Calculus Examples

$\frac{d y}{d x} = 8 x^{3} y - 8 x y$

Step 1

Separate the variables.

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Step 1.1

Factor.

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Step 1.1.1

Factor $8 x y$ out of $8 x^{3} y - 8 x y$ .

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Step 1.1.1.1

Factor $8 x y$ out of $8 x^{3} y$ .

$\frac{d y}{d x} = 8 x y (x^{2}) - 8 x y$

Step 1.1.1.2

Factor $8 x y$ out of $- 8 x y$ .

$\frac{d y}{d x} = 8 x y (x^{2}) + 8 x y (- 1)$

Step 1.1.1.3

Factor $8 x y$ out of $8 x y (x^{2}) + 8 x y (- 1)$ .

$\frac{d y}{d x} = 8 x y (x^{2} - 1)$

Step 1.1.2

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

$\frac{d y}{d x} = 8 x y (x^{2} - 1^{2})$

Step 1.1.3

Factor.

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Step 1.1.3.1

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

$\frac{d y}{d x} = 8 x y ((x + 1) (x - 1))$

Step 1.1.3.2

Remove unnecessary parentheses.

$\frac{d y}{d x} = 8 x y (x + 1) (x - 1)$

Step 1.2

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

$\frac{1}{y} \frac{d y}{d x} = \frac{1}{y} (8 x y (x + 1) (x - 1))$

Step 1.3

Simplify.

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Step 1.3.1

Rewrite using the commutative property of multiplication.

$\frac{1}{y} \frac{d y}{d x} = 8 \frac{1}{y} (x y (x + 1) (x - 1))$

Step 1.3.2

Combine $8$ and $\frac{1}{y}$ .

$\frac{1}{y} \frac{d y}{d x} = \frac{8}{y} (x y (x + 1) (x - 1))$

Step 1.3.3

Cancel the common factor of $y$ .

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Step 1.3.3.1

Factor $y$ out of $x y (x + 1) (x - 1)$ .

$\frac{1}{y} \frac{d y}{d x} = \frac{8}{y} (y ((x (x + 1)) (x - 1)))$

Step 1.3.3.2

Cancel the common factor.

$\frac{1}{y} \frac{d y}{d x} = \frac{8}{y} (y ((x (x + 1)) (x - 1)))$

Step 1.3.3.3

Rewrite the expression.

$\frac{1}{y} \frac{d y}{d x} = 8 ((x (x + 1)) (x - 1))$

Step 1.3.4

Apply the distributive property.

$\frac{1}{y} \frac{d y}{d x} = 8 ((x \cdot x + x \cdot 1) (x - 1))$

Step 1.3.5

Multiply $x$ by $x$ .

$\frac{1}{y} \frac{d y}{d x} = 8 ((x^{2} + x \cdot 1) (x - 1))$

Step 1.3.6

Multiply $x$ by $1$ .

$\frac{1}{y} \frac{d y}{d x} = 8 ((x^{2} + x) (x - 1))$

Step 1.3.7

Expand $(x^{2} + x) (x - 1)$ using the FOIL Method.

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Step 1.3.7.1

Apply the distributive property.

$\frac{1}{y} \frac{d y}{d x} = 8 (x^{2} (x - 1) + x (x - 1))$

Step 1.3.7.2

Apply the distributive property.

$\frac{1}{y} \frac{d y}{d x} = 8 (x^{2} x + x^{2} \cdot - 1 + x (x - 1))$

Step 1.3.7.3

Apply the distributive property.

$\frac{1}{y} \frac{d y}{d x} = 8 (x^{2} x + x^{2} \cdot - 1 + x \cdot x + x \cdot - 1)$

Step 1.3.8

Simplify and combine like terms.

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Step 1.3.8.1

Simplify each term.

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Step 1.3.8.1.1

Multiply $x^{2}$ by $x$ by adding the exponents.

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Step 1.3.8.1.1.1

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

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Step 1.3.8.1.1.1.1

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

$\frac{1}{y} \frac{d y}{d x} = 8 (x^{2} x^{1} + x^{2} \cdot - 1 + x \cdot x + x \cdot - 1)$

Step 1.3.8.1.1.1.2

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

$\frac{1}{y} \frac{d y}{d x} = 8 (x^{2 + 1} + x^{2} \cdot - 1 + x \cdot x + x \cdot - 1)$

Step 1.3.8.1.1.2

Add $2$ and $1$ .

$\frac{1}{y} \frac{d y}{d x} = 8 (x^{3} + x^{2} \cdot - 1 + x \cdot x + x \cdot - 1)$

Step 1.3.8.1.2

Move $- 1$ to the left of $x^{2}$ .

$\frac{1}{y} \frac{d y}{d x} = 8 (x^{3} - 1 \cdot x^{2} + x \cdot x + x \cdot - 1)$

Step 1.3.8.1.3

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

$\frac{1}{y} \frac{d y}{d x} = 8 (x^{3} - x^{2} + x \cdot x + x \cdot - 1)$

Step 1.3.8.1.4

Multiply $x$ by $x$ .

$\frac{1}{y} \frac{d y}{d x} = 8 (x^{3} - x^{2} + x^{2} + x \cdot - 1)$

Step 1.3.8.1.5

Move $- 1$ to the left of $x$ .

$\frac{1}{y} \frac{d y}{d x} = 8 (x^{3} - x^{2} + x^{2} - 1 \cdot x)$

Step 1.3.8.1.6

Rewrite $- 1 x$ as $- x$ .

$\frac{1}{y} \frac{d y}{d x} = 8 (x^{3} - x^{2} + x^{2} - x)$

Step 1.3.8.2

Add $- x^{2}$ and $x^{2}$ .

$\frac{1}{y} \frac{d y}{d x} = 8 (x^{3} + 0 - x)$

Step 1.3.8.3

Add $x^{3}$ and $0$ .

$\frac{1}{y} \frac{d y}{d x} = 8 (x^{3} - x)$

Step 1.3.9

Apply the distributive property.

$\frac{1}{y} \frac{d y}{d x} = 8 x^{3} + 8 (- x)$

Step 1.3.10

Multiply $- 1$ by $8$ .

$\frac{1}{y} \frac{d y}{d x} = 8 x^{3} - 8 x$

Step 1.4

Rewrite the equation.

$\frac{1}{y} d y = (8 x^{3} - 8 x) d x$

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

$\int \frac{1}{y} d y = \int 8 x^{3} - 8 x d x$

Step 2.2

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

$\ln (| y |) + C_{1} = \int 8 x^{3} - 8 x d x$

Step 2.3

Integrate the right side.

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Step 2.3.1

Split the single integral into multiple integrals.

$\ln (| y |) + C_{1} = \int 8 x^{3} d x + \int - 8 x d x$

Step 2.3.2

Since $8$ is constant with respect to $x$ , move $8$ out of the integral.

$\ln (| y |) + C_{1} = 8 \int x^{3} d x + \int - 8 x d x$

Step 2.3.3

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

$\ln (| y |) + C_{1} = 8 (\frac{1}{4} x^{4} + C_{2}) + \int - 8 x d x$

Step 2.3.4

Since $- 8$ is constant with respect to $x$ , move $- 8$ out of the integral.

$\ln (| y |) + C_{1} = 8 (\frac{1}{4} x^{4} + C_{2}) - 8 \int x d x$

Step 2.3.5

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

$\ln (| y |) + C_{1} = 8 (\frac{1}{4} x^{4} + C_{2}) - 8 (\frac{1}{2} x^{2} + C_{3})$

Step 2.3.6

Simplify.

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Step 2.3.6.1

Simplify.

$\ln (| y |) + C_{1} = 8 (\frac{1}{4}) x^{4} - 8 (\frac{1}{2}) x^{2} + C_{4}$

Step 2.3.6.2

Simplify.

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Step 2.3.6.2.1

Combine $8$ and $\frac{1}{4}$ .

$\ln (| y |) + C_{1} = \frac{8}{4} x^{4} - 8 (\frac{1}{2}) x^{2} + C_{4}$

Step 2.3.6.2.2

Cancel the common factor of $8$ and $4$ .

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Step 2.3.6.2.2.1

Factor $4$ out of $8$ .

$\ln (| y |) + C_{1} = \frac{4 \cdot 2}{4} x^{4} - 8 (\frac{1}{2}) x^{2} + C_{4}$

Step 2.3.6.2.2.2

Cancel the common factors.

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Step 2.3.6.2.2.2.1

Factor $4$ out of $4$ .

$\ln (| y |) + C_{1} = \frac{4 \cdot 2}{4 (1)} x^{4} - 8 (\frac{1}{2}) x^{2} + C_{4}$

Step 2.3.6.2.2.2.2

Cancel the common factor.

$\ln (| y |) + C_{1} = \frac{4 \cdot 2}{4 \cdot 1} x^{4} - 8 (\frac{1}{2}) x^{2} + C_{4}$

Step 2.3.6.2.2.2.3

Rewrite the expression.

$\ln (| y |) + C_{1} = \frac{2}{1} x^{4} - 8 (\frac{1}{2}) x^{2} + C_{4}$

Step 2.3.6.2.2.2.4

Divide $2$ by $1$ .

$\ln (| y |) + C_{1} = 2 x^{4} - 8 (\frac{1}{2}) x^{2} + C_{4}$

Step 2.3.6.2.3

Combine $- 8$ and $\frac{1}{2}$ .

$\ln (| y |) + C_{1} = 2 x^{4} + \frac{- 8}{2} x^{2} + C_{4}$

Step 2.3.6.2.4

Cancel the common factor of $- 8$ and $2$ .

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Step 2.3.6.2.4.1

Factor $2$ out of $- 8$ .

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

Step 2.3.6.2.4.2

Cancel the common factors.

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Step 2.3.6.2.4.2.1

Factor $2$ out of $2$ .

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

Step 2.3.6.2.4.2.2

Cancel the common factor.

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

Step 2.3.6.2.4.2.3

Rewrite the expression.

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

Step 2.3.6.2.4.2.4

Divide $- 4$ by $1$ .

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

Step 2.4

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

$\ln (| y |) = 2 x^{4} - 4 x^{2} + C$

Step 3

Solve for $y$ .

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Step 3.1

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

$e^{\ln (| y |)} = e^{2 x^{4} - 4 x^{2} + C}$

Step 3.2

Rewrite $\ln (| y |) = 2 x^{4} - 4 x^{2} + C$ 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 $b^{y} = x$ .

$e^{2 x^{4} - 4 x^{2} + C} = | y |$

Step 3.3

Solve for $y$ .

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Step 3.3.1

Rewrite the equation as $| y | = e^{2 x^{4} - 4 x^{2} + C}$ .

$| y | = e^{2 x^{4} - 4 x^{2} + C}$

Step 3.3.2

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

$y = \pm e^{2 x^{4} - 4 x^{2} + C}$

Step 4

Group the constant terms together.

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Step 4.1

Rewrite $e^{2 x^{4} - 4 x^{2} + C}$ as $e^{2 x^{4} - 4 x^{2}} e^{C}$ .

$y = \pm e^{2 x^{4} - 4 x^{2}} e^{C}$

Step 4.2

Reorder $e^{2 x^{4} - 4 x^{2}}$ and $e^{C}$ .

$y = \pm e^{C} e^{2 x^{4} - 4 x^{2}}$

Step 4.3

Combine constants with the plus or minus.

$y = C e^{2 x^{4} - 4 x^{2}}$