Calculus Examples

$6 x + \frac{1}{y} \cdot \frac{d y}{d x} = 12 x^{2}$

Step 1

Separate the variables.

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Step 1.1

Solve for $\frac{d y}{d x}$ .

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Step 1.1.1

Combine $\frac{1}{y}$ and $\frac{d y}{d x}$ .

$6 x + \frac{\frac{d y}{d x}}{y} = 12 x^{2}$

Step 1.1.2

Subtract $6 x$ from both sides of the equation.

$\frac{\frac{d y}{d x}}{y} = 12 x^{2} - 6 x$

Step 1.1.3

Multiply both sides by $y$ .

$\frac{\frac{d y}{d x}}{y} y = (12 x^{2} - 6 x) y$

Step 1.1.4

Simplify.

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Step 1.1.4.1

Simplify the left side.

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Step 1.1.4.1.1

Cancel the common factor of $y$ .

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Step 1.1.4.1.1.1

Cancel the common factor.

$\frac{\frac{d y}{d x}}{y} y = (12 x^{2} - 6 x) y$

Step 1.1.4.1.1.2

Rewrite the expression.

$\frac{d y}{d x} = (12 x^{2} - 6 x) y$

Step 1.1.4.2

Simplify the right side.

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Step 1.1.4.2.1

Apply the distributive property.

$\frac{d y}{d x} = 12 x^{2} y - 6 x y$

Step 1.2

Factor $6 x y$ out of $12 x^{2} y - 6 x y$ .

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Step 1.2.1

Factor $6 x y$ out of $12 x^{2} y$ .

$\frac{d y}{d x} = 6 x y (2 x) - 6 x y$

Step 1.2.2

Factor $6 x y$ out of $- 6 x y$ .

$\frac{d y}{d x} = 6 x y (2 x) + 6 x y (- 1)$

Step 1.2.3

Factor $6 x y$ out of $6 x y (2 x) + 6 x y (- 1)$ .

$\frac{d y}{d x} = 6 x y (2 x - 1)$

Step 1.3

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

$\frac{1}{y} \frac{d y}{d x} = \frac{1}{y} (6 x y (2 x - 1))$

Step 1.4

Simplify.

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Step 1.4.1

Rewrite using the commutative property of multiplication.

$\frac{1}{y} \frac{d y}{d x} = 6 \frac{1}{y} (x y (2 x - 1))$

Step 1.4.2

Combine $6$ and $\frac{1}{y}$ .

$\frac{1}{y} \frac{d y}{d x} = \frac{6}{y} (x y (2 x - 1))$

Step 1.4.3

Cancel the common factor of $y$ .

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Step 1.4.3.1

Factor $y$ out of $x y (2 x - 1)$ .

$\frac{1}{y} \frac{d y}{d x} = \frac{6}{y} (y (x (2 x - 1)))$

Step 1.4.3.2

Cancel the common factor.

$\frac{1}{y} \frac{d y}{d x} = \frac{6}{y} (y (x (2 x - 1)))$

Step 1.4.3.3

Rewrite the expression.

$\frac{1}{y} \frac{d y}{d x} = 6 (x (2 x - 1))$

Step 1.4.4

Apply the distributive property.

$\frac{1}{y} \frac{d y}{d x} = 6 (x (2 x) + x \cdot - 1)$

Step 1.4.5

Rewrite using the commutative property of multiplication.

$\frac{1}{y} \frac{d y}{d x} = 6 (2 x \cdot x + x \cdot - 1)$

Step 1.4.6

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

$\frac{1}{y} \frac{d y}{d x} = 6 (2 x \cdot x - 1 \cdot x)$

Step 1.4.7

Simplify each term.

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Step 1.4.7.1

Multiply $x$ by $x$ by adding the exponents.

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Step 1.4.7.1.1

Move $x$ .

$\frac{1}{y} \frac{d y}{d x} = 6 (2 (x \cdot x) - 1 \cdot x)$

Step 1.4.7.1.2

Multiply $x$ by $x$ .

$\frac{1}{y} \frac{d y}{d x} = 6 (2 x^{2} - 1 \cdot x)$

Step 1.4.7.2

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

$\frac{1}{y} \frac{d y}{d x} = 6 (2 x^{2} - x)$

Step 1.4.8

Apply the distributive property.

$\frac{1}{y} \frac{d y}{d x} = 6 (2 x^{2}) + 6 (- x)$

Step 1.4.9

Multiply $2$ by $6$ .

$\frac{1}{y} \frac{d y}{d x} = 12 x^{2} + 6 (- x)$

Step 1.4.10

Multiply $- 1$ by $6$ .

$\frac{1}{y} \frac{d y}{d x} = 12 x^{2} - 6 x$

Step 1.5

Rewrite the equation.

$\frac{1}{y} d y = (12 x^{2} - 6 x) d x$

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

$\int \frac{1}{y} d y = \int 12 x^{2} - 6 x d x$

Step 2.2

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

$\ln (| y |) + C_{1} = \int 12 x^{2} - 6 x d x$

Step 2.3

Integrate the right side.

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Step 2.3.1

Split the single integral into multiple integrals.

$\ln (| y |) + C_{1} = \int 12 x^{2} d x + \int - 6 x d x$

Step 2.3.2

Since $12$ is constant with respect to $x$ , move $12$ out of the integral.

$\ln (| y |) + C_{1} = 12 \int x^{2} d x + \int - 6 x d x$

Step 2.3.3

By the Power Rule, the integral of $x^{2}$ with respect to $x$ is $\frac{1}{3} x^{3}$ .

$\ln (| y |) + C_{1} = 12 (\frac{1}{3} x^{3} + C_{2}) + \int - 6 x d x$

Step 2.3.4

Since $- 6$ is constant with respect to $x$ , move $- 6$ out of the integral.

$\ln (| y |) + C_{1} = 12 (\frac{1}{3} x^{3} + C_{2}) - 6 \int x d x$

Step 2.3.5

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

$\ln (| y |) + C_{1} = 12 (\frac{1}{3} x^{3} + C_{2}) - 6 (\frac{1}{2} x^{2} + C_{3})$

Step 2.3.6

Simplify.

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Step 2.3.6.1

Simplify.

$\ln (| y |) + C_{1} = 12 (\frac{1}{3}) x^{3} - 6 (\frac{1}{2}) x^{2} + C_{4}$

Step 2.3.6.2

Simplify.

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Step 2.3.6.2.1

Combine $12$ and $\frac{1}{3}$ .

$\ln (| y |) + C_{1} = \frac{12}{3} x^{3} - 6 (\frac{1}{2}) x^{2} + C_{4}$

Step 2.3.6.2.2

Cancel the common factor of $12$ and $3$ .

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Step 2.3.6.2.2.1

Factor $3$ out of $12$ .

$\ln (| y |) + C_{1} = \frac{3 \cdot 4}{3} x^{3} - 6 (\frac{1}{2}) x^{2} + C_{4}$

Step 2.3.6.2.2.2

Cancel the common factors.

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Step 2.3.6.2.2.2.1

Factor $3$ out of $3$ .

$\ln (| y |) + C_{1} = \frac{3 \cdot 4}{3 (1)} x^{3} - 6 (\frac{1}{2}) x^{2} + C_{4}$

Step 2.3.6.2.2.2.2

Cancel the common factor.

$\ln (| y |) + C_{1} = \frac{3 \cdot 4}{3 \cdot 1} x^{3} - 6 (\frac{1}{2}) x^{2} + C_{4}$

Step 2.3.6.2.2.2.3

Rewrite the expression.

$\ln (| y |) + C_{1} = \frac{4}{1} x^{3} - 6 (\frac{1}{2}) x^{2} + C_{4}$

Step 2.3.6.2.2.2.4

Divide $4$ by $1$ .

$\ln (| y |) + C_{1} = 4 x^{3} - 6 (\frac{1}{2}) x^{2} + C_{4}$

Step 2.3.6.2.3

Combine $- 6$ and $\frac{1}{2}$ .

$\ln (| y |) + C_{1} = 4 x^{3} + \frac{- 6}{2} x^{2} + C_{4}$

Step 2.3.6.2.4

Cancel the common factor of $- 6$ and $2$ .

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Step 2.3.6.2.4.1

Factor $2$ out of $- 6$ .

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

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 $2$ .

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

Step 2.3.6.2.4.2.2

Cancel the common factor.

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

Step 2.3.6.2.4.2.3

Rewrite the expression.

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

Step 2.3.6.2.4.2.4

Divide $- 3$ by $1$ .

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

Step 2.4

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

$\ln (| y |) = 4 x^{3} - 3 x^{2} + C$

Step 3

Solve for $y$ .

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Step 3.1

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

$e^{\ln (| y |)} = e^{4 x^{3} - 3 x^{2} + C}$

Step 3.2

Rewrite $\ln (| y |) = 4 x^{3} - 3 x^{2} + C$ 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 $b^{y} = x$ .

$e^{4 x^{3} - 3 x^{2} + C} = | y |$

Step 3.3

Solve for $y$ .

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Step 3.3.1

Rewrite the equation as $| y | = e^{4 x^{3} - 3 x^{2} + C}$ .

$| y | = e^{4 x^{3} - 3 x^{2} + C}$

Step 3.3.2

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

$y = \pm e^{4 x^{3} - 3 x^{2} + C}$

Step 4

Group the constant terms together.

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Step 4.1

Rewrite $e^{4 x^{3} - 3 x^{2} + C}$ as $e^{4 x^{3} - 3 x^{2}} e^{C}$ .

$y = \pm e^{4 x^{3} - 3 x^{2}} e^{C}$

Step 4.2

Reorder $e^{4 x^{3} - 3 x^{2}}$ and $e^{C}$ .

$y = \pm e^{C} e^{4 x^{3} - 3 x^{2}}$

Step 4.3

Combine constants with the plus or minus.

$y = C e^{4 x^{3} - 3 x^{2}}$