Calculus Examples

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

Step 1

Separate the variables.

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Step 1.1

Regroup factors.

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

Step 1.2

Multiply both sides by $\frac{\ln (y)}{y}$ .

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

Step 1.3

Simplify.

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Step 1.3.1

Combine $x$ and $\frac{y}{\ln (y)}$ .

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

Step 1.3.2

Cancel the common factor of $\ln (y)$ .

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Step 1.3.2.1

Cancel the common factor.

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

Step 1.3.2.2

Rewrite the expression.

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

Step 1.3.3

Cancel the common factor of $y$ .

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Step 1.3.3.1

Factor $y$ out of $x y$ .

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

Step 1.3.3.2

Cancel the common factor.

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

Step 1.3.3.3

Rewrite the expression.

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

Step 1.4

Rewrite the equation.

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

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

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

Step 2.2

Integrate the left side.

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Step 2.2.1

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

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Step 2.2.1.1

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

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Step 2.2.1.1.1

Differentiate $\ln (y)$ .

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

Step 2.2.1.1.2

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

$\frac{1}{y}$

Step 2.2.1.2

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

$\int u d u = \int x d x$

Step 2.2.2

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

$\frac{1}{2} u^{2} + C_{1} = \int x d x$

Step 2.2.3

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

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

Step 2.3

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

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

Step 2.4

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

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

Step 3

Solve for $y$ .

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Step 3.1

Multiply both sides of the equation by $2$ .

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

Step 3.2

Simplify both sides of the equation.

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Step 3.2.1

Simplify the left side.

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Step 3.2.1.1

Simplify $2 (\frac{1}{2} \ln^{2} (y))$ .

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Step 3.2.1.1.1

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

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

Step 3.2.1.1.2

Cancel the common factor of $2$ .

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Step 3.2.1.1.2.1

Cancel the common factor.

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

Step 3.2.1.1.2.2

Rewrite the expression.

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

Step 3.2.2

Simplify the right side.

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Step 3.2.2.1

Simplify $2 (\frac{1}{2} x^{2} + K)$ .

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Step 3.2.2.1.1

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

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

Step 3.2.2.1.2

Apply the distributive property.

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

Step 3.2.2.1.3

Cancel the common factor of $2$ .

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Step 3.2.2.1.3.1

Cancel the common factor.

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

Step 3.2.2.1.3.2

Rewrite the expression.

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

Step 3.3

Solve for $y$ .

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Step 3.3.1

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

$\ln (y) = \pm \sqrt{x^{2} + 2 K}$

Step 3.3.2

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

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Step 3.3.2.1

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

$\ln (y) = \sqrt{x^{2} + 2 K}$

Step 3.3.2.2

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

$e^{\ln (y)} = e^{\sqrt{x^{2} + 2 K}}$

Step 3.3.2.3

Rewrite $\ln (y) = \sqrt{x^{2} + 2 K}$ 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 $b^{y} = x$ .

$e^{\sqrt{x^{2} + 2 K}} = y$

Step 3.3.2.4

Rewrite the equation as $y = e^{\sqrt{x^{2} + 2 K}}$ .

$y = e^{\sqrt{x^{2} + 2 K}}$

Step 3.3.2.5

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

$\ln (y) = - \sqrt{x^{2} + 2 K}$

Step 3.3.2.6

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

$e^{\ln (y)} = e^{- \sqrt{x^{2} + 2 K}}$

Step 3.3.2.7

Rewrite $\ln (y) = - \sqrt{x^{2} + 2 K}$ 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 $b^{y} = x$ .

$e^{- \sqrt{x^{2} + 2 K}} = y$

Step 3.3.2.8

Rewrite the equation as $y = e^{- \sqrt{x^{2} + 2 K}}$ .