Calculus Examples

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

Step 1

Rewrite the differential equation as a function of $\frac{y}{x}$ .

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Step 1.1

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

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

Step 1.2

Factor out $\frac{y}{x}$ from $\frac{y \ln (\frac{y}{x})}{x}$ .

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Step 1.2.1

Factor $\frac{y}{x}$ out of $\frac{y \ln (\frac{y}{x})}{x}$ .

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

Step 1.2.2

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

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

Step 2

Let $V = \frac{y}{x}$ . Substitute $V$ for $\frac{y}{x}$ .

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

Step 3

Solve $V = \frac{y}{x}$ for $y$ .

$y = V x$

Step 4

Use the product rule to find the derivative of $y = V x$ with respect to $x$ .

$\frac{d y}{d x} = x \frac{d V}{d x} + V$

Step 5

Substitute $x \frac{d V}{d x} + V$ for $\frac{d y}{d x}$ .

$x \frac{d V}{d x} + V = \ln (V) V$

Step 6

Solve the substituted differential equation.

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Step 6.1

Separate the variables.

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Step 6.1.1

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

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Step 6.1.1.1

Reorder factors in $\ln (V) V$ .

$x \frac{d V}{d x} + V = V \ln (V)$

Step 6.1.1.2

Subtract $V$ from both sides of the equation.

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

Step 6.1.1.3

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

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Step 6.1.1.3.1

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

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

Step 6.1.1.3.2

Simplify the left side.

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Step 6.1.1.3.2.1

Cancel the common factor of $x$ .

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Step 6.1.1.3.2.1.1

Cancel the common factor.

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

Step 6.1.1.3.2.1.2

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

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

Step 6.1.1.3.3

Simplify the right side.

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Step 6.1.1.3.3.1

Move the negative in front of the fraction.

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

Step 6.1.2

Factor.

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Step 6.1.2.1

Combine the numerators over the common denominator.

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

Step 6.1.2.2

Factor $V$ out of $V \ln (V) - V$ .

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Step 6.1.2.2.1

Factor $V$ out of $V \ln (V)$ .

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

Step 6.1.2.2.2

Factor $V$ out of $- V$ .

$\frac{d V}{d x} = \frac{V (\ln (V)) + V \cdot - 1}{x}$

Step 6.1.2.2.3

Factor $V$ out of $V (\ln (V)) + V \cdot - 1$ .

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

Step 6.1.3

Multiply both sides by $\frac{1}{V (\ln (V) - 1)}$ .

$\frac{1}{V (\ln (V) - 1)} \frac{d V}{d x} = \frac{1}{V (\ln (V) - 1)} \cdot \frac{V (\ln (V) - 1)}{x}$

Step 6.1.4

Cancel the common factor of $V (\ln (V) - 1)$ .

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Step 6.1.4.1

Cancel the common factor.

$\frac{1}{V (\ln (V) - 1)} \frac{d V}{d x} = \frac{1}{V (\ln (V) - 1)} \cdot \frac{V (\ln (V) - 1)}{x}$

Step 6.1.4.2

Rewrite the expression.

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

Step 6.1.5

Rewrite the equation.

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

Step 6.2

Integrate both sides.

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Step 6.2.1

Set up an integral on each side.

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

Step 6.2.2

Integrate the left side.

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Step 6.2.2.1

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

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Step 6.2.2.1.1

Let $u_{1} = \ln (V)$ . Find $\frac{d u_{1}}{d V}$ .

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Step 6.2.2.1.1.1

Differentiate $\ln (V)$ .

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

Step 6.2.2.1.1.2

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

$\frac{1}{V}$

Step 6.2.2.1.2

Rewrite the problem using $u_{1}$ and $d u_{1}$ .

$\int \frac{1}{u_{1} - 1} d u_{1} = \int \frac{1}{x} d x$

Step 6.2.2.2

Let $u_{2} = u_{1} - 1$ . Then $d u_{2} = d u_{1}$ . Rewrite using $u_{2}$ and $d$ $u_{2}$ .

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Step 6.2.2.2.1

Let $u_{2} = u_{1} - 1$ . Find $\frac{d u_{2}}{d u_{1}}$ .

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Step 6.2.2.2.1.1

Differentiate $u_{1} - 1$ .

$\frac{d}{d u_{1}} [u_{1} - 1]$

Step 6.2.2.2.1.2

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

$\frac{d}{d u_{1}} [u_{1}] + \frac{d}{d u_{1}} [- 1]$

Step 6.2.2.2.1.3

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

$1 + \frac{d}{d u_{1}} [- 1]$

Step 6.2.2.2.1.4

Since $- 1$ is constant with respect to $u_{1}$ , the derivative of $- 1$ with respect to $u_{1}$ is $0$ .

$1 + 0$

Step 6.2.2.2.1.5

Add $1$ and $0$ .

$1$

Step 6.2.2.2.2

Rewrite the problem using $u_{2}$ and $d u_{2}$ .

$\int \frac{1}{u_{2}} d u_{2} = \int \frac{1}{x} d x$

Step 6.2.2.3

The integral of $\frac{1}{u_{2}}$ with respect to $u_{2}$ is $\ln (| u_{2} |)$ .

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

Step 6.2.2.4

Substitute back in for each integration substitution variable.

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Step 6.2.2.4.1

Replace all occurrences of $u_{2}$ with $u_{1} - 1$ .

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

Step 6.2.2.4.2

Replace all occurrences of $u_{1}$ with $\ln (V)$ .

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

Step 6.2.3

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

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

Step 6.2.4

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

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

Step 6.3

Solve for $V$ .

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Step 6.3.1

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

$\ln (| \ln (V) - 1 |) - \ln (| x |) = C$

Step 6.3.2

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

$\ln (\frac{| \ln (V) - 1 |}{| x |}) = C$

Step 6.3.3

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

$e^{\ln (\frac{| \ln (V) - 1 |}{| x |})} = e^{C}$

Step 6.3.4

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

$e^{C} = \frac{| \ln (V) - 1 |}{| x |}$

Step 6.3.5

Solve for $V$ .

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Step 6.3.5.1

Rewrite the equation as $\frac{| \ln (V) - 1 |}{| x |} = e^{C}$ .

$\frac{| \ln (V) - 1 |}{| x |} = e^{C}$

Step 6.3.5.2

Multiply both sides by $| x |$ .

$\frac{| \ln (V) - 1 |}{| x |} | x | = e^{C} | x |$

Step 6.3.5.3

Simplify the left side.

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Step 6.3.5.3.1

Cancel the common factor of $| x |$ .

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Step 6.3.5.3.1.1

Cancel the common factor.

$\frac{| \ln (V) - 1 |}{| x |} | x | = e^{C} | x |$

Step 6.3.5.3.1.2

Rewrite the expression.

$| \ln (V) - 1 | = e^{C} | x |$

Step 6.3.5.4

Solve for $V$ .

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Step 6.3.5.4.1

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

$| \ln (V) - 1 | = | x | e^{C}$

Step 6.3.5.4.2

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

$\ln (V) - 1 = \pm | x | e^{C}$

Step 6.3.5.4.3

Add $1$ to both sides of the equation.

$\ln (V) = \pm | x | e^{C} + 1$

Step 6.3.5.4.4

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

$e^{\ln (V)} = e^{\pm | x | e^{C} + 1}$

Step 6.3.5.4.5

Rewrite $\ln (V) = \pm | x | e^{C} + 1$ 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 $b^{y} = x$ .

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

Step 6.3.5.4.6

Solve for $V$ .

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Step 6.3.5.4.6.1

Rewrite the equation as $V = e^{\pm | x | e^{C} + 1}$ .

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

Step 6.3.5.4.6.2

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

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

Step 6.4

Group the constant terms together.

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Step 6.4.1

Simplify the constant of integration.

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

Step 6.4.2

Combine constants with the plus or minus.

$V = e^{C | x | + 1}$

Step 7

Substitute $\frac{y}{x}$ for $V$ .

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

Step 8

Solve $\frac{y}{x} = e^{C | x | + 1}$ for $y$ .

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Step 8.1

Multiply both sides by $x$ .

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

Step 8.2

Simplify.

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Step 8.2.1

Simplify the left side.

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Step 8.2.1.1

Cancel the common factor of $x$ .

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Step 8.2.1.1.1

Cancel the common factor.

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

Step 8.2.1.1.2

Rewrite the expression.

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

Step 8.2.2

Simplify the right side.

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Step 8.2.2.1

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

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