Calculus Examples

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

Step 1

Separate the variables.

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Step 1.1

Regroup factors.

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

Step 1.2

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

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

Step 1.3

Simplify.

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Step 1.3.1

Multiply $\frac{y}{2}$ by $\frac{x + 1}{x}$ .

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

Step 1.3.2

Cancel the common factor of $2$ .

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Step 1.3.2.1

Factor $2$ out of $2 x$ .

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

Step 1.3.2.2

Cancel the common factor.

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

Step 1.3.2.3

Rewrite the expression.

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

Step 1.3.3

Cancel the common factor of $y$ .

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Step 1.3.3.1

Cancel the common factor.

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

Step 1.3.3.2

Rewrite the expression.

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

Step 1.4

Rewrite the equation.

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

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

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

Step 2.2

Integrate the left side.

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Step 2.2.1

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

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

Step 2.2.2

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

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

Step 2.2.3

Simplify.

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

Step 2.3

Integrate the right side.

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Step 2.3.1

Split the fraction into multiple fractions.

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

Step 2.3.2

Split the single integral into multiple integrals.

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

Step 2.3.3

Cancel the common factor of $x$ .

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Step 2.3.3.1

Cancel the common factor.

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

Step 2.3.3.2

Rewrite the expression.

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

Step 2.3.4

Apply the constant rule.

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

Step 2.3.5

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

$2 \ln (| y |) + C_{1} = x + C_{2} + \ln (| x |) + C_{3}$

Step 2.3.6

Simplify.

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

Step 2.4

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

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

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.

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

Step 3.2

Simplify the left side.

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Step 3.2.1

Simplify $2 \ln (| y |) - \ln (| x |)$ .

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Step 3.2.1.1

Simplify each term.

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Step 3.2.1.1.1

Simplify $2 \ln (| y |)$ by moving $2$ inside the logarithm.

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

Step 3.2.1.1.2

Remove the absolute value in ${| y |}^{2}$ because exponentiations with even powers are always positive.

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

Step 3.2.1.2

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

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

Step 3.3

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

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

Step 3.4

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

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

Step 3.5

Solve for $y$ .

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Step 3.5.1

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

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

Step 3.5.2

Multiply both sides by $| x |$ .

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

Step 3.5.3

Simplify the left side.

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Step 3.5.3.1

Cancel the common factor of $| x |$ .

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Step 3.5.3.1.1

Cancel the common factor.

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

Step 3.5.3.1.2

Rewrite the expression.

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

Step 3.5.4

Solve for $y$ .

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Step 3.5.4.1

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

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

Step 3.5.4.2

The complete solution is the result of both the positive and negative portions of the solution.

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Step 3.5.4.2.1

First, use the positive value of the $\pm$ to find the first solution.

$y = \sqrt{e^{x + C} | x |}$

Step 3.5.4.2.2

Next, use the negative value of the $\pm$ to find the second solution.

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

Step 3.5.4.2.3

The complete solution is the result of both the positive and negative portions of the solution.

$y = \sqrt{e^{x + C} | x |}$

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

$y = \sqrt{e^{x + C} | x |}$

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

$y = \sqrt{e^{x + C} | x |}$

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

$y = \sqrt{e^{x + C} | x |}$

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

$y = \sqrt{e^{x + C} | x |}$

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

Step 4

Group the constant terms together.

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Step 4.1

Rewrite $e^{x + C}$ as $e^{x} e^{C}$ .

$y = \sqrt{e^{x} e^{C} | x |}$

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

Step 4.2

Reorder $e^{x}$ and $e^{C}$ .

$y = \sqrt{e^{C} e^{x} | x |}$

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

Step 4.3

Rewrite $e^{x + C}$ as $e^{x} e^{C}$ .

$y = \sqrt{e^{C} e^{x} | x |}$

$y = - \sqrt{e^{x} e^{C} | x |}$

Step 4.4

Reorder $e^{x}$ and $e^{C}$ .

$y = \sqrt{e^{C} e^{x} | x |}$

$y = - \sqrt{e^{C} e^{x} | x |}$

$y = \sqrt{e^{C} e^{x} | x |}$

$y = - \sqrt{e^{C} e^{x} | x |}$