Calculus Examples

$(\frac{x}{y}) d y + (1 + \ln (y)) d x = 0$

Step 1

Subtract $(1 + \ln (y)) d x$ from both sides of the equation.

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

Step 2

Multiply both sides by $\frac{1}{x (1 + \ln (y))}$ .

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

Step 3

Simplify.

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Step 3.1

Cancel the common factor of $x$ .

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Step 3.1.1

Cancel the common factor.

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

Step 3.1.2

Rewrite the expression.

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

Step 3.2

Multiply $\frac{1}{1 + \ln (y)}$ by $\frac{1}{y}$ .

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

Step 3.3

Reorder factors in $\frac{1}{(1 + \ln (y)) y}$ .

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

Step 3.4

Rewrite using the commutative property of multiplication.

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

Step 3.5

Cancel the common factor of $1 + \ln (y)$ .

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Step 3.5.1

Move the leading negative in $- \frac{1}{x (1 + \ln (y))}$ into the numerator.

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

Step 3.5.2

Factor $1 + \ln (y)$ out of $x (1 + \ln (y))$ .

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

Step 3.5.3

Cancel the common factor.

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

Step 3.5.4

Rewrite the expression.

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

Step 3.6

Move the negative in front of the fraction.

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

Step 4

Integrate both sides.

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Step 4.1

Set up an integral on each side.

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

Step 4.2

Integrate the left side.

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Step 4.2.1

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

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Step 4.2.1.1

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

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Step 4.2.1.1.1

Differentiate $\ln (y)$ .

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

Step 4.2.1.1.2

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

$\frac{1}{y}$

Step 4.2.1.2

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

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

Step 4.2.2

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

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Step 4.2.2.1

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

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Step 4.2.2.1.1

Differentiate $1 + u_{1}$ .

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

Step 4.2.2.1.2

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

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

Step 4.2.2.1.3

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

$0 + \frac{d}{d u_{1}} [u_{1}]$

Step 4.2.2.1.4

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

$0 + 1$

Step 4.2.2.1.5

Add $0$ and $1$ .

$1$

Step 4.2.2.2

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 4.2.3

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 4.2.4

Substitute back in for each integration substitution variable.

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Step 4.2.4.1

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

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

Step 4.2.4.2

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

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

Step 4.3

Integrate the right side.

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Step 4.3.1

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

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

Step 4.3.2

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

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

Step 4.3.3

Simplify.

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

Step 4.4

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

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

Step 5

Solve for $y$ .

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Step 5.1

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

$\ln (| 1 + \ln (y) |) + \ln (| x |) = C$

Step 5.2

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

$\ln (| 1 + \ln (y) | | x |) = C$

Step 5.3

To multiply absolute values, multiply the terms inside each absolute value.

$\ln (| (1 + \ln (y)) x |) = C$

Step 5.4

Apply the distributive property.

$\ln (| 1 x + \ln (y) x |) = C$

Step 5.5

Simplify the expression.

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Step 5.5.1

Multiply $x$ by $1$ .

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

Step 5.5.2

Reorder factors in $\ln (| x + \ln (y) x |)$ .

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

Step 5.6

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

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

Step 5.7

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

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

Step 5.8

Solve for $y$ .

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Step 5.8.1

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

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

Step 5.8.2

Solve for $y$ .

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Step 5.8.2.1

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

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

Step 5.8.2.2

Subtract $x$ from both sides of the equation.

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

Step 5.8.2.3

Divide each term in $x \ln (y) = \pm e^{C} - x$ by $x$ and simplify.

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Step 5.8.2.3.1

Divide each term in $x \ln (y) = \pm e^{C} - x$ by $x$ .

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

Step 5.8.2.3.2

Simplify the left side.

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Step 5.8.2.3.2.1

Cancel the common factor of $x$ .

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Step 5.8.2.3.2.1.1

Cancel the common factor.

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

Step 5.8.2.3.2.1.2

Divide $\ln (y)$ by $1$ .

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

Step 5.8.2.3.3

Simplify the right side.

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Step 5.8.2.3.3.1

Cancel the common factor of $x$ .

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Step 5.8.2.3.3.1.1

Cancel the common factor.

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

Step 5.8.2.3.3.1.2

Divide $- 1$ by $1$ .

$\ln (y) = \frac{\pm e^{C}}{x} - 1$

Step 5.8.2.4

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

$e^{\ln (y)} = e^{\frac{\pm e^{C}}{x} - 1}$

Step 5.8.2.5

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

$e^{\frac{\pm e^{C}}{x} - 1} = y$

Step 5.8.2.6

Rewrite the equation as $y = e^{\frac{\pm e^{C}}{x} - 1}$ .

$y = e^{\frac{\pm e^{C}}{x} - 1}$

Step 6

Simplify the constant of integration.

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