Calculus Examples

$\frac{d y}{d x} = \frac{2 y}{2 x + 1}$

Step 1

Separate the variables.

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Step 1.1

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

$\frac{1}{y} \frac{d y}{d x} = \frac{1}{y} \cdot \frac{2 y}{2 x + 1}$

Step 1.2

Cancel the common factor of $y$ .

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Step 1.2.1

Factor $y$ out of $2 y$ .

$\frac{1}{y} \frac{d y}{d x} = \frac{1}{y} \cdot \frac{y \cdot 2}{2 x + 1}$

Step 1.2.2

Cancel the common factor.

$\frac{1}{y} \frac{d y}{d x} = \frac{1}{y} \cdot \frac{y \cdot 2}{2 x + 1}$

Step 1.2.3

Rewrite the expression.

$\frac{1}{y} \frac{d y}{d x} = \frac{2}{2 x + 1}$

Step 1.3

Rewrite the equation.

$\frac{1}{y} d y = \frac{2}{2 x + 1} d x$

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

$\int \frac{1}{y} d y = \int \frac{2}{2 x + 1} d x$

Step 2.2

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

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

Step 2.3

Integrate the right side.

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Step 2.3.1

Since $2$ is constant with respect to $x$ , move $2$ out of the integral.

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

Step 2.3.2

Let $u = 2 x + 1$ . 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 2.3.2.1

Let $u = 2 x + 1$ . Find $\frac{d u}{d x}$ .

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Step 2.3.2.1.1

Differentiate $2 x + 1$ .

$\frac{d}{d x} [2 x + 1]$

Step 2.3.2.1.2

By the Sum Rule, the derivative of $2 x + 1$ with respect to $x$ is $\frac{d}{d x} [2 x] + \frac{d}{d x} [1]$ .

$\frac{d}{d x} [2 x] + \frac{d}{d x} [1]$

Step 2.3.2.1.3

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

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Step 2.3.2.1.3.1

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

$2 \frac{d}{d x} [x] + \frac{d}{d x} [1]$

Step 2.3.2.1.3.2

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

$2 \cdot 1 + \frac{d}{d x} [1]$

Step 2.3.2.1.3.3

Multiply $2$ by $1$ .

$2 + \frac{d}{d x} [1]$

Step 2.3.2.1.4

Differentiate using the Constant Rule.

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Step 2.3.2.1.4.1

Since $1$ is constant with respect to $x$ , the derivative of $1$ with respect to $x$ is $0$ .

$2 + 0$

Step 2.3.2.1.4.2

Add $2$ and $0$ .

$2$

Step 2.3.2.2

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

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

Step 2.3.3

Simplify.

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Step 2.3.3.1

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

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

Step 2.3.3.2

Move $2$ to the left of $u$ .

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

Step 2.3.4

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

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

Step 2.3.5

Simplify.

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Step 2.3.5.1

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

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

Step 2.3.5.2

Cancel the common factor of $2$ .

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Step 2.3.5.2.1

Cancel the common factor.

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

Step 2.3.5.2.2

Rewrite the expression.

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

Step 2.3.5.3

Multiply $\int \frac{1}{u} d u$ by $1$ .

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

Step 2.3.6

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

$\ln (| y |) + C_{1} = \ln (| u |) + C_{2}$

Step 2.3.7

Replace all occurrences of $u$ with $2 x + 1$ .

$\ln (| y |) + C_{1} = \ln (| 2 x + 1 |) + C_{2}$

Step 2.4

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

$\ln (| y |) = \ln (| 2 x + 1 |) + C$

Step 3

Solve for $y$ .

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Step 3.1

Move all the terms containing a logarithm to the left side of the equation.

$\ln (| y |) - \ln (| 2 x + 1 |) = C$

Step 3.2

Use the quotient property of logarithms, $\log_{b} (x) - \log_{b} (y) = \log_{b} (\frac{x}{y})$ .

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

Step 3.3

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

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

Step 3.4

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

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

Step 3.5

Solve for $y$ .

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Step 3.5.1

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

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

Step 3.5.2

Multiply both sides by $| 2 x + 1 |$ .

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

Step 3.5.3

Simplify the left side.

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Step 3.5.3.1

Cancel the common factor of $| 2 x + 1 |$ .

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Step 3.5.3.1.1

Cancel the common factor.

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

Step 3.5.3.1.2

Rewrite the expression.

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

Step 3.5.4

Solve for $y$ .

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Step 3.5.4.1

Reorder factors in $e^{C} | 2 x + 1 |$ .

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

Step 3.5.4.2

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

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

Step 4

Group the constant terms together.

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Step 4.1

Simplify the constant of integration.

$y = \pm | 2 x + 1 | C$

Step 4.2

Combine constants with the plus or minus.

$y = C | 2 x + 1 |$