Calculus Examples

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

Step 1

Separate the variables.

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Step 1.1

Solve for $\frac{d y}{d x}$ .

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Step 1.1.1

Rewrite the equation as $x - 3 \frac{d y}{d x} = \frac{1}{e^{x}} + 2$ .

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

Step 1.1.2

Subtract $x$ from both sides of the equation.

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

Step 1.1.3

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

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Step 1.1.3.1

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

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

Step 1.1.3.2

Simplify the left side.

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Step 1.1.3.2.1

Cancel the common factor of $- 3$ .

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Step 1.1.3.2.1.1

Cancel the common factor.

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

Step 1.1.3.2.1.2

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

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

Step 1.1.3.3

Simplify the right side.

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Step 1.1.3.3.1

Simplify each term.

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Step 1.1.3.3.1.1

Multiply the numerator by the reciprocal of the denominator.

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

Step 1.1.3.3.1.2

Combine.

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

Step 1.1.3.3.1.3

Multiply $1$ by $1$ .

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

Step 1.1.3.3.1.4

Move $- 3$ to the left of $e^{x}$ .

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

Step 1.1.3.3.1.5

Move the negative in front of the fraction.

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

Step 1.1.3.3.1.6

Move the negative in front of the fraction.

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

Step 1.1.3.3.1.7

Dividing two negative values results in a positive value.

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

Step 1.2

Rewrite the equation.

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

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

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

Step 2.2

Apply the constant rule.

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

Step 2.3

Integrate the right side.

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Step 2.3.1

Split the single integral into multiple integrals.

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

Step 2.3.2

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

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

Step 2.3.3

Since $\frac{1}{3}$ is constant with respect to $x$ , move $\frac{1}{3}$ out of the integral.

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

Step 2.3.4

Simplify the expression.

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Step 2.3.4.1

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

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

Step 2.3.4.2

Simplify.

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Step 2.3.4.2.1

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

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Step 2.3.4.2.1.1

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

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

Step 2.3.4.2.1.2

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

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

Step 2.3.4.2.1.3

Rewrite $- 1 x$ as $- x$ .

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

Step 2.3.4.2.2

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

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

Step 2.3.5

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

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Step 2.3.5.1

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

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Step 2.3.5.1.1

Differentiate $- x$ .

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

Step 2.3.5.1.2

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

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

Step 2.3.5.1.3

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

$- 1 \cdot 1$

Step 2.3.5.1.4

Multiply $- 1$ by $1$ .

$- 1$

Step 2.3.5.2

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

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

Step 2.3.6

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

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

Step 2.3.7

Simplify.

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Step 2.3.7.1

Multiply $- 1$ by $- 1$ .

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

Step 2.3.7.2

Multiply $\frac{1}{3}$ by $1$ .

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

Step 2.3.8

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

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

Step 2.3.9

Apply the constant rule.

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

Step 2.3.10

Since $\frac{1}{3}$ is constant with respect to $x$ , move $\frac{1}{3}$ out of the integral.

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

Step 2.3.11

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

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

Step 2.3.12

Simplify.

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Step 2.3.12.1

Simplify.

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

Step 2.3.12.2

Simplify.

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Step 2.3.12.2.1

Multiply $\frac{1}{3}$ by $\frac{1}{2}$ .

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

Step 2.3.12.2.2

Multiply $3$ by $2$ .

$y + C_{1} = \frac{1}{3} e^{u} - \frac{2}{3} x + \frac{1}{6} x^{2} + C_{5}$

Step 2.3.13

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

$y + C_{1} = \frac{1}{3} e^{- x} - \frac{2}{3} x + \frac{1}{6} x^{2} + C_{5}$

Step 2.4

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

$y = \frac{1}{3} e^{- x} - \frac{2}{3} x + \frac{1}{6} x^{2} + K$