Calculus Examples

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

Step 1

Separate the variables.

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Step 1.1

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

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

Step 1.2

Cancel the common factor of $y$ .

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Step 1.2.1

Cancel the common factor.

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

Step 1.2.2

Rewrite the expression.

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

Step 1.3

Rewrite the equation.

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

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

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

Step 2.2

The integral of $\frac{1}{y}$ with respect to $y$ is $\ln (| y |)$ .

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

Step 2.3

Integrate the right side.

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Step 2.3.1

Split the single integral into multiple integrals.

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

Step 2.3.2

Apply the constant rule.

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

Step 2.3.3

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

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

Step 2.3.4

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

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Step 2.3.4.1

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

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Step 2.3.4.1.1

Differentiate $2 x$ .

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

Step 2.3.4.1.2

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

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

Step 2.3.4.1.3

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

$2 \cdot 1$

Step 2.3.4.1.4

Multiply $2$ by $1$ .

$2$

Step 2.3.4.2

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

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

Step 2.3.5

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

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

Step 2.3.6

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

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

Step 2.3.7

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

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

Step 2.3.8

Simplify.

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

Step 2.3.9

Replace all occurrences of $u$ with $2 x$ .

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

Step 2.3.10

Reorder terms.

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

Step 2.4

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

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

Step 3

Solve for $y$ .

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Step 3.1

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

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

Step 3.2

Rewrite $\ln (| y |) = - \frac{1}{2} e^{2 x} + x + C$ 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 $b^{y} = x$ .

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

Step 3.3

Solve for $y$ .

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Step 3.3.1

Rewrite the equation as $| y | = e^{- \frac{1}{2} \cdot e^{2 x} + x + C}$ .

$| y | = e^{- \frac{1}{2} \cdot e^{2 x} + x + C}$

Step 3.3.2

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

$| y | = e^{- \frac{e^{2 x}}{2} + x + C}$

Step 3.3.3

Remove the absolute value term. This creates a $\pm$ on the right side of the equation because $| x | = \pm x$ .

$y = \pm e^{- \frac{e^{2 x}}{2} + x + C}$

Step 4

Group the constant terms together.

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Step 4.1

Rewrite $e^{- \frac{e^{2 x}}{2} + x + C}$ as $e^{- \frac{e^{2 x}}{2} + x} e^{C}$ .

$y = \pm e^{- \frac{e^{2 x}}{2} + x} e^{C}$

Step 4.2

Reorder $e^{- \frac{e^{2 x}}{2} + x}$ and $e^{C}$ .

$y = \pm e^{C} e^{- \frac{e^{2 x}}{2} + x}$

Step 4.3

Combine constants with the plus or minus.

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