Calculus Examples

$x \frac{d y}{d x} = \frac{1}{y^{3}}$

Step 1

Separate the variables.

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Step 1.1

Divide each term in $x \frac{d y}{d x} = \frac{1}{y^{3}}$ by $x$ and simplify.

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Step 1.1.1

Divide each term in $x \frac{d y}{d x} = \frac{1}{y^{3}}$ by $x$ .

$\frac{x \frac{d y}{d x}}{x} = \frac{\frac{1}{y^{3}}}{x}$

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.

$\frac{x \frac{d y}{d x}}{x} = \frac{\frac{1}{y^{3}}}{x}$

Step 1.1.2.1.2

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

$\frac{d y}{d x} = \frac{\frac{1}{y^{3}}}{x}$

Step 1.1.3

Simplify the right side.

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Step 1.1.3.1

Multiply the numerator by the reciprocal of the denominator.

$\frac{d y}{d x} = \frac{1}{y^{3}} \cdot \frac{1}{x}$

Step 1.1.3.2

Combine.

$\frac{d y}{d x} = \frac{1 \cdot 1}{y^{3} x}$

Step 1.1.3.3

Multiply $1$ by $1$ .

$\frac{d y}{d x} = \frac{1}{y^{3} x}$

Step 1.2

Regroup factors.

$\frac{d y}{d x} = \frac{1}{y^{3}} \cdot \frac{1}{x}$

Step 1.3

Multiply both sides by $y^{3}$ .

$y^{3} \frac{d y}{d x} = y^{3} (\frac{1}{y^{3}} \cdot \frac{1}{x})$

Step 1.4

Simplify.

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Step 1.4.1

Combine.

$y^{3} \frac{d y}{d x} = y^{3} \frac{1 \cdot 1}{y^{3} x}$

Step 1.4.2

Cancel the common factor of $y^{3}$ .

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Step 1.4.2.1

Factor $y^{3}$ out of $y^{3} x$ .

$y^{3} \frac{d y}{d x} = y^{3} \frac{1 \cdot 1}{y^{3} (x)}$

Step 1.4.2.2

Cancel the common factor.

$y^{3} \frac{d y}{d x} = y^{3} \frac{1 \cdot 1}{y^{3} x}$

Step 1.4.2.3

Rewrite the expression.

$y^{3} \frac{d y}{d x} = \frac{1 \cdot 1}{x}$

Step 1.4.3

Multiply $1$ by $1$ .

$y^{3} \frac{d y}{d x} = \frac{1}{x}$

Step 1.5

Rewrite the equation.

$y^{3} d y = \frac{1}{x} d x$

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

$\int y^{3} d y = \int \frac{1}{x} d x$

Step 2.2

By the Power Rule, the integral of $y^{3}$ with respect to $y$ is $\frac{1}{4} y^{4}$ .

$\frac{1}{4} y^{4} + C_{1} = \int \frac{1}{x} d x$

Step 2.3

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

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

Step 2.4

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

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

Step 3

Solve for $y$ .

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Step 3.1

Multiply both sides of the equation by $4$ .

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

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 $4 (\frac{1}{4} y^{4})$ .

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Step 3.2.1.1.1

Combine $\frac{1}{4}$ and $y^{4}$ .

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

Step 3.2.1.1.2

Cancel the common factor of $4$ .

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Step 3.2.1.1.2.1

Cancel the common factor.

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

Step 3.2.1.1.2.2

Rewrite the expression.

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

Step 3.2.2

Simplify the right side.

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Step 3.2.2.1

Apply the distributive property.

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

Step 3.3

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

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

Step 3.4

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

$y = \pm \sqrt[4]{\ln ({| x |}^{4}) + 4 C}$

Step 3.5

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

$y = \pm \sqrt[4]{\ln (x^{4}) + 4 C}$

Step 3.6

The complete solution is the result of both the positive and negative portions of the solution.

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Step 3.6.1

First, use the positive value of the $\pm$ to find the first solution.

$y = \sqrt[4]{\ln (x^{4}) + 4 C}$

Step 3.6.2

Next, use the negative value of the $\pm$ to find the second solution.

$y = - \sqrt[4]{\ln (x^{4}) + 4 C}$

Step 3.6.3

The complete solution is the result of both the positive and negative portions of the solution.

$y = \sqrt[4]{\ln (x^{4}) + 4 C}$

$y = - \sqrt[4]{\ln (x^{4}) + 4 C}$

$y = \sqrt[4]{\ln (x^{4}) + 4 C}$

$y = - \sqrt[4]{\ln (x^{4}) + 4 C}$

$y = \sqrt[4]{\ln (x^{4}) + 4 C}$

$y = - \sqrt[4]{\ln (x^{4}) + 4 C}$

Step 4

Simplify the constant of integration.

$y = \sqrt[4]{\ln (x^{4}) + C}$

$y = - \sqrt[4]{\ln (x^{4}) + C}$