Calculus Examples

$\frac{d y}{d x} = - \frac{y}{x^{2} - 4 x}$

Step 1

Separate the variables.

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Step 1.1

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

$\frac{1}{y} \frac{d y}{d x} = \frac{1}{y} (- \frac{y}{x^{2} - 4 x})$

Step 1.2

Simplify.

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Step 1.2.1

Rewrite using the commutative property of multiplication.

$\frac{1}{y} \frac{d y}{d x} = - \frac{1}{y} \cdot \frac{y}{x^{2} - 4 x}$

Step 1.2.2

Cancel the common factor of $y$ .

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Step 1.2.2.1

Move the leading negative in $- \frac{1}{y}$ into the numerator.

$\frac{1}{y} \frac{d y}{d x} = \frac{- 1}{y} \cdot \frac{y}{x^{2} - 4 x}$

Step 1.2.2.2

Cancel the common factor.

$\frac{1}{y} \frac{d y}{d x} = \frac{- 1}{y} \cdot \frac{y}{x^{2} - 4 x}$

Step 1.2.2.3

Rewrite the expression.

$\frac{1}{y} \frac{d y}{d x} = - \frac{1}{x^{2} - 4 x}$

Step 1.2.3

Factor $x$ out of $x^{2} - 4 x$ .

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Step 1.2.3.1

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

$\frac{1}{y} \frac{d y}{d x} = - \frac{1}{x \cdot x - 4 x}$

Step 1.2.3.2

Factor $x$ out of $- 4 x$ .

$\frac{1}{y} \frac{d y}{d x} = - \frac{1}{x \cdot x + x \cdot - 4}$

Step 1.2.3.3

Factor $x$ out of $x \cdot x + x \cdot - 4$ .

$\frac{1}{y} \frac{d y}{d x} = - \frac{1}{x (x - 4)}$

Step 1.3

Rewrite the equation.

$\frac{1}{y} d y = - \frac{1}{x (x - 4)} d x$

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

$\int \frac{1}{y} d y = \int - \frac{1}{x (x - 4)} d x$

Step 2.2

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

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

Step 2.3

Integrate the right side.

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Step 2.3.1

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

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

Step 2.3.2

Write the fraction using partial fraction decomposition.

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Step 2.3.2.1

Decompose the fraction and multiply through by the common denominator.

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Step 2.3.2.1.1

For each factor in the denominator, create a new fraction using the factor as the denominator, and an unknown value as the numerator. Since the factor in the denominator is linear, put a single variable in its place $B$ .

$\frac{A}{x} + \frac{B}{x - 4}$

Step 2.3.2.1.2

Multiply each fraction in the equation by the denominator of the original expression. In this case, the denominator is $x (x - 4)$ .

$\frac{1 (x (x - 4))}{x (x - 4)} = \frac{A (x (x - 4))}{x} + \frac{(B) (x (x - 4))}{x - 4}$

Step 2.3.2.1.3

Cancel the common factor of $x$ .

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Step 2.3.2.1.3.1

Cancel the common factor.

$\frac{1 (x (x - 4))}{x (x - 4)} = \frac{A (x (x - 4))}{x} + \frac{(B) (x (x - 4))}{x - 4}$

Step 2.3.2.1.3.2

Rewrite the expression.

$\frac{1 (x - 4)}{x - 4} = \frac{A (x (x - 4))}{x} + \frac{(B) (x (x - 4))}{x - 4}$

Step 2.3.2.1.4

Cancel the common factor of $x - 4$ .

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Step 2.3.2.1.4.1

Cancel the common factor.

$\frac{1 (x - 4)}{x - 4} = \frac{A (x (x - 4))}{x} + \frac{(B) (x (x - 4))}{x - 4}$

Step 2.3.2.1.4.2

Rewrite the expression.

$1 = \frac{A (x (x - 4))}{x} + \frac{(B) (x (x - 4))}{x - 4}$

Step 2.3.2.1.5

Simplify each term.

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Step 2.3.2.1.5.1

Cancel the common factor of $x$ .

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Step 2.3.2.1.5.1.1

Cancel the common factor.

$1 = \frac{A (x (x - 4))}{x} + \frac{(B) (x (x - 4))}{x - 4}$

Step 2.3.2.1.5.1.2

Divide $A (x - 4)$ by $1$ .

$1 = A (x - 4) + \frac{(B) (x (x - 4))}{x - 4}$

Step 2.3.2.1.5.2

Apply the distributive property.

$1 = A x + A \cdot - 4 + \frac{(B) (x (x - 4))}{x - 4}$

Step 2.3.2.1.5.3

Move $- 4$ to the left of $A$ .

$1 = A x - 4 \cdot A + \frac{(B) (x (x - 4))}{x - 4}$

Step 2.3.2.1.5.4

Cancel the common factor of $x - 4$ .

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Step 2.3.2.1.5.4.1

Cancel the common factor.

$1 = A x - 4 A + \frac{B (x (x - 4))}{x - 4}$

Step 2.3.2.1.5.4.2

Divide $(B) (x)$ by $1$ .

$1 = A x - 4 A + (B) (x)$

$1 = A x - 4 A + B x$

Step 2.3.2.1.6

Move $- 4 A$ .

$1 = A x + B x - 4 A$

Step 2.3.2.2

Create equations for the partial fraction variables and use them to set up a system of equations.

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Step 2.3.2.2.1

Create an equation for the partial fraction variables by equating the coefficients of $x$ from each side of the equation. For the equation to be equal, the equivalent coefficients on each side of the equation must be equal.

$0 = A + B$

Step 2.3.2.2.2

Create an equation for the partial fraction variables by equating the coefficients of the terms not containing $x$ . For the equation to be equal, the equivalent coefficients on each side of the equation must be equal.

$1 = - 4 A$

Step 2.3.2.2.3

Set up the system of equations to find the coefficients of the partial fractions.

$0 = A + B$

$1 = - 4 A$

$0 = A + B$

$1 = - 4 A$

Step 2.3.2.3

Solve the system of equations.

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Step 2.3.2.3.1

Solve for $A$ in $1 = - 4 A$ .

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Step 2.3.2.3.1.1

Rewrite the equation as $- 4 A = 1$ .

$- 4 A = 1$

$0 = A + B$

Step 2.3.2.3.1.2

Divide each term in $- 4 A = 1$ by $- 4$ and simplify.

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Step 2.3.2.3.1.2.1

Divide each term in $- 4 A = 1$ by $- 4$ .

$\frac{- 4 A}{- 4} = \frac{1}{- 4}$

$0 = A + B$

Step 2.3.2.3.1.2.2

Simplify the left side.

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Step 2.3.2.3.1.2.2.1

Cancel the common factor of $- 4$ .

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Step 2.3.2.3.1.2.2.1.1

Cancel the common factor.

$\frac{- 4 A}{- 4} = \frac{1}{- 4}$

$0 = A + B$

Step 2.3.2.3.1.2.2.1.2

Divide $A$ by $1$ .

$A = \frac{1}{- 4}$

$0 = A + B$

$A = \frac{1}{- 4}$

$0 = A + B$

$A = \frac{1}{- 4}$

$0 = A + B$

Step 2.3.2.3.1.2.3

Simplify the right side.

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Step 2.3.2.3.1.2.3.1

Move the negative in front of the fraction.

$A = - \frac{1}{4}$

$0 = A + B$

$A = - \frac{1}{4}$

$0 = A + B$

$A = - \frac{1}{4}$

$0 = A + B$

$A = - \frac{1}{4}$

$0 = A + B$

Step 2.3.2.3.2

Replace all occurrences of $A$ with $- \frac{1}{4}$ in each equation.

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Step 2.3.2.3.2.1

Replace all occurrences of $A$ in $0 = A + B$ with $- \frac{1}{4}$ .

$0 = (- \frac{1}{4}) + B$

$A = - \frac{1}{4}$

Step 2.3.2.3.2.2

Simplify the right side.

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Step 2.3.2.3.2.2.1

Remove parentheses.

$0 = - \frac{1}{4} + B$

$A = - \frac{1}{4}$

$0 = - \frac{1}{4} + B$

$A = - \frac{1}{4}$

$0 = - \frac{1}{4} + B$

$A = - \frac{1}{4}$

Step 2.3.2.3.3

Solve for $B$ in $0 = - \frac{1}{4} + B$ .

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Step 2.3.2.3.3.1

Rewrite the equation as $- \frac{1}{4} + B = 0$ .

$- \frac{1}{4} + B = 0$

$A = - \frac{1}{4}$

Step 2.3.2.3.3.2

Add $\frac{1}{4}$ to both sides of the equation.

$B = \frac{1}{4}$

$A = - \frac{1}{4}$

$B = \frac{1}{4}$

$A = - \frac{1}{4}$

Step 2.3.2.3.4

Solve the system of equations.

$B = \frac{1}{4} A = - \frac{1}{4}$

Step 2.3.2.3.5

List all of the solutions.

$B = \frac{1}{4}, A = - \frac{1}{4}$

Step 2.3.2.4

Replace each of the partial fraction coefficients in $\frac{A}{x} + \frac{B}{x - 4}$ with the values found for $A$ and $B$ .

$- \frac{\frac{1}{4}}{x} + \frac{\frac{1}{4}}{x - 4}$

Step 2.3.2.5

Simplify.

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Step 2.3.2.5.1

Multiply the numerator by the reciprocal of the denominator.

$- \frac{1}{4} \cdot \frac{1}{x} + \frac{\frac{1}{4}}{x - 4}$

Step 2.3.2.5.2

Multiply $\frac{1}{x}$ by $\frac{1}{4}$ .

$- \frac{1}{x \cdot 4} + \frac{\frac{1}{4}}{x - 4}$

Step 2.3.2.5.3

Move $4$ to the left of $x$ .

$- \frac{1}{4 x} + \frac{\frac{1}{4}}{x - 4}$

Step 2.3.2.5.4

Multiply the numerator by the reciprocal of the denominator.

$- \frac{1}{4 x} + \frac{1}{4} \cdot \frac{1}{x - 4}$

Step 2.3.2.5.5

Multiply $\frac{1}{4}$ by $\frac{1}{x - 4}$ .

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

Step 2.3.3

Split the single integral into multiple integrals.

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

Step 2.3.4

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

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

Step 2.3.5

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

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

Step 2.3.6

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

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

Step 2.3.7

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

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

Step 2.3.8

Let $u = x - 4$ . Then $d u = d x$ . Rewrite using $u$ and $d$ $u$ .

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Step 2.3.8.1

Let $u = x - 4$ . Find $\frac{d u}{d x}$ .

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Step 2.3.8.1.1

Differentiate $x - 4$ .

$\frac{d}{d x} [x - 4]$

Step 2.3.8.1.2

By the Sum Rule, the derivative of $x - 4$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [- 4]$ .

$\frac{d}{d x} [x] + \frac{d}{d x} [- 4]$

Step 2.3.8.1.3

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

$1 + \frac{d}{d x} [- 4]$

Step 2.3.8.1.4

Since $- 4$ is constant with respect to $x$ , the derivative of $- 4$ with respect to $x$ is $0$ .

$1 + 0$

Step 2.3.8.1.5

Add $1$ and $0$ .

$1$

Step 2.3.8.2

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

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

Step 2.3.9

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

$\ln (| y |) + C_{1} = - (- \frac{1}{4} (\ln (| x |) + C_{2}) + \frac{1}{4} (\ln (| u |) + C_{3}))$

Step 2.3.10

Simplify.

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

Step 2.3.11

Replace all occurrences of $u$ with $x - 4$ .

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

Step 2.3.12

Simplify.

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Step 2.3.12.1

Simplify each term.

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Step 2.3.12.1.1

Combine $\ln (| x |)$ and $\frac{1}{4}$ .

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

Step 2.3.12.1.2

Combine $\frac{1}{4}$ and $\ln (| x - 4 |)$ .

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

Step 2.3.12.2

Combine the numerators over the common denominator.

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

Step 2.3.12.3

Factor $- 1$ out of $- \ln (| x |)$ .

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

Step 2.3.12.4

Factor $- 1$ out of $\ln (| x - 4 |)$ .

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

Step 2.3.12.5

Factor $- 1$ out of $- (\ln (| x |)) - 1 (- \ln (| x - 4 |))$ .

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

Step 2.3.12.6

Rewrite $- (\ln (| x |) - \ln (| x - 4 |))$ as $- 1 (\ln (| x |) - \ln (| x - 4 |))$ .

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

Step 2.3.12.7

Move the negative in front of the fraction.

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

Step 2.3.12.8

Multiply $- 1$ by $- 1$ .

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

Step 2.3.12.9

Multiply $\frac{\ln (| x |) - \ln (| x - 4 |)}{4}$ by $1$ .

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

Step 2.3.12.10

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

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

Step 2.3.13

Reorder terms.

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

Step 2.4

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

$\ln (| y |) = \frac{1}{4} \ln (\frac{| x |}{| x - 4 |}) + C$

Step 3

Solve for $y$ .

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Step 3.1

Simplify the right side.

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Step 3.1.1

Combine $\frac{1}{4}$ and $\ln (\frac{| x |}{| x - 4 |})$ .

$\ln (| y |) = \frac{\ln (\frac{| x |}{| x - 4 |})}{4} + C$

Step 3.2

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

$\ln (| y |) - \frac{\ln (\frac{| x |}{| x - 4 |})}{4} = C$

Step 3.3

To write $\ln (| y |)$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

$\ln (| y |) \cdot \frac{4}{4} - \frac{\ln (\frac{| x |}{| x - 4 |})}{4} = C$

Step 3.4

Simplify terms.

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Step 3.4.1

Combine $\ln (| y |)$ and $\frac{4}{4}$ .

$\frac{\ln (| y |) \cdot 4}{4} - \frac{\ln (\frac{| x |}{| x - 4 |})}{4} = C$

Step 3.4.2

Combine the numerators over the common denominator.

$\frac{\ln (| y |) \cdot 4 - \ln (\frac{| x |}{| x - 4 |})}{4} = C$

Step 3.5

Move $4$ to the left of $\ln (| y |)$ .

$\frac{4 \ln (| y |) - \ln (\frac{| x |}{| x - 4 |})}{4} = C$

Step 3.6

Simplify the left side.

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Step 3.6.1

Simplify $\frac{4 \ln (| y |) - \ln (\frac{| x |}{| x - 4 |})}{4}$ .

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Step 3.6.1.1

Simplify the numerator.

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Step 3.6.1.1.1

Simplify $4 \ln (| y |)$ by moving $4$ inside the logarithm.

$\frac{\ln ({| y |}^{4}) - \ln (\frac{| x |}{| x - 4 |})}{4} = C$

Step 3.6.1.1.2

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

$\frac{\ln (y^{4}) - \ln (\frac{| x |}{| x - 4 |})}{4} = C$

Step 3.6.1.1.3

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

$\frac{\ln (\frac{y^{4}}{\frac{| x |}{| x - 4 |}})}{4} = C$

Step 3.6.1.1.4

Multiply the numerator by the reciprocal of the denominator.

$\frac{\ln (y^{4} \frac{| x - 4 |}{| x |})}{4} = C$

Step 3.6.1.1.5

Combine $y^{4}$ and $\frac{| x - 4 |}{| x |}$ .

$\frac{\ln (\frac{y^{4} | x - 4 |}{| x |})}{4} = C$

Step 3.6.1.2

Rewrite $\frac{\ln (\frac{y^{4} | x - 4 |}{| x |})}{4}$ as $\frac{1}{4} \ln (\frac{y^{4} | x - 4 |}{| x |})$ .

$\frac{1}{4} \ln (\frac{y^{4} | x - 4 |}{| x |}) = C$

Step 3.6.1.3

Simplify $\frac{1}{4} \ln (\frac{y^{4} | x - 4 |}{| x |})$ by moving $\frac{1}{4}$ inside the logarithm.

$\ln ({(\frac{y^{4} | x - 4 |}{| x |})}^{\frac{1}{4}}) = C$

Step 3.6.1.4

Use the power rule ${(a b)}^{n} = a^{n} b^{n}$ to distribute the exponent.

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Step 3.6.1.4.1

Apply the product rule to $\frac{y^{4} | x - 4 |}{| x |}$ .

$\ln (\frac{{(y^{4} | x - 4 |)}^{\frac{1}{4}}}{{| x |}^{\frac{1}{4}}}) = C$

Step 3.6.1.4.2

Apply the product rule to $y^{4} | x - 4 |$ .

$\ln (\frac{{(y^{4})}^{\frac{1}{4}} {| x - 4 |}^{\frac{1}{4}}}{{| x |}^{\frac{1}{4}}}) = C$

Step 3.6.1.5

Simplify the numerator.

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Step 3.6.1.5.1

Multiply the exponents in ${(y^{4})}^{\frac{1}{4}}$ .

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Step 3.6.1.5.1.1

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

$\ln (\frac{y^{4 (\frac{1}{4})} {| x - 4 |}^{\frac{1}{4}}}{{| x |}^{\frac{1}{4}}}) = C$

Step 3.6.1.5.1.2

Cancel the common factor of $4$ .

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Step 3.6.1.5.1.2.1

Cancel the common factor.

$\ln (\frac{y^{4 (\frac{1}{4})} {| x - 4 |}^{\frac{1}{4}}}{{| x |}^{\frac{1}{4}}}) = C$

Step 3.6.1.5.1.2.2

Rewrite the expression.

$\ln (\frac{y^{1} {| x - 4 |}^{\frac{1}{4}}}{{| x |}^{\frac{1}{4}}}) = C$

Step 3.6.1.5.2

Simplify.

$\ln (\frac{y {| x - 4 |}^{\frac{1}{4}}}{{| x |}^{\frac{1}{4}}}) = C$

Step 3.7

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

$e^{\ln (\frac{y {| x - 4 |}^{\frac{1}{4}}}{{| x |}^{\frac{1}{4}}})} = e^{C}$

Step 3.8

Rewrite $\ln (\frac{y {| x - 4 |}^{\frac{1}{4}}}{{| x |}^{\frac{1}{4}}}) = C$ 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 $b^{y} = x$ .

$e^{C} = \frac{y {| x - 4 |}^{\frac{1}{4}}}{{| x |}^{\frac{1}{4}}}$

Step 3.9

Solve for $y$ .

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Step 3.9.1

Rewrite the equation as $\frac{y {| x - 4 |}^{\frac{1}{4}}}{{| x |}^{\frac{1}{4}}} = e^{C}$ .

$\frac{y {| x - 4 |}^{\frac{1}{4}}}{{| x |}^{\frac{1}{4}}} = e^{C}$

Step 3.9.2

Multiply both sides by ${| x |}^{\frac{1}{4}}$ .

$\frac{y {| x - 4 |}^{\frac{1}{4}}}{{| x |}^{\frac{1}{4}}} {| x |}^{\frac{1}{4}} = e^{C} {| x |}^{\frac{1}{4}}$

Step 3.9.3

Simplify the left side.

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Step 3.9.3.1

Cancel the common factor of ${| x |}^{\frac{1}{4}}$ .

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Step 3.9.3.1.1

Cancel the common factor.

$\frac{y {| x - 4 |}^{\frac{1}{4}}}{{| x |}^{\frac{1}{4}}} {| x |}^{\frac{1}{4}} = e^{C} {| x |}^{\frac{1}{4}}$

Step 3.9.3.1.2

Rewrite the expression.

$y {| x - 4 |}^{\frac{1}{4}} = e^{C} {| x |}^{\frac{1}{4}}$

Step 3.9.4

Divide each term in $y {| x - 4 |}^{\frac{1}{4}} = e^{C} {| x |}^{\frac{1}{4}}$ by ${| x - 4 |}^{\frac{1}{4}}$ and simplify.

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Step 3.9.4.1

Divide each term in $y {| x - 4 |}^{\frac{1}{4}} = e^{C} {| x |}^{\frac{1}{4}}$ by ${| x - 4 |}^{\frac{1}{4}}$ .

$\frac{y {| x - 4 |}^{\frac{1}{4}}}{{| x - 4 |}^{\frac{1}{4}}} = \frac{e^{C} {| x |}^{\frac{1}{4}}}{{| x - 4 |}^{\frac{1}{4}}}$

Step 3.9.4.2

Simplify the left side.

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Step 3.9.4.2.1

Cancel the common factor.

$\frac{y {| x - 4 |}^{\frac{1}{4}}}{{| x - 4 |}^{\frac{1}{4}}} = \frac{e^{C} {| x |}^{\frac{1}{4}}}{{| x - 4 |}^{\frac{1}{4}}}$

Step 3.9.4.2.2

Divide $y$ by $1$ .

$y = \frac{e^{C} {| x |}^{\frac{1}{4}}}{{| x - 4 |}^{\frac{1}{4}}}$

Step 4

Simplify the constant of integration.

$y = \frac{C {| x |}^{\frac{1}{4}}}{{| x - 4 |}^{\frac{1}{4}}}$