Calculus Examples

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

Step 1

Separate the variables.

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Step 1.1

Divide each term in $x \frac{d y}{d x} = (1 + x) (1 + y)$ by $x$ and simplify.

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Step 1.1.1

Divide each term in $x \frac{d y}{d x} = (1 + x) (1 + y)$ by $x$ .

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

Step 1.1.2

Simplify the left side.

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Step 1.1.2.1

Cancel the common factor of $x$ .

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Step 1.1.2.1.1

Cancel the common factor.

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

Step 1.1.2.1.2

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

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

Step 1.2

Regroup factors.

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

Step 1.3

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

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

Step 1.4

Cancel the common factor of $1 + y$ .

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Step 1.4.1

Factor $1 + y$ out of $\frac{1 + x}{x} (1 + y)$ .

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

Step 1.4.2

Cancel the common factor.

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

Step 1.4.3

Rewrite the expression.

$\frac{1}{1 + y} \frac{d y}{d x} = \frac{1 + x}{x}$

Step 1.5

Rewrite the equation.

$\frac{1}{1 + y} d y = \frac{1 + x}{x} d x$

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

$\int \frac{1}{1 + y} d y = \int \frac{1 + x}{x} d x$

Step 2.2

Integrate the left side.

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Step 2.2.1

Let $u = 1 + y$ . Then $d u = d y$ . Rewrite using $u$ and $d$ $u$ .

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Step 2.2.1.1

Let $u = 1 + y$ . Find $\frac{d u}{d y}$ .

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Step 2.2.1.1.1

Differentiate $1 + y$ .

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

Step 2.2.1.1.2

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

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

Step 2.2.1.1.3

Since $1$ is constant with respect to $y$ , the derivative of $1$ with respect to $y$ is $0$ .

$0 + \frac{d}{d y} [y]$

Step 2.2.1.1.4

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

$0 + 1$

Step 2.2.1.1.5

Add $0$ and $1$ .

$1$

Step 2.2.1.2

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

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

Step 2.2.2

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

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

Step 2.2.3

Replace all occurrences of $u$ with $1 + y$ .

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

Step 2.3

Integrate the right side.

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Step 2.3.1

Split the fraction into multiple fractions.

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

Step 2.3.2

Split the single integral into multiple integrals.

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

Step 2.3.3

Cancel the common factor of $x$ .

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Step 2.3.3.1

Cancel the common factor.

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

Step 2.3.3.2

Rewrite the expression.

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

Step 2.3.4

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

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

Step 2.3.5

Apply the constant rule.

$\ln (| 1 + y |) + C_{1} = \ln (| x |) + C_{2} + x + C_{3}$

Step 2.3.6

Simplify.

$\ln (| 1 + y |) + C_{1} = \ln (| x |) + x + C_{4}$

Step 2.3.7

Reorder terms.

$\ln (| 1 + y |) + C_{1} = x + \ln (| x |) + C_{4}$

Step 2.4

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

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

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.

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

Step 3.2

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

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

Step 3.3

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

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

Step 3.4

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

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

Step 3.5

Solve for $y$ .

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Step 3.5.1

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

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

Step 3.5.2

Multiply both sides by $| x |$ .

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

Step 3.5.3

Simplify the left side.

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Step 3.5.3.1

Cancel the common factor of $| x |$ .

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Step 3.5.3.1.1

Cancel the common factor.

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

Step 3.5.3.1.2

Rewrite the expression.

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

Step 3.5.4

Solve for $y$ .

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Step 3.5.4.1

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

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

Step 3.5.4.2

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

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

Step 3.5.4.3

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

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

Step 3.5.4.4

Subtract $1$ from both sides of the equation.

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

Step 4

Group the constant terms together.

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Step 4.1

Rewrite $e^{x + C}$ as $e^{x} e^{C}$ .

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

Step 4.2

Reorder $e^{x}$ and $e^{C}$ .

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

Step 4.3

Combine constants with the plus or minus.

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