Calculus Examples

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

Step 1

Separate the variables.

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Step 1.1

Divide each term in $x y \frac{d y}{d x} = (1 + x^{2}) (1 + y^{2})$ by $x y$ and simplify.

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Step 1.1.1

Divide each term in $x y \frac{d y}{d x} = (1 + x^{2}) (1 + y^{2})$ by $x y$ .

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

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.

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

Step 1.1.2.1.2

Rewrite the expression.

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

Step 1.1.2.2

Cancel the common factor of $y$ .

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Step 1.1.2.2.1

Cancel the common factor.

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

Step 1.1.2.2.2

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

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

Step 1.2

Regroup factors.

$\frac{d y}{d x} = \frac{1 + x^{2}}{x} \cdot \frac{1 + y^{2}}{y}$

Step 1.3

Multiply both sides by $\frac{y}{1 + y^{2}}$ .

$\frac{y}{1 + y^{2}} \frac{d y}{d x} = \frac{y}{1 + y^{2}} (\frac{1 + x^{2}}{x} \cdot \frac{1 + y^{2}}{y})$

Step 1.4

Simplify.

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Step 1.4.1

Multiply $\frac{1 + x^{2}}{x}$ by $\frac{1 + y^{2}}{y}$ .

$\frac{y}{1 + y^{2}} \frac{d y}{d x} = \frac{y}{1 + y^{2}} \cdot \frac{(1 + x^{2}) (1 + y^{2})}{x y}$

Step 1.4.2

Cancel the common factor of $y$ .

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Step 1.4.2.1

Factor $y$ out of $x y$ .

$\frac{y}{1 + y^{2}} \frac{d y}{d x} = \frac{y}{1 + y^{2}} \cdot \frac{(1 + x^{2}) (1 + y^{2})}{y x}$

Step 1.4.2.2

Cancel the common factor.

$\frac{y}{1 + y^{2}} \frac{d y}{d x} = \frac{y}{1 + y^{2}} \cdot \frac{(1 + x^{2}) (1 + y^{2})}{y x}$

Step 1.4.2.3

Rewrite the expression.

$\frac{y}{1 + y^{2}} \frac{d y}{d x} = \frac{1}{1 + y^{2}} \cdot \frac{(1 + x^{2}) (1 + y^{2})}{x}$

Step 1.4.3

Cancel the common factor of $1 + y^{2}$ .

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Step 1.4.3.1

Factor $1 + y^{2}$ out of $(1 + x^{2}) (1 + y^{2})$ .

$\frac{y}{1 + y^{2}} \frac{d y}{d x} = \frac{1}{1 + y^{2}} \cdot \frac{(1 + y^{2}) (1 + x^{2})}{x}$

Step 1.4.3.2

Cancel the common factor.

$\frac{y}{1 + y^{2}} \frac{d y}{d x} = \frac{1}{1 + y^{2}} \cdot \frac{(1 + y^{2}) (1 + x^{2})}{x}$

Step 1.4.3.3

Rewrite the expression.

$\frac{y}{1 + y^{2}} \frac{d y}{d x} = \frac{1 + x^{2}}{x}$

Step 1.5

Rewrite the equation.

$\frac{y}{1 + y^{2}} d y = \frac{1 + x^{2}}{x} d x$

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

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

Step 2.2

Integrate the left side.

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Step 2.2.1

Let $u = 1 + y^{2}$ . Then $d u = 2 y d y$ , so $\frac{1}{2} d u = y d y$ . Rewrite using $u$ and $d$ $u$ .

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Step 2.2.1.1

Let $u = 1 + y^{2}$ . Find $\frac{d u}{d y}$ .

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Step 2.2.1.1.1

Differentiate $1 + y^{2}$ .

$\frac{d}{d y} [1 + y^{2}]$

Step 2.2.1.1.2

By the Sum Rule, the derivative of $1 + y^{2}$ with respect to $y$ is $\frac{d}{d y} [1] + \frac{d}{d y} [y^{2}]$ .

$\frac{d}{d y} [1] + \frac{d}{d y} [y^{2}]$

Step 2.2.1.1.3

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

$0 + \frac{d}{d y} [y^{2}]$

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 $n = 2$ .

$0 + 2 y$

Step 2.2.1.1.5

Add $0$ and $2 y$ .

$2 y$

Step 2.2.1.2

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

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

Step 2.2.2

Simplify.

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Step 2.2.2.1

Multiply $\frac{1}{u}$ by $\frac{1}{2}$ .

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

Step 2.2.2.2

Move $2$ to the left of $u$ .

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

Step 2.2.3

Since $\frac{1}{2}$ is constant with respect to $u$ , move $\frac{1}{2}$ out of the integral.

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

Step 2.2.4

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

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

Step 2.2.5

Simplify.

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

Step 2.2.6

Replace all occurrences of $u$ with $1 + y^{2}$ .

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

Step 2.3

Integrate the right side.

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Step 2.3.1

Split the fraction into multiple fractions.

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

Step 2.3.2

Split the single integral into multiple integrals.

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

Step 2.3.3

Cancel the common factor of $x^{2}$ and $x$ .

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Step 2.3.3.1

Factor $x$ out of $x^{2}$ .

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

Step 2.3.3.2

Cancel the common factors.

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Step 2.3.3.2.1

Raise $x$ to the power of $1$ .

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

Step 2.3.3.2.2

Factor $x$ out of $x^{1}$ .

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

Step 2.3.3.2.3

Cancel the common factor.

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

Step 2.3.3.2.4

Rewrite the expression.

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

Step 2.3.3.2.5

Divide $x$ by $1$ .

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

Step 2.3.4

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

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

Step 2.3.5

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

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

Step 2.3.6

Simplify.

$\frac{1}{2} \ln (| 1 + y^{2} |) + C_{1} = \ln (| x |) + \frac{1}{2} x^{2} + C_{4}$

Step 2.3.7

Reorder terms.

$\frac{1}{2} \ln (| 1 + y^{2} |) + C_{1} = \frac{1}{2} x^{2} + \ln (| x |) + C_{4}$

Step 2.4

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

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

Step 3

Solve for $y$ .

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Step 3.1

Multiply both sides of the equation by $2$ .

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

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

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Step 3.2.1.1.1

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

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

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.

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

Step 3.2.1.1.2.2

Rewrite the expression.

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

Step 3.2.2

Simplify the right side.

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Step 3.2.2.1

Simplify $2 (\frac{1}{2} x^{2} + \ln (| x |) + C)$ .

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Step 3.2.2.1.1

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

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

Step 3.2.2.1.2

Apply the distributive property.

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

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.

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

Step 3.2.2.1.3.2

Rewrite the expression.

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

Step 3.3

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

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

Step 3.4

Simplify the left side.

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Step 3.4.1

Simplify $\ln (| 1 + y^{2} |) - 2 \ln (| x |)$ .

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Step 3.4.1.1

Simplify each term.

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Step 3.4.1.1.1

Simplify $- 2 \ln (| x |)$ by moving $2$ inside the logarithm.

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

Step 3.4.1.1.2

Remove the absolute value in ${| x |}^{2}$ because exponentiations with even powers are always positive.

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

Step 3.4.1.2

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

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

Step 3.5

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

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

Step 3.6

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

$e^{x^{2} + 2 C} = \frac{| 1 + y^{2} |}{x^{2}}$

Step 3.7

Solve for $y$ .

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Step 3.7.1

Rewrite the equation as $\frac{| 1 + y^{2} |}{x^{2}} = e^{x^{2} + 2 C}$ .

$\frac{| 1 + y^{2} |}{x^{2}} = e^{x^{2} + 2 C}$

Step 3.7.2

Multiply both sides by $x^{2}$ .

$\frac{| 1 + y^{2} |}{x^{2}} x^{2} = e^{x^{2} + 2 C} x^{2}$

Step 3.7.3

Simplify.

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Step 3.7.3.1

Simplify the left side.

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Step 3.7.3.1.1

Cancel the common factor of $x^{2}$ .

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Step 3.7.3.1.1.1

Cancel the common factor.

$\frac{| 1 + y^{2} |}{x^{2}} x^{2} = e^{x^{2} + 2 C} x^{2}$

Step 3.7.3.1.1.2

Rewrite the expression.

$| 1 + y^{2} | = e^{x^{2} + 2 C} x^{2}$

Step 3.7.3.2

Simplify the right side.

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Step 3.7.3.2.1

Reorder factors in $e^{x^{2} + 2 C} x^{2}$ .

$| 1 + y^{2} | = x^{2} e^{x^{2} + 2 C}$

Step 3.7.4

Solve for $y$ .

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Step 3.7.4.1

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

$1 + y^{2} = \pm x^{2} e^{x^{2} + 2 C}$

Step 3.7.4.2

Subtract $1$ from both sides of the equation.

$y^{2} = \pm x^{2} e^{x^{2} + 2 C} - 1$

Step 3.7.4.3

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

$y = \pm \sqrt{\pm x^{2} e^{x^{2} + 2 C} - 1}$

Step 4

Group the constant terms together.

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Step 4.1

Simplify the constant of integration.

$y = \pm \sqrt{\pm x^{2} e^{x^{2} + C} - 1}$

Step 4.2

Rewrite $e^{x^{2} + C}$ as $e^{x^{2}} e^{C}$ .

$y = \pm \sqrt{\pm x^{2} (e^{x^{2}} e^{C}) - 1}$

Step 4.3

Reorder $e^{x^{2}}$ and $e^{C}$ .

$y = \pm \sqrt{\pm x^{2} (e^{C} e^{x^{2}}) - 1}$

Step 4.4

Combine constants with the plus or minus.

$y = \pm \sqrt{C x^{2} e^{x^{2}} - 1}$