Calculus Examples

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

Step 1

Separate the variables.

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Step 1.1

Multiply both sides by $\frac{1}{e^{y}}$ .

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

Step 1.2

Simplify.

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Step 1.2.1

Cancel the common factor of $e^{y}$ .

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Step 1.2.1.1

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

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

Step 1.2.1.2

Cancel the common factor.

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

Step 1.2.1.3

Rewrite the expression.

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

Step 1.2.2

Rewrite ${(x + 2)}^{2}$ as $(x + 2) (x + 2)$ .

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

Step 1.2.3

Expand $(x + 2) (x + 2)$ using the FOIL Method.

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Step 1.2.3.1

Apply the distributive property.

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

Step 1.2.3.2

Apply the distributive property.

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

Step 1.2.3.3

Apply the distributive property.

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

Step 1.2.4

Simplify and combine like terms.

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Step 1.2.4.1

Simplify each term.

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Step 1.2.4.1.1

Multiply $x$ by $x$ .

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

Step 1.2.4.1.2

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

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

Step 1.2.4.1.3

Multiply $2$ by $2$ .

$\frac{1}{e^{y}} \frac{d y}{d x} = x^{2} + 2 x + 2 x + 4$

Step 1.2.4.2

Add $2 x$ and $2 x$ .

$\frac{1}{e^{y}} \frac{d y}{d x} = x^{2} + 4 x + 4$

Step 1.3

Rewrite the equation.

$\frac{1}{e^{y}} d y = (x^{2} + 4 x + 4) d x$

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

$\int \frac{1}{e^{y}} d y = \int x^{2} + 4 x + 4 d x$

Step 2.2

Integrate the left side.

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Step 2.2.1

Simplify the expression.

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Step 2.2.1.1

Negate the exponent of $e^{y}$ and move it out of the denominator.

$\int 1 {(e^{y})}^{- 1} d y = \int x^{2} + 4 x + 4 d x$

Step 2.2.1.2

Simplify.

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Step 2.2.1.2.1

Multiply the exponents in ${(e^{y})}^{- 1}$ .

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Step 2.2.1.2.1.1

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

$\int 1 e^{y \cdot - 1} d y = \int x^{2} + 4 x + 4 d x$

Step 2.2.1.2.1.2

Move $- 1$ to the left of $y$ .

$\int 1 e^{- 1 \cdot y} d y = \int x^{2} + 4 x + 4 d x$

Step 2.2.1.2.1.3

Rewrite $- 1 y$ as $- y$ .

$\int 1 e^{- y} d y = \int x^{2} + 4 x + 4 d x$

Step 2.2.1.2.2

Multiply $e^{- y}$ by $1$ .

$\int e^{- y} d y = \int x^{2} + 4 x + 4 d x$

Step 2.2.2

Let $u = - y$ . Then $d u = - d y$ , so $- d u = d y$ . Rewrite using $u$ and $d$ $u$ .

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Step 2.2.2.1

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

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Step 2.2.2.1.1

Differentiate $- y$ .

$\frac{d}{d y} [- y]$

Step 2.2.2.1.2

Since $- 1$ is constant with respect to $y$ , the derivative of $- y$ with respect to $y$ is $- \frac{d}{d y} [y]$ .

$- \frac{d}{d y} [y]$

Step 2.2.2.1.3

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

$- 1 \cdot 1$

Step 2.2.2.1.4

Multiply $- 1$ by $1$ .

$- 1$

Step 2.2.2.2

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

$\int - e^{u} d u = \int x^{2} + 4 x + 4 d x$

Step 2.2.3

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

$- \int e^{u} d u = \int x^{2} + 4 x + 4 d x$

Step 2.2.4

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

$- (e^{u} + C_{1}) = \int x^{2} + 4 x + 4 d x$

Step 2.2.5

Simplify.

$- e^{u} + C_{1} = \int x^{2} + 4 x + 4 d x$

Step 2.2.6

Replace all occurrences of $u$ with $- y$ .

$- e^{- y} + C_{1} = \int x^{2} + 4 x + 4 d x$

Step 2.3

Integrate the right side.

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Step 2.3.1

Split the single integral into multiple integrals.

$- e^{- y} + C_{1} = \int x^{2} d x + \int 4 x d x + \int 4 d x$

Step 2.3.2

By the Power Rule, the integral of $x^{2}$ with respect to $x$ is $\frac{1}{3} x^{3}$ .

$- e^{- y} + C_{1} = \frac{1}{3} x^{3} + C_{2} + \int 4 x d x + \int 4 d x$

Step 2.3.3

Since $4$ is constant with respect to $x$ , move $4$ out of the integral.

$- e^{- y} + C_{1} = \frac{1}{3} x^{3} + C_{2} + 4 \int x d x + \int 4 d x$

Step 2.3.4

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

$- e^{- y} + C_{1} = \frac{1}{3} x^{3} + C_{2} + 4 (\frac{1}{2} x^{2} + C_{3}) + \int 4 d x$

Step 2.3.5

Apply the constant rule.

$- e^{- y} + C_{1} = \frac{1}{3} x^{3} + C_{2} + 4 (\frac{1}{2} x^{2} + C_{3}) + 4 x + C_{4}$

Step 2.3.6

Simplify.

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Step 2.3.6.1

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

$- e^{- y} + C_{1} = \frac{1}{3} x^{3} + C_{2} + 4 (\frac{x^{2}}{2} + C_{3}) + 4 x + C_{4}$

Step 2.3.6.2

Simplify.

$- e^{- y} + C_{1} = \frac{x^{3}}{3} + 2 x^{2} + 4 x + C_{5}$

Step 2.3.6.3

Reorder terms.

$- e^{- y} + C_{1} = \frac{1}{3} x^{3} + 2 x^{2} + 4 x + C_{5}$

Step 2.3.7

Reorder terms.

$- e^{- y} + C_{1} = 2 x^{2} + \frac{1}{3} x^{3} + 4 x + C_{5}$

Step 2.4

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

$- e^{- y} = 2 x^{2} + \frac{1}{3} x^{3} + 4 x + K$

Step 3

Solve for $y$ .

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Step 3.1

Divide each term in $- e^{- y} = 2 x^{2} + \frac{1}{3} x^{3} + 4 x + K$ by $- 1$ and simplify.

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Step 3.1.1

Divide each term in $- e^{- y} = 2 x^{2} + \frac{1}{3} x^{3} + 4 x + K$ by $- 1$ .

$\frac{- e^{- y}}{- 1} = \frac{2 x^{2}}{- 1} + \frac{\frac{1}{3} x^{3}}{- 1} + \frac{4 x}{- 1} + \frac{K}{- 1}$

Step 3.1.2

Simplify the left side.

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Step 3.1.2.1

Dividing two negative values results in a positive value.

$\frac{e^{- y}}{1} = \frac{2 x^{2}}{- 1} + \frac{\frac{1}{3} x^{3}}{- 1} + \frac{4 x}{- 1} + \frac{K}{- 1}$

Step 3.1.2.2

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

$e^{- y} = \frac{2 x^{2}}{- 1} + \frac{\frac{1}{3} x^{3}}{- 1} + \frac{4 x}{- 1} + \frac{K}{- 1}$

Step 3.1.3

Simplify the right side.

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Step 3.1.3.1

Simplify each term.

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Step 3.1.3.1.1

Move the negative one from the denominator of $\frac{2 x^{2}}{- 1}$ .

$e^{- y} = - 1 \cdot (2 x^{2}) + \frac{\frac{1}{3} x^{3}}{- 1} + \frac{4 x}{- 1} + \frac{K}{- 1}$

Step 3.1.3.1.2

Rewrite $- 1 \cdot (2 x^{2})$ as $- (2 x^{2})$ .

$e^{- y} = - (2 x^{2}) + \frac{\frac{1}{3} x^{3}}{- 1} + \frac{4 x}{- 1} + \frac{K}{- 1}$

Step 3.1.3.1.3

Multiply $2$ by $- 1$ .

$e^{- y} = - 2 x^{2} + \frac{\frac{1}{3} x^{3}}{- 1} + \frac{4 x}{- 1} + \frac{K}{- 1}$

Step 3.1.3.1.4

Move the negative one from the denominator of $\frac{\frac{1}{3} x^{3}}{- 1}$ .

$e^{- y} = - 2 x^{2} - 1 \cdot (\frac{1}{3} x^{3}) + \frac{4 x}{- 1} + \frac{K}{- 1}$

Step 3.1.3.1.5

Rewrite $- 1 \cdot (\frac{1}{3} x^{3})$ as $- (\frac{1}{3} x^{3})$ .

$e^{- y} = - 2 x^{2} - (\frac{1}{3} x^{3}) + \frac{4 x}{- 1} + \frac{K}{- 1}$

Step 3.1.3.1.6

Combine $\frac{1}{3}$ and $x^{3}$ .

$e^{- y} = - 2 x^{2} - \frac{x^{3}}{3} + \frac{4 x}{- 1} + \frac{K}{- 1}$

Step 3.1.3.1.7

Move the negative one from the denominator of $\frac{4 x}{- 1}$ .

$e^{- y} = - 2 x^{2} - \frac{x^{3}}{3} - 1 \cdot (4 x) + \frac{K}{- 1}$

Step 3.1.3.1.8

Rewrite $- 1 \cdot (4 x)$ as $- (4 x)$ .

$e^{- y} = - 2 x^{2} - \frac{x^{3}}{3} - (4 x) + \frac{K}{- 1}$

Step 3.1.3.1.9

Multiply $4$ by $- 1$ .

$e^{- y} = - 2 x^{2} - \frac{x^{3}}{3} - 4 x + \frac{K}{- 1}$

Step 3.1.3.1.10

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

$e^{- y} = - 2 x^{2} - \frac{x^{3}}{3} - 4 x - 1 \cdot K$

Step 3.1.3.1.11

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

$e^{- y} = - 2 x^{2} - \frac{x^{3}}{3} - 4 x - K$

Step 3.2

Take the natural logarithm of both sides of the equation to remove the variable from the exponent.

$\ln (e^{- y}) = \ln (- 2 x^{2} - \frac{x^{3}}{3} - 4 x - K)$

Step 3.3

Expand the left side.

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Step 3.3.1

Expand $\ln (e^{- y})$ by moving $- y$ outside the logarithm.

$- y \ln (e) = \ln (- 2 x^{2} - \frac{x^{3}}{3} - 4 x - K)$

Step 3.3.2

The natural logarithm of $e$ is $1$ .

$- y \cdot 1 = \ln (- 2 x^{2} - \frac{x^{3}}{3} - 4 x - K)$

Step 3.3.3

Multiply $- 1$ by $1$ .

$- y = \ln (- 2 x^{2} - \frac{x^{3}}{3} - 4 x - K)$

Step 3.4

Divide each term in $- y = \ln (- 2 x^{2} - \frac{x^{3}}{3} - 4 x - K)$ by $- 1$ and simplify.

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Step 3.4.1

Divide each term in $- y = \ln (- 2 x^{2} - \frac{x^{3}}{3} - 4 x - K)$ by $- 1$ .

$\frac{- y}{- 1} = \frac{\ln (- 2 x^{2} - \frac{x^{3}}{3} - 4 x - K)}{- 1}$

Step 3.4.2

Simplify the left side.

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Step 3.4.2.1

Dividing two negative values results in a positive value.

$\frac{y}{1} = \frac{\ln (- 2 x^{2} - \frac{x^{3}}{3} - 4 x - K)}{- 1}$

Step 3.4.2.2

Divide $y$ by $1$ .

$y = \frac{\ln (- 2 x^{2} - \frac{x^{3}}{3} - 4 x - K)}{- 1}$

Step 3.4.3

Simplify the right side.

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Step 3.4.3.1

Move the negative one from the denominator of $\frac{\ln (- 2 x^{2} - \frac{x^{3}}{3} - 4 x - K)}{- 1}$ .

$y = - 1 \cdot \ln (- 2 x^{2} - \frac{x^{3}}{3} - 4 x - K)$

Step 3.4.3.2

Rewrite $- 1 \cdot \ln (- 2 x^{2} - \frac{x^{3}}{3} - 4 x - K)$ as $- \ln (- 2 x^{2} - \frac{x^{3}}{3} - 4 x - K)$ .

$y = - \ln (- 2 x^{2} - \frac{x^{3}}{3} - 4 x - K)$

Step 4

Simplify the constant of integration.

$y = - \ln (- 2 x^{2} - \frac{x^{3}}{3} - 4 x + K)$

Step 5

Use the initial condition to find the value of $K$ by substituting $1$ for $x$ and $0$ for $y$ in $y = - \ln (- 2 x^{2} - \frac{x^{3}}{3} - 4 x + K)$ .

$0 = - \ln (- 2 \cdot 1^{2} - \frac{1^{3}}{3} - 4 \cdot 1 + K)$

Step 6

Solve for $K$ .

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Step 6.1

Rewrite the equation as $- \ln (- 2 \cdot 1^{2} - \frac{1^{3}}{3} - 4 \cdot 1 + K) = 0$ .

$- \ln (- 2 \cdot 1^{2} - \frac{1^{3}}{3} - 4 \cdot 1 + K) = 0$

Step 6.2

Divide each term in $- \ln (- 2 \cdot 1^{2} - \frac{1^{3}}{3} - 4 \cdot 1 + K) = 0$ by $- 1$ and simplify.

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Step 6.2.1

Divide each term in $- \ln (- 2 \cdot 1^{2} - \frac{1^{3}}{3} - 4 \cdot 1 + K) = 0$ by $- 1$ .

$\frac{- \ln (- 2 \cdot 1^{2} - \frac{1^{3}}{3} - 4 \cdot 1 + K)}{- 1} = \frac{0}{- 1}$

Step 6.2.2

Simplify the left side.

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Step 6.2.2.1

Reduce the expression by cancelling the common factors.

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Step 6.2.2.1.1

Dividing two negative values results in a positive value.

$\frac{\ln (- 2 \cdot 1^{2} - \frac{1^{3}}{3} - 4 + K)}{1} = \frac{0}{- 1}$

Step 6.2.2.1.2

Divide $\ln (- 2 \cdot 1^{2} - \frac{1^{3}}{3} - 4 + K)$ by $1$ .

$\ln (- 2 \cdot 1^{2} - \frac{1^{3}}{3} - 4 + K) = \frac{0}{- 1}$

Step 6.2.2.2

Simplify each term.

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Step 6.2.2.2.1

One to any power is one.

$\ln (- 2 \cdot 1 - \frac{1^{3}}{3} - 4 + K) = \frac{0}{- 1}$

Step 6.2.2.2.2

Multiply $- 2$ by $1$ .

$\ln (- 2 - \frac{1^{3}}{3} - 4 + K) = \frac{0}{- 1}$

Step 6.2.2.2.3

One to any power is one.

$\ln (- 2 - \frac{1}{3} - 4 + K) = \frac{0}{- 1}$

Step 6.2.2.3

To write $- 2$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$\ln (- 2 \cdot \frac{3}{3} - \frac{1}{3} - 4 + K) = \frac{0}{- 1}$

Step 6.2.2.4

Combine $- 2$ and $\frac{3}{3}$ .

$\ln (\frac{- 2 \cdot 3}{3} - \frac{1}{3} - 4 + K) = \frac{0}{- 1}$

Step 6.2.2.5

Combine the numerators over the common denominator.

$\ln (\frac{- 2 \cdot 3 - 1}{3} - 4 + K) = \frac{0}{- 1}$

Step 6.2.2.6

Simplify the numerator.

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Step 6.2.2.6.1

Multiply $- 2$ by $3$ .

$\ln (\frac{- 6 - 1}{3} - 4 + K) = \frac{0}{- 1}$

Step 6.2.2.6.2

Subtract $1$ from $- 6$ .

$\ln (\frac{- 7}{3} - 4 + K) = \frac{0}{- 1}$

Step 6.2.2.7

Move the negative in front of the fraction.

$\ln (- \frac{7}{3} - 4 + K) = \frac{0}{- 1}$

Step 6.2.2.8

To write $- 4$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$\ln (- \frac{7}{3} - 4 \cdot \frac{3}{3} + K) = \frac{0}{- 1}$

Step 6.2.2.9

Combine $- 4$ and $\frac{3}{3}$ .

$\ln (- \frac{7}{3} + \frac{- 4 \cdot 3}{3} + K) = \frac{0}{- 1}$

Step 6.2.2.10

Combine the numerators over the common denominator.

$\ln (\frac{- 7 - 4 \cdot 3}{3} + K) = \frac{0}{- 1}$

Step 6.2.2.11

Simplify the numerator.

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Step 6.2.2.11.1

Multiply $- 4$ by $3$ .

$\ln (\frac{- 7 - 12}{3} + K) = \frac{0}{- 1}$

Step 6.2.2.11.2

Subtract $12$ from $- 7$ .

$\ln (\frac{- 19}{3} + K) = \frac{0}{- 1}$

Step 6.2.2.12

Move the negative in front of the fraction.

$\ln (- \frac{19}{3} + K) = \frac{0}{- 1}$

Step 6.2.3

Simplify the right side.

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Step 6.2.3.1

Divide $0$ by $- 1$ .

$\ln (- \frac{19}{3} + K) = 0$

Step 6.3

To solve for $K$ , rewrite the equation using properties of logarithms.

$e^{\ln (- \frac{19}{3} + K)} = e^{0}$

Step 6.4

Rewrite $\ln (- \frac{19}{3} + K) = 0$ in exponential form using the definition of a logarithm. If $x$ and $b$ are positive real numbers and $b \neq 1$ , then $\log_{b} (x) = y$ is equivalent to $b^{y} = x$ .

$e^{0} = - \frac{19}{3} + K$

Step 6.5

Solve for $K$ .

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Step 6.5.1

Rewrite the equation as $- \frac{19}{3} + K = e^{0}$ .

$- \frac{19}{3} + K = e^{0}$

Step 6.5.2

Anything raised to $0$ is $1$ .

$- \frac{19}{3} + K = 1$

Step 6.5.3

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

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Step 6.5.3.1

Add $\frac{19}{3}$ to both sides of the equation.

$K = 1 + \frac{19}{3}$

Step 6.5.3.2

Write $1$ as a fraction with a common denominator.

$K = \frac{3}{3} + \frac{19}{3}$

Step 6.5.3.3

Combine the numerators over the common denominator.

$K = \frac{3 + 19}{3}$

Step 6.5.3.4

Add $3$ and $19$ .

$K = \frac{22}{3}$

Step 7

Substitute $\frac{22}{3}$ for $K$ in $y = - \ln (- 2 x^{2} - \frac{x^{3}}{3} - 4 x + K)$ and simplify.

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Step 7.1

Substitute $\frac{22}{3}$ for $K$ .

$y = - \ln (- 2 x^{2} - \frac{x^{3}}{3} - 4 x + \frac{22}{3})$