Calculus Examples

$y \ln (x) - x \frac{d y}{d x} = 0$

Step 1

Separate the variables.

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Step 1.1

Solve for $\frac{d y}{d x}$ .

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Step 1.1.1

Subtract $y \ln (x)$ from both sides of the equation.

$- x \frac{d y}{d x} = - y \ln (x)$

Step 1.1.2

Divide each term in $- x \frac{d y}{d x} = - y \ln (x)$ by $- x$ and simplify.

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Step 1.1.2.1

Divide each term in $- x \frac{d y}{d x} = - y \ln (x)$ by $- x$ .

$\frac{- x \frac{d y}{d x}}{- x} = \frac{- y \ln (x)}{- x}$

Step 1.1.2.2

Simplify the left side.

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Step 1.1.2.2.1

Dividing two negative values results in a positive value.

$\frac{x \frac{d y}{d x}}{x} = \frac{- y \ln (x)}{- x}$

Step 1.1.2.2.2

Cancel the common factor of $x$ .

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Step 1.1.2.2.2.1

Cancel the common factor.

$\frac{x \frac{d y}{d x}}{x} = \frac{- y \ln (x)}{- x}$

Step 1.1.2.2.2.2

Divide $\frac{d y}{d x}$ by $1$ .

$\frac{d y}{d x} = \frac{- y \ln (x)}{- x}$

Step 1.1.2.3

Simplify the right side.

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Step 1.1.2.3.1

Dividing two negative values results in a positive value.

$\frac{d y}{d x} = \frac{y \ln (x)}{x}$

Step 1.2

Regroup factors.

$\frac{d y}{d x} = y \frac{\ln (x)}{x}$

Step 1.3

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

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

Step 1.4

Cancel the common factor of $y$ .

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Step 1.4.1

Cancel the common factor.

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

Step 1.4.2

Rewrite the expression.

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

Step 1.5

Rewrite the equation.

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

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

$\int \frac{1}{y} d y = \int \frac{\ln (x)}{x} d x$

Step 2.2

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

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

Step 2.3

Integrate the right side.

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Step 2.3.1

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

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Step 2.3.1.1

Let $u = \ln (x)$ . Find $\frac{d u}{d x}$ .

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Step 2.3.1.1.1

Differentiate $\ln (x)$ .

$\frac{d}{d x} [\ln (x)]$

Step 2.3.1.1.2

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

$\frac{1}{x}$

Step 2.3.1.2

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

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

Step 2.3.2

By the Power Rule, the integral of $u$ with respect to $u$ is $\frac{1}{2} u^{2}$ .

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

Step 2.3.3

Replace all occurrences of $u$ with $\ln (x)$ .

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

Step 2.4

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

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

Step 3

Solve for $y$ .

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Step 3.1

Simplify the right side.

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Step 3.1.1

Combine $\frac{1}{2}$ and $\ln^{2} (x)$ .

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

Step 3.2

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

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

Step 3.3

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

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

Step 3.4

Solve for $y$ .

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Step 3.4.1

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

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

Step 3.4.2

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

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

Step 4

Group the constant terms together.

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Step 4.1

Rewrite $e^{\frac{\ln^{2} (x)}{2} + C}$ as $e^{\frac{\ln^{2} (x)}{2}} e^{C}$ .

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

Step 4.2

Reorder $e^{\frac{\ln^{2} (x)}{2}}$ and $e^{C}$ .

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

Step 4.3

Combine constants with the plus or minus.

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