Calculus Examples

$2 x (1 + y^{2}) d x - y (1 + 2 x^{2}) d y = 0$

Step 1

Subtract $2 x (1 + y^{2}) d x$ from both sides of the equation.

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

Step 2

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

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

Step 3

Simplify.

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Step 3.1

Rewrite using the commutative property of multiplication.

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

Step 3.2

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

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Step 3.2.1

Move the leading negative in $- \frac{1}{(1 + 2 x^{2}) (1 + y^{2})}$ into the numerator.

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

Step 3.2.2

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

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

Step 3.2.3

Cancel the common factor.

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

Step 3.2.4

Rewrite the expression.

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

Step 3.3

Combine $\frac{- 1}{1 + y^{2}}$ and $y$ .

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

Step 3.4

Move the negative in front of the fraction.

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

Step 3.5

Rewrite using the commutative property of multiplication.

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

Step 3.6

Combine $- 2$ and $\frac{1}{(1 + 2 x^{2}) (1 + y^{2})}$ .

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

Step 3.7

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

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Step 3.7.1

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

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

Step 3.7.2

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

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

Step 3.7.3

Cancel the common factor.

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

Step 3.7.4

Rewrite the expression.

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

Step 3.8

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

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

Step 3.9

Move the negative in front of the fraction.

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

Step 4

Integrate both sides.

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Step 4.1

Set up an integral on each side.

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

Step 4.2

Integrate the left side.

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Step 4.2.1

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

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

Step 4.2.2

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

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Step 4.2.2.1

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

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Step 4.2.2.1.1

Differentiate $1 + y^{2}$ .

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

Step 4.2.2.1.2

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 4.2.2.1.3

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 4.2.2.1.4

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 4.2.2.1.5

Add $0$ and $2 y$ .

$2 y$

Step 4.2.2.2

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

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

Step 4.2.3

Simplify.

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Step 4.2.3.1

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

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

Step 4.2.3.2

Move $2$ to the left of $u_{1}$ .

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

Step 4.2.4

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

$- (\frac{1}{2} \int \frac{1}{u_{1}} d u_{1}) = \int - \frac{2 x}{1 + 2 x^{2}} d x$

Step 4.2.5

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

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

Step 4.2.6

Simplify.

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

Step 4.2.7

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

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

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.

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

Step 4.3.2

Since $2$ is constant with respect to $x$ , move $2$ out of the integral.

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

Step 4.3.3

Multiply $2$ by $- 1$ .

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

Step 4.3.4

Let $u_{2} = 1 + 2 x^{2}$ . Then $d u_{2} = 4 x d x$ , so $\frac{1}{4} d u_{2} = x d x$ . Rewrite using $u_{2}$ and $d$ $u_{2}$ .

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Step 4.3.4.1

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

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Step 4.3.4.1.1

Differentiate $1 + 2 x^{2}$ .

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

Step 4.3.4.1.2

Differentiate.

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Step 4.3.4.1.2.1

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

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

Step 4.3.4.1.2.2

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

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

Step 4.3.4.1.3

Evaluate $\frac{d}{d x} [2 x^{2}]$ .

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Step 4.3.4.1.3.1

Since $2$ is constant with respect to $x$ , the derivative of $2 x^{2}$ with respect to $x$ is $2 \frac{d}{d x} [x^{2}]$ .

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

Step 4.3.4.1.3.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 2$ .

$0 + 2 (2 x)$

Step 4.3.4.1.3.3

Multiply $2$ by $2$ .

$0 + 4 x$

Step 4.3.4.1.4

Add $0$ and $4 x$ .

$4 x$

Step 4.3.4.2

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

$- \frac{1}{2} \ln (| 1 + y^{2} |) + C_{1} = - 2 \int \frac{1}{u_{2}} \cdot \frac{1}{4} d u_{2}$

Step 4.3.5

Simplify.

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Step 4.3.5.1

Multiply $\frac{1}{u_{2}}$ by $\frac{1}{4}$ .

$- \frac{1}{2} \ln (| 1 + y^{2} |) + C_{1} = - 2 \int \frac{1}{u_{2} \cdot 4} d u_{2}$

Step 4.3.5.2

Move $4$ to the left of $u_{2}$ .

$- \frac{1}{2} \ln (| 1 + y^{2} |) + C_{1} = - 2 \int \frac{1}{4 u_{2}} d u_{2}$

Step 4.3.6

Since $\frac{1}{4}$ is constant with respect to $u_{2}$ , move $\frac{1}{4}$ out of the integral.

$- \frac{1}{2} \ln (| 1 + y^{2} |) + C_{1} = - 2 (\frac{1}{4} \int \frac{1}{u_{2}} d u_{2})$

Step 4.3.7

Simplify.

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Step 4.3.7.1

Combine $\frac{1}{4}$ and $- 2$ .

$- \frac{1}{2} \ln (| 1 + y^{2} |) + C_{1} = \frac{- 2}{4} \int \frac{1}{u_{2}} d u_{2}$

Step 4.3.7.2

Cancel the common factor of $- 2$ and $4$ .

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Step 4.3.7.2.1

Factor $2$ out of $- 2$ .

$- \frac{1}{2} \ln (| 1 + y^{2} |) + C_{1} = \frac{2 (- 1)}{4} \int \frac{1}{u_{2}} d u_{2}$

Step 4.3.7.2.2

Cancel the common factors.

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Step 4.3.7.2.2.1

Factor $2$ out of $4$ .

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

Step 4.3.7.2.2.2

Cancel the common factor.

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

Step 4.3.7.2.2.3

Rewrite the expression.

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

Step 4.3.7.3

Move the negative in front of the fraction.

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

Step 4.3.8

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

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

Step 4.3.9

Simplify.

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

Step 4.3.10

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

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

Step 4.4

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

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

Step 5

Solve for $y$ .

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Step 5.1

Multiply both sides of the equation by $- 2$ .

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

Step 5.2

Simplify both sides of the equation.

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Step 5.2.1

Simplify the left side.

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Step 5.2.1.1

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

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Step 5.2.1.1.1

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

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

Step 5.2.1.1.2

Cancel the common factor of $2$ .

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Step 5.2.1.1.2.1

Move the leading negative in $- \frac{\ln (| 1 + y^{2} |)}{2}$ into the numerator.

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

Step 5.2.1.1.2.2

Factor $2$ out of $- 2$ .

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

Step 5.2.1.1.2.3

Cancel the common factor.

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

Step 5.2.1.1.2.4

Rewrite the expression.

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

Step 5.2.1.1.3

Multiply.

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Step 5.2.1.1.3.1

Multiply $- 1$ by $- 1$ .

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

Step 5.2.1.1.3.2

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

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

Step 5.2.2

Simplify the right side.

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Step 5.2.2.1

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

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Step 5.2.2.1.1

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

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

Step 5.2.2.1.2

To write $C$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

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

Step 5.2.2.1.3

Simplify terms.

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Step 5.2.2.1.3.1

Combine $C$ and $\frac{2}{2}$ .

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

Step 5.2.2.1.3.2

Combine the numerators over the common denominator.

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

Step 5.2.2.1.3.3

Cancel the common factor of $2$ .

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Step 5.2.2.1.3.3.1

Factor $2$ out of $- 2$ .

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

Step 5.2.2.1.3.3.2

Cancel the common factor.

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

Step 5.2.2.1.3.3.3

Rewrite the expression.

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

Step 5.2.2.1.4

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

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

Step 5.2.2.1.5

Apply the distributive property.

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

Step 5.2.2.1.6

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

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Step 5.2.2.1.6.1

Multiply $- 1$ by $- 1$ .

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

Step 5.2.2.1.6.2

Multiply $\ln (| 1 + 2 x^{2} |)$ by $1$ .

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

Step 5.2.2.1.7

Multiply $2$ by $- 1$ .

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

Step 5.3

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

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

Step 5.4

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

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

Step 5.5

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

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

Step 5.6

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

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

Step 5.7

Solve for $y$ .

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Step 5.7.1

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

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

Step 5.7.2

Multiply both sides by $| 1 + 2 x^{2} |$ .

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

Step 5.7.3

Simplify the left side.

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Step 5.7.3.1

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

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Step 5.7.3.1.1

Cancel the common factor.

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

Step 5.7.3.1.2

Rewrite the expression.

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

Step 5.7.4

Solve for $y$ .

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Step 5.7.4.1

Reorder factors in $e^{- 2 C} | 1 + 2 x^{2} |$ .

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

Step 5.7.4.2

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

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

Step 5.7.4.3

Reorder factors in $\pm | 1 + 2 x^{2} | e^{- 2 C}$ .

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

Step 5.7.4.4

Subtract $1$ from both sides of the equation.

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

Step 5.7.4.5

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

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

Step 6

Group the constant terms together.

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Step 6.1

Simplify the constant of integration.

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

Step 6.2

Combine constants with the plus or minus.

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