Calculus Examples

$\frac{d y}{d x} = (7 - 3 x) e^{- y}$

Step 1

Separate the variables.

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Step 1.1

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

$\frac{1}{e^{- y}} \frac{d y}{d x} = \frac{1}{e^{- y}} ((7 - 3 x) e^{- y})$

Step 1.2

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

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Step 1.2.1

Factor $e^{- y}$ out of $(7 - 3 x) e^{- y}$ .

$\frac{1}{e^{- y}} \frac{d y}{d x} = \frac{1}{e^{- y}} (e^{- y} (7 - 3 x))$

Step 1.2.2

Cancel the common factor.

$\frac{1}{e^{- y}} \frac{d y}{d x} = \frac{1}{e^{- y}} (e^{- y} (7 - 3 x))$

Step 1.2.3

Rewrite the expression.

$\frac{1}{e^{- y}} \frac{d y}{d x} = 7 - 3 x$

Step 1.3

Rewrite the equation.

$\frac{1}{e^{- y}} d y = (7 - 3 x) d x$

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

$\int \frac{1}{e^{- y}} d y = \int 7 - 3 x d x$

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.

$\int 1 {(e^{- y})}^{- 1} d y = \int 7 - 3 x d x$

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, ${(a^{m})}^{n} = a^{m n}$ .

$\int 1 e^{- y \cdot - 1} d y = \int 7 - 3 x d x$

Step 2.2.1.2.1.2

Multiply $- y \cdot - 1$ .

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Step 2.2.1.2.1.2.1

Multiply $- 1$ by $- 1$ .

$\int 1 e^{1 y} d y = \int 7 - 3 x d x$

Step 2.2.1.2.1.2.2

Multiply $y$ by $1$ .

$\int 1 e^{y} d y = \int 7 - 3 x d x$

Step 2.2.1.2.2

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

$\int e^{y} d y = \int 7 - 3 x d x$

Step 2.2.2

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

$e^{y} + C_{1} = \int 7 - 3 x d x$

Step 2.3

Integrate the right side.

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Step 2.3.1

Split the single integral into multiple integrals.

$e^{y} + C_{1} = \int 7 d x + \int - 3 x d x$

Step 2.3.2

Apply the constant rule.

$e^{y} + C_{1} = 7 x + C_{2} + \int - 3 x d x$

Step 2.3.3

Since $- 3$ is constant with respect to $x$ , move $- 3$ out of the integral.

$e^{y} + C_{1} = 7 x + C_{2} - 3 \int x d x$

Step 2.3.4

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

$e^{y} + C_{1} = 7 x + C_{2} - 3 (\frac{1}{2} x^{2} + C_{3})$

Step 2.3.5

Simplify.

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Step 2.3.5.1

Simplify.

$e^{y} + C_{1} = 7 x - 3 (\frac{1}{2}) x^{2} + C_{4}$

Step 2.3.5.2

Simplify.

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Step 2.3.5.2.1

Combine $- 3$ and $\frac{1}{2}$ .

$e^{y} + C_{1} = 7 x + \frac{- 3}{2} x^{2} + C_{4}$

Step 2.3.5.2.2

Move the negative in front of the fraction.

$e^{y} + C_{1} = 7 x - \frac{3}{2} x^{2} + C_{4}$

Step 2.3.6

Reorder terms.

$e^{y} + C_{1} = - \frac{3}{2} x^{2} + 7 x + C_{4}$

Step 2.4

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

$e^{y} = - \frac{3}{2} x^{2} + 7 x + K$

Step 3

Solve for $y$ .

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Step 3.1

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

$\ln (e^{y}) = \ln (- \frac{3 x^{2}}{2} + 7 x + K)$

Step 3.2

Expand the left side.

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Step 3.2.1

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

$y \ln (e) = \ln (- \frac{3 x^{2}}{2} + 7 x + K)$

Step 3.2.2

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

$y \cdot 1 = \ln (- \frac{3 x^{2}}{2} + 7 x + K)$

Step 3.2.3

Multiply $y$ by $1$ .

$y = \ln (- \frac{3 x^{2}}{2} + 7 x + K)$