Calculus Examples

$10 x y y' = 1 - y^{2}$

Step 1

Rewrite the differential equation.

$10 x y \frac{d y}{d x} = 1 - y^{2}$

Step 2

Separate the variables.

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Step 2.1

Divide each term in $10 x y \frac{d y}{d x} = 1 - y^{2}$ by $10 x y$ and simplify.

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Step 2.1.1

Divide each term in $10 x y \frac{d y}{d x} = 1 - y^{2}$ by $10 x y$ .

$\frac{10 x y \frac{d y}{d x}}{10 x y} = \frac{1}{10 x y} + \frac{- y^{2}}{10 x y}$

Step 2.1.2

Simplify the left side.

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Step 2.1.2.1

Cancel the common factor of $10$ .

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Step 2.1.2.1.1

Cancel the common factor.

$\frac{10 x y \frac{d y}{d x}}{10 x y} = \frac{1}{10 x y} + \frac{- y^{2}}{10 x y}$

Step 2.1.2.1.2

Rewrite the expression.

$\frac{x y \frac{d y}{d x}}{x y} = \frac{1}{10 x y} + \frac{- y^{2}}{10 x y}$

Step 2.1.2.2

Cancel the common factor of $x$ .

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Step 2.1.2.2.1

Cancel the common factor.

$\frac{x y \frac{d y}{d x}}{x y} = \frac{1}{10 x y} + \frac{- y^{2}}{10 x y}$

Step 2.1.2.2.2

Rewrite the expression.

$\frac{y \frac{d y}{d x}}{y} = \frac{1}{10 x y} + \frac{- y^{2}}{10 x y}$

Step 2.1.2.3

Cancel the common factor of $y$ .

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Step 2.1.2.3.1

Cancel the common factor.

$\frac{y \frac{d y}{d x}}{y} = \frac{1}{10 x y} + \frac{- y^{2}}{10 x y}$

Step 2.1.2.3.2

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

$\frac{d y}{d x} = \frac{1}{10 x y} + \frac{- y^{2}}{10 x y}$

Step 2.1.3

Simplify the right side.

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Step 2.1.3.1

Simplify each term.

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Step 2.1.3.1.1

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

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Step 2.1.3.1.1.1

Factor $y$ out of $- y^{2}$ .

$\frac{d y}{d x} = \frac{1}{10 x y} + \frac{y (- y)}{10 x y}$

Step 2.1.3.1.1.2

Cancel the common factors.

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Step 2.1.3.1.1.2.1

Factor $y$ out of $10 x y$ .

$\frac{d y}{d x} = \frac{1}{10 x y} + \frac{y (- y)}{y (10 x)}$

Step 2.1.3.1.1.2.2

Cancel the common factor.

$\frac{d y}{d x} = \frac{1}{10 x y} + \frac{y (- y)}{y (10 x)}$

Step 2.1.3.1.1.2.3

Rewrite the expression.

$\frac{d y}{d x} = \frac{1}{10 x y} + \frac{- y}{10 x}$

Step 2.1.3.1.2

Move the negative in front of the fraction.

$\frac{d y}{d x} = \frac{1}{10 x y} - \frac{y}{10 x}$

Step 2.2

Factor.

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Step 2.2.1

To write $- \frac{y}{10 x}$ as a fraction with a common denominator, multiply by $\frac{y}{y}$ .

$\frac{d y}{d x} = \frac{1}{10 x y} - \frac{y}{10 x} \cdot \frac{y}{y}$

Step 2.2.2

Multiply $\frac{y}{10 x}$ by $\frac{y}{y}$ .

$\frac{d y}{d x} = \frac{1}{10 x y} - \frac{y \cdot y}{10 x y}$

Step 2.2.3

Combine the numerators over the common denominator.

$\frac{d y}{d x} = \frac{1 - y \cdot y}{10 x y}$

Step 2.2.4

Simplify the numerator.

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Step 2.2.4.1

Rewrite $1$ as $1^{2}$ .

$\frac{d y}{d x} = \frac{1^{2} - y \cdot y}{10 x y}$

Step 2.2.4.2

Rewrite $y \cdot y$ as $y^{2}$ .

$\frac{d y}{d x} = \frac{1^{2} - y^{2}}{10 x y}$

Step 2.2.4.3

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 1$ and $b = y$ .

$\frac{d y}{d x} = \frac{(1 + y) (1 - y)}{10 x y}$

Step 2.3

Regroup factors.

$\frac{d y}{d x} = \frac{(1 + y) (1 - y)}{10 y} \cdot \frac{1}{x}$

Step 2.4

Multiply both sides by $\frac{10 y}{(1 + y) (1 - y)}$ .

$\frac{10 y}{(1 + y) (1 - y)} \frac{d y}{d x} = \frac{10 y}{(1 + y) (1 - y)} (\frac{(1 + y) (1 - y)}{10 y} \cdot \frac{1}{x})$

Step 2.5

Simplify.

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Step 2.5.1

Multiply $\frac{(1 + y) (1 - y)}{10 y}$ by $\frac{1}{x}$ .

$\frac{10 y}{(1 + y) (1 - y)} \frac{d y}{d x} = \frac{10 y}{(1 + y) (1 - y)} \cdot \frac{(1 + y) (1 - y)}{10 y x}$

Step 2.5.2

Cancel the common factor of $10 y$ .

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Step 2.5.2.1

Factor $10 y$ out of $10 y x$ .

$\frac{10 y}{(1 + y) (1 - y)} \frac{d y}{d x} = \frac{10 y}{(1 + y) (1 - y)} \cdot \frac{(1 + y) (1 - y)}{10 y (x)}$

Step 2.5.2.2

Cancel the common factor.

$\frac{10 y}{(1 + y) (1 - y)} \frac{d y}{d x} = \frac{10 y}{(1 + y) (1 - y)} \cdot \frac{(1 + y) (1 - y)}{10 y x}$

Step 2.5.2.3

Rewrite the expression.

$\frac{10 y}{(1 + y) (1 - y)} \frac{d y}{d x} = \frac{1}{(1 + y) (1 - y)} \cdot \frac{(1 + y) (1 - y)}{x}$

Step 2.5.3

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

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Step 2.5.3.1

Cancel the common factor.

$\frac{10 y}{(1 + y) (1 - y)} \frac{d y}{d x} = \frac{1}{(1 + y) (1 - y)} \cdot \frac{(1 + y) (1 - y)}{x}$

Step 2.5.3.2

Rewrite the expression.

$\frac{10 y}{(1 + y) (1 - y)} \frac{d y}{d x} = \frac{1}{x}$

Step 2.6

Rewrite the equation.

$\frac{10 y}{(1 + y) (1 - y)} d y = \frac{1}{x} d x$

Step 3

Integrate both sides.

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Step 3.1

Set up an integral on each side.

$\int \frac{10 y}{(1 + y) (1 - y)} d y = \int \frac{1}{x} d x$

Step 3.2

Integrate the left side.

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Step 3.2.1

Since $10$ is constant with respect to $y$ , move $10$ out of the integral.

$10 \int \frac{y}{(1 + y) (1 - y)} d y = \int \frac{1}{x} d x$

Step 3.2.2

Let $u = (1 + y) (1 - y)$ . 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 3.2.2.1

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

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Step 3.2.2.1.1

Differentiate $(1 + y) (1 - y)$ .

$\frac{d}{d y} [(1 + y) (1 - y)]$

Step 3.2.2.1.2

Differentiate using the Product Rule which states that $\frac{d}{d y} [f (y) g (y)]$ is $f (y) \frac{d}{d y} [g (y)] + g (y) \frac{d}{d y} [f (y)]$ where $f (y) = 1 + y$ and $g (y) = 1 - y$ .

$(1 + y) \frac{d}{d y} [1 - y] + (1 - y) \frac{d}{d y} [1 + y]$

Step 3.2.2.1.3

Differentiate.

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Step 3.2.2.1.3.1

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

$(1 + y) (\frac{d}{d y} [1] + \frac{d}{d y} [- y]) + (1 - y) \frac{d}{d y} [1 + y]$

Step 3.2.2.1.3.2

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

$(1 + y) (0 + \frac{d}{d y} [- y]) + (1 - y) \frac{d}{d y} [1 + y]$

Step 3.2.2.1.3.3

Add $0$ and $\frac{d}{d y} [- y]$ .

$(1 + y) \frac{d}{d y} [- y] + (1 - y) \frac{d}{d y} [1 + y]$

Step 3.2.2.1.3.4

Since $- 1$ is constant with respect to $y$ , the derivative of $- y$ with respect to $y$ is $- \frac{d}{d y} [y]$ .

$(1 + y) (- \frac{d}{d y} [y]) + (1 - y) \frac{d}{d y} [1 + y]$

Step 3.2.2.1.3.5

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

$(1 + y) (- 1 \cdot 1) + (1 - y) \frac{d}{d y} [1 + y]$

Step 3.2.2.1.3.6

Simplify the expression.

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Step 3.2.2.1.3.6.1

Multiply $- 1$ by $1$ .

$(1 + y) \cdot - 1 + (1 - y) \frac{d}{d y} [1 + y]$

Step 3.2.2.1.3.6.2

Move $- 1$ to the left of $1 + y$ .

$- 1 \cdot (1 + y) + (1 - y) \frac{d}{d y} [1 + y]$

Step 3.2.2.1.3.6.3

Rewrite $- 1 (1 + y)$ as $- (1 + y)$ .

$- (1 + y) + (1 - y) \frac{d}{d y} [1 + y]$

Step 3.2.2.1.3.7

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

$- (1 + y) + (1 - y) (\frac{d}{d y} [1] + \frac{d}{d y} [y])$

Step 3.2.2.1.3.8

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

$- (1 + y) + (1 - y) (0 + \frac{d}{d y} [y])$

Step 3.2.2.1.3.9

Add $0$ and $\frac{d}{d y} [y]$ .

$- (1 + y) + (1 - y) \frac{d}{d y} [y]$

Step 3.2.2.1.3.10

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

$- (1 + y) + (1 - y) \cdot 1$

Step 3.2.2.1.3.11

Multiply $1 - y$ by $1$ .

$- (1 + y) + 1 - y$

Step 3.2.2.1.4

Simplify.

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Step 3.2.2.1.4.1

Apply the distributive property.

$- 1 \cdot 1 - y + 1 - y$

Step 3.2.2.1.4.2

Combine terms.

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Step 3.2.2.1.4.2.1

Multiply $- 1$ by $1$ .

$- 1 - y + 1 - y$

Step 3.2.2.1.4.2.2

Add $- 1$ and $1$ .

$- y + 0 - y$

Step 3.2.2.1.4.2.3

Add $- y$ and $0$ .

$- y - y$

Step 3.2.2.1.4.2.4

Subtract $y$ from $- y$ .

$- 2 y$

Step 3.2.2.2

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

$10 \int \frac{1}{u} \cdot \frac{1}{- 2} d u = \int \frac{1}{x} d x$

Step 3.2.3

Simplify.

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Step 3.2.3.1

Move the negative in front of the fraction.

$10 \int \frac{1}{u} (- \frac{1}{2}) d u = \int \frac{1}{x} d x$

Step 3.2.3.2

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

$10 \int - \frac{1}{u \cdot 2} d u = \int \frac{1}{x} d x$

Step 3.2.3.3

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

$10 \int - \frac{1}{2 u} d u = \int \frac{1}{x} d x$

Step 3.2.4

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

$10 (- \int \frac{1}{2 u} d u) = \int \frac{1}{x} d x$

Step 3.2.5

Multiply $- 1$ by $10$ .

$- 10 \int \frac{1}{2 u} d u = \int \frac{1}{x} d x$

Step 3.2.6

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

$- 10 (\frac{1}{2} \int \frac{1}{u} d u) = \int \frac{1}{x} d x$

Step 3.2.7

Simplify.

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Step 3.2.7.1

Combine $\frac{1}{2}$ and $- 10$ .

$\frac{- 10}{2} \int \frac{1}{u} d u = \int \frac{1}{x} d x$

Step 3.2.7.2

Cancel the common factor of $- 10$ and $2$ .

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Step 3.2.7.2.1

Factor $2$ out of $- 10$ .

$\frac{2 \cdot - 5}{2} \int \frac{1}{u} d u = \int \frac{1}{x} d x$

Step 3.2.7.2.2

Cancel the common factors.

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Step 3.2.7.2.2.1

Factor $2$ out of $2$ .

$\frac{2 \cdot - 5}{2 (1)} \int \frac{1}{u} d u = \int \frac{1}{x} d x$

Step 3.2.7.2.2.2

Cancel the common factor.

$\frac{2 \cdot - 5}{2 \cdot 1} \int \frac{1}{u} d u = \int \frac{1}{x} d x$

Step 3.2.7.2.2.3

Rewrite the expression.

$\frac{- 5}{1} \int \frac{1}{u} d u = \int \frac{1}{x} d x$

Step 3.2.7.2.2.4

Divide $- 5$ by $1$ .

$- 5 \int \frac{1}{u} d u = \int \frac{1}{x} d x$

Step 3.2.8

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

$- 5 (\ln (| u |) + C_{1}) = \int \frac{1}{x} d x$

Step 3.2.9

Simplify.

$- 5 \ln (| u |) + C_{1} = \int \frac{1}{x} d x$

Step 3.2.10

Replace all occurrences of $u$ with $(1 + y) (1 - y)$ .

$- 5 \ln (| (1 + y) (1 - y) |) + C_{1} = \int \frac{1}{x} d x$

Step 3.3

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

$- 5 \ln (| (1 + y) (1 - y) |) + C_{1} = \ln (| x |) + C_{2}$

Step 3.4

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

$- 5 \ln (| (1 + y) (1 - y) |) = \ln (| x |) + C$

Step 4

Solve for $y$ .

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Step 4.1

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

$- 5 \ln (| (1 + y) (1 - y) |) - \ln (| x |) = C$

Step 4.2

Simplify each term.

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Step 4.2.1

Expand $(1 + y) (1 - y)$ using the FOIL Method.

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Step 4.2.1.1

Apply the distributive property.

$- 5 \ln (| 1 (1 - y) + y (1 - y) |) - \ln (| x |) = C$

Step 4.2.1.2

Apply the distributive property.

$- 5 \ln (| 1 \cdot 1 + 1 (- y) + y (1 - y) |) - \ln (| x |) = C$

Step 4.2.1.3

Apply the distributive property.

$- 5 \ln (| 1 \cdot 1 + 1 (- y) + y \cdot 1 + y (- y) |) - \ln (| x |) = C$

Step 4.2.2

Simplify and combine like terms.

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Step 4.2.2.1

Simplify each term.

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Step 4.2.2.1.1

Multiply $1$ by $1$ .

$- 5 \ln (| 1 + 1 (- y) + y \cdot 1 + y (- y) |) - \ln (| x |) = C$

Step 4.2.2.1.2

Multiply $- y$ by $1$ .

$- 5 \ln (| 1 - y + y \cdot 1 + y (- y) |) - \ln (| x |) = C$

Step 4.2.2.1.3

Multiply $y$ by $1$ .

$- 5 \ln (| 1 - y + y + y (- y) |) - \ln (| x |) = C$

Step 4.2.2.1.4

Rewrite using the commutative property of multiplication.

$- 5 \ln (| 1 - y + y - y \cdot y |) - \ln (| x |) = C$

Step 4.2.2.1.5

Multiply $y$ by $y$ by adding the exponents.

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Step 4.2.2.1.5.1

Move $y$ .

$- 5 \ln (| 1 - y + y - (y \cdot y) |) - \ln (| x |) = C$

Step 4.2.2.1.5.2

Multiply $y$ by $y$ .

$- 5 \ln (| 1 - y + y - y^{2} |) - \ln (| x |) = C$

Step 4.2.2.2

Add $- y$ and $y$ .

$- 5 \ln (| 1 + 0 - y^{2} |) - \ln (| x |) = C$

Step 4.2.2.3

Add $1$ and $0$ .

$- 5 \ln (| 1 - y^{2} |) - \ln (| x |) = C$

Step 4.3

Simplify the left side.

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Step 4.3.1

Simplify $- 5 \ln (| 1 - y^{2} |)$ by moving $5$ inside the logarithm.

$- \ln ({| 1 - y^{2} |}^{5}) - \ln (| x |) = C$

Step 4.4

Add $\ln (| x |)$ to both sides of the equation.

$- \ln ({| 1 - y^{2} |}^{5}) = C + \ln (| x |)$

Step 4.5

Divide each term in $- \ln ({| 1 - y^{2} |}^{5}) = C + \ln (| x |)$ by $- 1$ and simplify.

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Step 4.5.1

Divide each term in $- \ln ({| 1 - y^{2} |}^{5}) = C + \ln (| x |)$ by $- 1$ .

$\frac{- \ln ({| 1 - y^{2} |}^{5})}{- 1} = \frac{C}{- 1} + \frac{\ln (| x |)}{- 1}$

Step 4.5.2

Simplify the left side.

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Step 4.5.2.1

Dividing two negative values results in a positive value.

$\frac{\ln ({| 1 - y^{2} |}^{5})}{1} = \frac{C}{- 1} + \frac{\ln (| x |)}{- 1}$

Step 4.5.2.2

Divide $\ln ({| 1 - y^{2} |}^{5})$ by $1$ .

$\ln ({| 1 - y^{2} |}^{5}) = \frac{C}{- 1} + \frac{\ln (| x |)}{- 1}$

Step 4.5.3

Simplify the right side.

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Step 4.5.3.1

Simplify each term.

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Step 4.5.3.1.1

Move the negative one from the denominator of $\frac{C}{- 1}$ .

$\ln ({| 1 - y^{2} |}^{5}) = - 1 \cdot C + \frac{\ln (| x |)}{- 1}$

Step 4.5.3.1.2

Rewrite $- 1 \cdot C$ as $- C$ .

$\ln ({| 1 - y^{2} |}^{5}) = - C + \frac{\ln (| x |)}{- 1}$

Step 4.5.3.1.3

Move the negative one from the denominator of $\frac{\ln (| x |)}{- 1}$ .

$\ln ({| 1 - y^{2} |}^{5}) = - C - 1 \cdot \ln (| x |)$

Step 4.5.3.1.4

Rewrite $- 1 \cdot \ln (| x |)$ as $- \ln (| x |)$ .

$\ln ({| 1 - y^{2} |}^{5}) = - C - \ln (| x |)$

Step 4.6

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

$\ln ({| 1 - y^{2} |}^{5}) + \ln (| x |) = - C$

Step 4.7

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

$\ln ({| 1 - y^{2} |}^{5} | x |) = - C$

Step 4.8

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

$e^{\ln ({| 1 - y^{2} |}^{5} | x |)} = e^{- C}$

Step 4.9

Rewrite $\ln ({| 1 - y^{2} |}^{5} | x |) = - C$ 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 $b^{y} = x$ .

$e^{- C} = {| 1 - y^{2} |}^{5} | x |$

Step 4.10

Solve for $y$ .

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Step 4.10.1

Rewrite the equation as ${| 1 - y^{2} |}^{5} | x | = e^{- C}$ .

${| 1 - y^{2} |}^{5} | x | = e^{- C}$

Step 4.10.2

Divide each term in ${| 1 - y^{2} |}^{5} | x | = e^{- C}$ by $| x |$ and simplify.

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Step 4.10.2.1

Divide each term in ${| 1 - y^{2} |}^{5} | x | = e^{- C}$ by $| x |$ .

$\frac{{| 1 - y^{2} |}^{5} | x |}{| x |} = \frac{e^{- C}}{| x |}$

Step 4.10.2.2

Simplify the left side.

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Step 4.10.2.2.1

Cancel the common factor of $| x |$ .

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Step 4.10.2.2.1.1

Cancel the common factor.

$\frac{{| 1 - y^{2} |}^{5} | x |}{| x |} = \frac{e^{- C}}{| x |}$

Step 4.10.2.2.1.2

Divide ${| 1 - y^{2} |}^{5}$ by $1$ .

${| 1 - y^{2} |}^{5} = \frac{e^{- C}}{| x |}$

Step 4.10.3

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

$| 1 - y^{2} | = \sqrt[5]{\frac{e^{- C}}{| x |}}$

Step 4.10.4

Simplify $\sqrt[5]{\frac{e^{- C}}{| x |}}$ .

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Step 4.10.4.1

Rewrite $\sqrt[5]{\frac{e^{- C}}{| x |}}$ as $\frac{\sqrt[5]{e^{- C}}}{\sqrt[5]{| x |}}$ .

$| 1 - y^{2} | = \frac{\sqrt[5]{e^{- C}}}{\sqrt[5]{| x |}}$

Step 4.10.4.2

Multiply $\frac{\sqrt[5]{e^{- C}}}{\sqrt[5]{| x |}}$ by $\frac{{\sqrt[5]{| x |}}^{4}}{{\sqrt[5]{| x |}}^{4}}$ .

$| 1 - y^{2} | = \frac{\sqrt[5]{e^{- C}}}{\sqrt[5]{| x |}} \cdot \frac{{\sqrt[5]{| x |}}^{4}}{{\sqrt[5]{| x |}}^{4}}$

Step 4.10.4.3

Combine and simplify the denominator.

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Step 4.10.4.3.1

Multiply $\frac{\sqrt[5]{e^{- C}}}{\sqrt[5]{| x |}}$ by $\frac{{\sqrt[5]{| x |}}^{4}}{{\sqrt[5]{| x |}}^{4}}$ .

$| 1 - y^{2} | = \frac{\sqrt[5]{e^{- C}} {\sqrt[5]{| x |}}^{4}}{\sqrt[5]{| x |} {\sqrt[5]{| x |}}^{4}}$

Step 4.10.4.3.2

Raise $\sqrt[5]{| x |}$ to the power of $1$ .

$| 1 - y^{2} | = \frac{\sqrt[5]{e^{- C}} {\sqrt[5]{| x |}}^{4}}{{\sqrt[5]{| x |}}^{1} {\sqrt[5]{| x |}}^{4}}$

Step 4.10.4.3.3

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$| 1 - y^{2} | = \frac{\sqrt[5]{e^{- C}} {\sqrt[5]{| x |}}^{4}}{{\sqrt[5]{| x |}}^{1 + 4}}$

Step 4.10.4.3.4

Add $1$ and $4$ .

$| 1 - y^{2} | = \frac{\sqrt[5]{e^{- C}} {\sqrt[5]{| x |}}^{4}}{{\sqrt[5]{| x |}}^{5}}$

Step 4.10.4.3.5

Rewrite ${\sqrt[5]{| x |}}^{5}$ as $| x |$ .

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Step 4.10.4.3.5.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt[5]{| x |}$ as ${| x |}^{\frac{1}{5}}$ .

$| 1 - y^{2} | = \frac{\sqrt[5]{e^{- C}} {\sqrt[5]{| x |}}^{4}}{{({| x |}^{\frac{1}{5}})}^{5}}$

Step 4.10.4.3.5.2

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$| 1 - y^{2} | = \frac{\sqrt[5]{e^{- C}} {\sqrt[5]{| x |}}^{4}}{{| x |}^{\frac{1}{5} \cdot 5}}$

Step 4.10.4.3.5.3

Combine $\frac{1}{5}$ and $5$ .

$| 1 - y^{2} | = \frac{\sqrt[5]{e^{- C}} {\sqrt[5]{| x |}}^{4}}{{| x |}^{\frac{5}{5}}}$

Step 4.10.4.3.5.4

Cancel the common factor of $5$ .

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Step 4.10.4.3.5.4.1

Cancel the common factor.

$| 1 - y^{2} | = \frac{\sqrt[5]{e^{- C}} {\sqrt[5]{| x |}}^{4}}{{| x |}^{\frac{5}{5}}}$

Step 4.10.4.3.5.4.2

Rewrite the expression.

$| 1 - y^{2} | = \frac{\sqrt[5]{e^{- C}} {\sqrt[5]{| x |}}^{4}}{{| x |}^{1}}$

Step 4.10.4.3.5.5

Simplify.

$| 1 - y^{2} | = \frac{\sqrt[5]{e^{- C}} {\sqrt[5]{| x |}}^{4}}{| x |}$

Step 4.10.4.4

Rewrite ${\sqrt[5]{| x |}}^{4}$ as $\sqrt[5]{{| x |}^{4}}$ .

$| 1 - y^{2} | = \frac{\sqrt[5]{e^{- C}} \sqrt[5]{{| x |}^{4}}}{| x |}$

Step 4.10.4.5

Combine using the product rule for radicals.

$| 1 - y^{2} | = \frac{\sqrt[5]{e^{- C} {| x |}^{4}}}{| x |}$

Step 4.10.4.6

Reorder factors in $\frac{\sqrt[5]{e^{- C} {| x |}^{4}}}{| x |}$ .

$| 1 - y^{2} | = \frac{\sqrt[5]{{| x |}^{4} e^{- C}}}{| x |}$

Step 4.10.5

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

$1 - y^{2} = \pm \frac{\sqrt[5]{{| x |}^{4} e^{- C}}}{| x |}$

Step 4.10.6

Remove the absolute value in ${| x |}^{4}$ because exponentiations with even powers are always positive.

$1 - y^{2} = \pm \frac{\sqrt[5]{x^{4} e^{- C}}}{| x |}$

Step 4.10.7

Subtract $1$ from both sides of the equation.

$- y^{2} = \pm \frac{\sqrt[5]{x^{4} e^{- C}}}{| x |} - 1$

Step 4.10.8

Divide each term in $- y^{2} = \pm \frac{\sqrt[5]{x^{4} e^{- C}}}{| x |} - 1$ by $- 1$ and simplify.

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Step 4.10.8.1

Divide each term in $- y^{2} = \pm \frac{\sqrt[5]{x^{4} e^{- C}}}{| x |} - 1$ by $- 1$ .

$\frac{- y^{2}}{- 1} = \frac{\pm \frac{\sqrt[5]{x^{4} e^{- C}}}{| x |}}{- 1} + \frac{- 1}{- 1}$

Step 4.10.8.2

Simplify the left side.

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Step 4.10.8.2.1

Dividing two negative values results in a positive value.

$\frac{y^{2}}{1} = \frac{\pm \frac{\sqrt[5]{x^{4} e^{- C}}}{| x |}}{- 1} + \frac{- 1}{- 1}$

Step 4.10.8.2.2

Divide $y^{2}$ by $1$ .

$y^{2} = \frac{\pm \frac{\sqrt[5]{x^{4} e^{- C}}}{| x |}}{- 1} + \frac{- 1}{- 1}$

Step 4.10.8.3

Simplify the right side.

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Step 4.10.8.3.1

Simplify each term.

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Step 4.10.8.3.1.1

Move the negative one from the denominator of $\frac{\pm \frac{\sqrt[5]{x^{4} e^{- C}}}{| x |}}{- 1}$ .

$y^{2} = - 1 \cdot (\pm \frac{\sqrt[5]{x^{4} e^{- C}}}{| x |}) + \frac{- 1}{- 1}$

Step 4.10.8.3.1.2

Rewrite $- 1 \cdot (\pm \frac{\sqrt[5]{x^{4} e^{- C}}}{| x |})$ as $- (\pm \frac{\sqrt[5]{x^{4} e^{- C}}}{| x |})$ .

$y^{2} = - (\pm \frac{\sqrt[5]{x^{4} e^{- C}}}{| x |}) + \frac{- 1}{- 1}$

Step 4.10.8.3.1.3

Divide $- 1$ by $- 1$ .

$y^{2} = - (\pm \frac{\sqrt[5]{x^{4} e^{- C}}}{| x |}) + 1$

Step 4.10.9

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

$y = \pm \sqrt{- (\pm \frac{\sqrt[5]{x^{4} e^{- C}}}{| x |}) + 1}$

Step 5

Simplify the constant of integration.

$y = \pm \sqrt{- (\pm \frac{\sqrt[5]{x^{4} C}}{| x |}) + 1}$