Calculus Examples

$(e^{2 x} + 5) d y + y e^{2 x} d x = 0$

Step 1

Subtract $y e^{2 x} d x$ from both sides of the equation.

$(e^{2 x} + 5) d y = - y e^{2 x} d x$

Step 2

Multiply both sides by $\frac{1}{(e^{2 x} + 5) y}$ .

$\frac{1}{(e^{2 x} + 5) y} (e^{2 x} + 5) d y = \frac{1}{(e^{2 x} + 5) y} (- y e^{2 x}) d x$

Step 3

Simplify.

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Step 3.1

Cancel the common factor of $e^{2 x} + 5$ .

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Step 3.1.1

Factor $e^{2 x} + 5$ out of $(e^{2 x} + 5) y$ .

$\frac{1}{(e^{2 x} + 5) (y)} (e^{2 x} + 5) d y = \frac{1}{(e^{2 x} + 5) y} (- y e^{2 x}) d x$

Step 3.1.2

Cancel the common factor.

$\frac{1}{(e^{2 x} + 5) y} (e^{2 x} + 5) d y = \frac{1}{(e^{2 x} + 5) y} (- y e^{2 x}) d x$

Step 3.1.3

Rewrite the expression.

$\frac{1}{y} d y = \frac{1}{(e^{2 x} + 5) y} (- y e^{2 x}) d x$

Step 3.2

Rewrite using the commutative property of multiplication.

$\frac{1}{y} d y = - \frac{1}{(e^{2 x} + 5) y} (y e^{2 x}) d x$

Step 3.3

Cancel the common factor of $y$ .

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Step 3.3.1

Move the leading negative in $- \frac{1}{(e^{2 x} + 5) y}$ into the numerator.

$\frac{1}{y} d y = \frac{- 1}{(e^{2 x} + 5) y} (y e^{2 x}) d x$

Step 3.3.2

Factor $y$ out of $(e^{2 x} + 5) y$ .

$\frac{1}{y} d y = \frac{- 1}{y (e^{2 x} + 5)} (y e^{2 x}) d x$

Step 3.3.3

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

$\frac{1}{y} d y = \frac{- 1}{y (e^{2 x} + 5)} (y (e^{2 x})) d x$

Step 3.3.4

Cancel the common factor.

$\frac{1}{y} d y = \frac{- 1}{y (e^{2 x} + 5)} (y e^{2 x}) d x$

Step 3.3.5

Rewrite the expression.

$\frac{1}{y} d y = \frac{- 1}{e^{2 x} + 5} e^{2 x} d x$

Step 3.4

Combine $\frac{- 1}{e^{2 x} + 5}$ and $e^{2 x}$ .

$\frac{1}{y} d y = \frac{- e^{2 x}}{e^{2 x} + 5} d x$

Step 3.5

Move the negative in front of the fraction.

$\frac{1}{y} d y = - \frac{e^{2 x}}{e^{2 x} + 5} d x$

Step 4

Integrate both sides.

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Step 4.1

Set up an integral on each side.

$\int \frac{1}{y} d y = \int - \frac{e^{2 x}}{e^{2 x} + 5} d x$

Step 4.2

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

$\ln (| y |) + C_{1} = \int - \frac{e^{2 x}}{e^{2 x} + 5} d x$

Step 4.3

Integrate the right side.

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Step 4.3.1

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

$\ln (| y |) + C_{1} = - \int \frac{e^{2 x}}{e^{2 x} + 5} d x$

Step 4.3.2

Let $u_{2} = e^{2 x} + 5$ . Then $d u_{2} = 2 e^{2 x} d x$ , so $\frac{1}{2} d u_{2} = e^{2 x} d x$ . Rewrite using $u_{2}$ and $d$ $u_{2}$ .

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Step 4.3.2.1

Let $u_{2} = e^{2 x} + 5$ . Find $\frac{d u_{2}}{d x}$ .

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Step 4.3.2.1.1

Differentiate $e^{2 x} + 5$ .

$\frac{d}{d x} [e^{2 x} + 5]$

Step 4.3.2.1.2

By the Sum Rule, the derivative of $e^{2 x} + 5$ with respect to $x$ is $\frac{d}{d x} [e^{2 x}] + \frac{d}{d x} [5]$ .

$\frac{d}{d x} [e^{2 x}] + \frac{d}{d x} [5]$

Step 4.3.2.1.3

Evaluate $\frac{d}{d x} [e^{2 x}]$ .

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Step 4.3.2.1.3.1

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = e^{x}$ and $g (x) = 2 x$ .

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Step 4.3.2.1.3.1.1

To apply the Chain Rule, set $u_{1}$ as $2 x$ .

$\frac{d}{d u_{1}} [e^{u_{1}}] \frac{d}{d x} [2 x] + \frac{d}{d x} [5]$

Step 4.3.2.1.3.1.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u_{1}} [a^{u_{1}}]$ is $a^{u_{1}} \ln (a)$ where $a$ = $e$ .

$e^{u_{1}} \frac{d}{d x} [2 x] + \frac{d}{d x} [5]$

Step 4.3.2.1.3.1.3

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

$e^{2 x} \frac{d}{d x} [2 x] + \frac{d}{d x} [5]$

Step 4.3.2.1.3.2

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

$e^{2 x} (2 \frac{d}{d x} [x]) + \frac{d}{d x} [5]$

Step 4.3.2.1.3.3

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

$e^{2 x} (2 \cdot 1) + \frac{d}{d x} [5]$

Step 4.3.2.1.3.4

Multiply $2$ by $1$ .

$e^{2 x} \cdot 2 + \frac{d}{d x} [5]$

Step 4.3.2.1.3.5

Move $2$ to the left of $e^{2 x}$ .

$2 e^{2 x} + \frac{d}{d x} [5]$

Step 4.3.2.1.4

Differentiate using the Constant Rule.

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Step 4.3.2.1.4.1

Since $5$ is constant with respect to $x$ , the derivative of $5$ with respect to $x$ is $0$ .

$2 e^{2 x} + 0$

Step 4.3.2.1.4.2

Add $2 e^{2 x}$ and $0$ .

$2 e^{2 x}$

Step 4.3.2.2

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

$\ln (| y |) + C_{1} = - \int \frac{1}{u_{2}} \cdot \frac{1}{2} d u_{2}$

Step 4.3.3

Simplify.

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Step 4.3.3.1

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

$\ln (| y |) + C_{1} = - \int \frac{1}{u_{2} \cdot 2} d u_{2}$

Step 4.3.3.2

Move $2$ to the left of $u_{2}$ .

$\ln (| y |) + C_{1} = - \int \frac{1}{2 u_{2}} d u_{2}$

Step 4.3.4

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

$\ln (| y |) + C_{1} = - (\frac{1}{2} \int \frac{1}{u_{2}} d u_{2})$

Step 4.3.5

The integral of $\frac{1}{u_{2}}$ with respect to $u_{2}$ is $\ln (| u_{2} |)$ .

$\ln (| y |) + C_{1} = - \frac{1}{2} (\ln (| u_{2} |) + C_{2})$

Step 4.3.6

Simplify.

$\ln (| y |) + C_{1} = - \frac{1}{2} \ln (| u_{2} |) + C_{2}$

Step 4.3.7

Replace all occurrences of $u_{2}$ with $e^{2 x} + 5$ .

$\ln (| y |) + C_{1} = - \frac{1}{2} \ln (| e^{2 x} + 5 |) + C_{2}$

Step 4.4

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

$\ln (| y |) = - \frac{1}{2} \ln (| e^{2 x} + 5 |) + C$