Calculus Examples

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

Step 1

Separate the variables.

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Step 1.1

Multiply both sides by $\frac{1}{e^{2 y}}$ .

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

Step 1.2

Simplify.

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Step 1.2.1

Cancel the common factor of $e^{2 y}$ .

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Step 1.2.1.1

Factor $e^{2 y}$ out of $e^{3 x} \cdot e^{2 y}$ .

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

Step 1.2.1.2

Cancel the common factor.

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

Step 1.2.1.3

Rewrite the expression.

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

Step 1.2.2

Multiply $e^{3 x - 2 y}$ by $e^{2 y}$ by adding the exponents.

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Step 1.2.2.1

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

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

Step 1.2.2.2

Combine the opposite terms in $3 x - 2 y + 2 y$ .

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Step 1.2.2.2.1

Add $- 2 y$ and $2 y$ .

$\frac{1}{e^{2 y}} \frac{d y}{d x} = e^{3 x + 0}$

Step 1.2.2.2.2

Add $3 x$ and $0$ .

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

Step 1.3

Rewrite the equation.

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

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

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

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 $e^{2 y}$ and move it out of the denominator.

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

Step 2.2.1.2

Simplify.

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Step 2.2.1.2.1

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

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Step 2.2.1.2.1.1

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

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

Step 2.2.1.2.1.2

Multiply $- 1$ by $2$ .

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

Step 2.2.1.2.2

Multiply $e^{- 2 y}$ by $1$ .

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

Step 2.2.2

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

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Step 2.2.2.1

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

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Step 2.2.2.1.1

Differentiate $- 2 y$ .

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

Step 2.2.2.1.2

Since $- 2$ is constant with respect to $y$ , the derivative of $- 2 y$ with respect to $y$ is $- 2 \frac{d}{d y} [y]$ .

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

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 $n = 1$ .

$- 2 \cdot 1$

Step 2.2.2.1.4

Multiply $- 2$ by $1$ .

$- 2$

Step 2.2.2.2

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

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

Step 2.2.3

Simplify.

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Step 2.2.3.1

Move the negative in front of the fraction.

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

Step 2.2.3.2

Combine $e^{u_{1}}$ and $\frac{1}{2}$ .

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

Step 2.2.4

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

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

Step 2.2.5

Since $\frac{1}{2}$ is constant with respect to $u_{1}$ , move $\frac{1}{2}$ out of the integral.

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

Step 2.2.6

The integral of $e^{u_{1}}$ with respect to $u_{1}$ is $e^{u_{1}}$ .

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

Step 2.2.7

Simplify.

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

Step 2.2.8

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

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

Step 2.3

Integrate the right side.

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Step 2.3.1

Let $u_{2} = 3 x$ . Then $d u_{2} = 3 d x$ , so $\frac{1}{3} d u_{2} = d x$ . Rewrite using $u_{2}$ and $d$ $u_{2}$ .

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Step 2.3.1.1

Let $u_{2} = 3 x$ . Find $\frac{d u_{2}}{d x}$ .

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Step 2.3.1.1.1

Differentiate $3 x$ .

$\frac{d}{d x} [3 x]$

Step 2.3.1.1.2

Since $3$ is constant with respect to $x$ , the derivative of $3 x$ with respect to $x$ is $3 \frac{d}{d x} [x]$ .

$3 \frac{d}{d x} [x]$

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 $n = 1$ .

$3 \cdot 1$

Step 2.3.1.1.4

Multiply $3$ by $1$ .

$3$

Step 2.3.1.2

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

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

Step 2.3.2

Combine $e^{u_{2}}$ and $\frac{1}{3}$ .

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

Step 2.3.3

Since $\frac{1}{3}$ is constant with respect to $u_{2}$ , move $\frac{1}{3}$ out of the integral.

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

Step 2.3.4

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

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

Step 2.3.5

Simplify.

$- \frac{1}{2} e^{- 2 y} + C_{1} = \frac{1}{3} e^{u_{2}} + C_{2}$

Step 2.3.6

Replace all occurrences of $u_{2}$ with $3 x$ .

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

Step 2.4

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

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

Step 3

Solve for $y$ .

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Step 3.1

Multiply both sides of the equation by $- 2$ .

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

Step 3.2

Simplify both sides of the equation.

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Step 3.2.1

Simplify the left side.

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Step 3.2.1.1

Simplify $- 2 (- \frac{1}{2} e^{- 2 y})$ .

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Step 3.2.1.1.1

Combine $e^{- 2 y}$ and $\frac{1}{2}$ .

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

Step 3.2.1.1.2

Cancel the common factor of $2$ .

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Step 3.2.1.1.2.1

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

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

Step 3.2.1.1.2.2

Factor $2$ out of $- 2$ .

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

Step 3.2.1.1.2.3

Cancel the common factor.

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

Step 3.2.1.1.2.4

Rewrite the expression.

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

Step 3.2.1.1.3

Multiply.

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Step 3.2.1.1.3.1

Multiply $- 1$ by $- 1$ .

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

Step 3.2.1.1.3.2

Multiply $e^{- 2 y}$ by $1$ .

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

Step 3.2.2

Simplify the right side.

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Step 3.2.2.1

Simplify $- 2 (\frac{1}{3} e^{3 x} + K)$ .

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Step 3.2.2.1.1

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

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

Step 3.2.2.1.2

Apply the distributive property.

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

Step 3.2.2.1.3

Combine $- 2$ and $\frac{e^{3 x}}{3}$ .

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

Step 3.2.2.1.4

Move the negative in front of the fraction.

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

Step 3.3

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

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

Step 3.4

Expand the left side.

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Step 3.4.1

Expand $\ln (e^{- 2 y})$ by moving $- 2 y$ outside the logarithm.

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

Step 3.4.2

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

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

Step 3.4.3

Multiply $- 2$ by $1$ .

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

Step 3.5

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

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Step 3.5.1

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

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

Step 3.5.2

Simplify the left side.

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Step 3.5.2.1

Cancel the common factor of $- 2$ .

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Step 3.5.2.1.1

Cancel the common factor.

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

Step 3.5.2.1.2

Divide $y$ by $1$ .

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

Step 3.5.3

Simplify the right side.

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Step 3.5.3.1

Move the negative in front of the fraction.

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

Step 4

Simplify the constant of integration.

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