Calculus Examples

$\frac{d y}{d x} = \frac{- 2 y}{x}$

Step 1

Separate the variables.

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Step 1.1

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

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

Step 1.2

Simplify.

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Step 1.2.1

Cancel the common factor of $y$ .

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Step 1.2.1.1

Factor $y$ out of $- 2 y$ .

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

Step 1.2.1.2

Cancel the common factor.

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

Step 1.2.1.3

Rewrite the expression.

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

Step 1.2.2

Move the negative in front of the fraction.

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

Step 1.3

Rewrite the equation.

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

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

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

Step 2.2

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

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

Step 2.3

Integrate the right side.

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Step 2.3.1

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

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

Step 2.3.2

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

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

Step 2.3.3

Multiply $2$ by $- 1$ .

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

Step 2.3.4

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

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

Step 2.3.5

Simplify.

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

Step 2.4

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

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

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.

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

Step 3.2

Simplify the left side.

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Step 3.2.1

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

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Step 3.2.1.1

Simplify each term.

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Step 3.2.1.1.1

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

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

Step 3.2.1.1.2

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

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

Step 3.2.1.2

Use the product property of logarithms, $\log_{b} (x) + \log_{b} (y) = \log_{b} (x y)$ .

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

Step 3.2.1.3

Reorder factors in $\ln (| y | x^{2})$ .

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

Step 3.3

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

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

Step 3.4

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

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

Step 3.5

Solve for $y$ .

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Step 3.5.1

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

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

Step 3.5.2

Divide each term in $x^{2} | y | = e^{C}$ by $x^{2}$ and simplify.

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Step 3.5.2.1

Divide each term in $x^{2} | y | = e^{C}$ by $x^{2}$ .

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

Step 3.5.2.2

Simplify the left side.

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Step 3.5.2.2.1

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

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Step 3.5.2.2.1.1

Cancel the common factor.

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

Step 3.5.2.2.1.2

Divide $| y |$ by $1$ .

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

Step 3.5.3

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

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

Step 4

Group the constant terms together.

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Step 4.1

Simplify the constant of integration.

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

Step 4.2

Combine constants with the plus or minus.

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