2.2.7

Replace all occurrences of $u_{1}$ with $2 - y$ .

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

Step 2.3

Integrate the right side.

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Step 2.3.1

Let $u_{2} = x + 2$ . Then $d u_{2} = d x$ . Rewrite using $u_{2}$ and $d$ $u_{2}$ .

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Step 2.3.1.1

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

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Step 2.3.1.1.1

Differentiate $x + 2$ .

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

Step 2.3.1.1.2

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

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

Step 2.3.1.1.3

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

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

Step 2.3.1.1.4

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

$1 + 0$

Step 2.3.1.1.5

Add $1$ and $0$ .

$1$

Step 2.3.1.2

Rewrite the problem using $u_{2}$ and $d u_{2}$ .

$\frac{1}{2 - y} + C_{1} = \int e^{u_{2}} d u_{2}$

Step 2.3.2

The integral of $e^{u_{2}}$ with respect to $u_{2}$ is $e^{u_{2}}$ .

$\frac{1}{2 - y} + C_{1} = e^{u_{2}} + C_{2}$

Step 2.3.3

Replace all occurrences of $u_{2}$ with $x + 2$ .

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

Step 2.4

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

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

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 $\frac{1}{2 - y} = e^{x + 2} + K$ by $2 - y$ to eliminate the fractions.

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Step 3.2.1

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

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

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

Cancel the common factor.

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

Step 3.2.2.1.2

Rewrite the expression.

$1 = e^{x + 2} (2 - y) + K (2 - y)$

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.

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

Step 3.2.3.1.2

Move $2$ to the left of $e^{x + 2}$ .

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

Step 3.2.3.1.3

Rewrite using the commutative property of multiplication.

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

Step 3.2.3.1.4

Apply the distributive property.

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

Step 3.2.3.1.5

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

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

Step 3.2.3.1.6

Rewrite using the commutative property of multiplication.

$1 = 2 e^{x + 2} - e^{x + 2} y + 2 K - K y$

Step 3.2.3.2

Reorder factors in $2 e^{x + 2} - e^{x + 2} y + 2 K - K y$ .

$1 = 2 e^{x + 2} - y e^{x + 2} + 2 K - K y$

Step 3.3

Solve the equation.

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Step 3.3.1

Rewrite the equation as $2 e^{x + 2} - y e^{x + 2} + 2 K - K y = 1$ .

$2 e^{x + 2} - y e^{x + 2} + 2 K - K y = 1$

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 $2 e^{x + 2}$ from both sides of the equation.

$- y e^{x + 2} + 2 K - K y = 1 - 2 e^{x + 2}$

Step 3.3.2.2

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

$- y e^{x + 2} - K y = 1 - 2 e^{x + 2} - 2 K$

Step 3.3.3

Factor $y$ out of $- y e^{x + 2} - K y$ .

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Step 3.3.3.1

Factor $y$ out of $- y e^{x + 2}$ .

$y (- 1 e^{x + 2}) - K y = 1 - 2 e^{x + 2} - 2 K$

Step 3.3.3.2

Factor $y$ out of $- K y$ .

$y (- 1 e^{x + 2}) + y (- K) = 1 - 2 e^{x + 2} - 2 K$

Step 3.3.3.3

Factor $y$ out of $y (- 1 e^{x + 2}) + y (- K)$ .

$y (- 1 e^{x + 2} - K) = 1 - 2 e^{x + 2} - 2 K$

Step 3.3.4

Rewrite $- 1 e^{x + 2}$ as $- e^{x + 2}$ .

$y (- e^{x + 2} - K) = 1 - 2 e^{x + 2} - 2 K$

Step 3.3.5

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

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Step 3.3.5.1

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

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

Step 3.3.5.2

Simplify the left side.

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Step 3.3.5.2.1

Cancel the common factor of $- e^{x + 2} - K$ .

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Step 3.3.5.2.1.1

Cancel the common factor.

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

Step 3.3.5.2.1.2

Divide $y$ by $1$ .

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

Step 3.3.5.3

Simplify the right side.

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Step 3.3.5.3.1

Simplify each term.

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Step 3.3.5.3.1.1

Move the negative in front of the fraction.

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

Step 3.3.5.3.1.2

Move the negative in front of the fraction.

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

Step 3.3.5.3.2

Combine the numerators over the common denominator.

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

Step 4

Simplify the constant of integration.

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