Calculus Examples

$\frac{d y}{d x} = \frac{3 y + x^{2} y}{x - 4 x y}$

Step 1

Separate the variables.

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Step 1.1

Factor $y$ out of $3 y + x^{2} y$ .

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Step 1.1.1

Factor $y$ out of $3 y$ .

$\frac{d y}{d x} = \frac{y \cdot 3 + x^{2} y}{x - 4 x y}$

Step 1.1.2

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

$\frac{d y}{d x} = \frac{y \cdot 3 + y x^{2}}{x - 4 x y}$

Step 1.1.3

Factor $y$ out of $y \cdot 3 + y x^{2}$ .

$\frac{d y}{d x} = \frac{y \cdot (3 + x^{2})}{x - 4 x y}$

Step 1.1.4

Multiply $y$ by $3 + x^{2}$ .

$\frac{d y}{d x} = \frac{y (3 + x^{2})}{x - 4 x y}$

Step 1.2

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

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Step 1.2.1

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

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

Step 1.2.2

Factor $x$ out of $x^{1}$ .

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

Step 1.2.3

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

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

Step 1.2.4

Factor $x$ out of $x \cdot 1 + x (- 4 y)$ .

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

Step 1.2.5

Multiply $x$ by $1 - 4 y$ .

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

Step 1.3

Regroup factors.

$\frac{d y}{d x} = \frac{y}{1 - 4 y} \cdot \frac{3 + x^{2}}{x}$

Step 1.4

Multiply both sides by $\frac{1 - 4 y}{y}$ .

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

Step 1.5

Simplify.

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Step 1.5.1

Multiply $\frac{y}{1 - 4 y}$ by $\frac{3 + x^{2}}{x}$ .

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

Step 1.5.2

Cancel the common factor of $1 - 4 y$ .

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Step 1.5.2.1

Factor $1 - 4 y$ out of $(1 - 4 y) x$ .

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

Step 1.5.2.2

Cancel the common factor.

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

Step 1.5.2.3

Rewrite the expression.

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

Step 1.5.3

Cancel the common factor of $y$ .

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Step 1.5.3.1

Cancel the common factor.

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

Step 1.5.3.2

Rewrite the expression.

$\frac{1 - 4 y}{y} \frac{d y}{d x} = \frac{3 + x^{2}}{x}$

Step 1.6

Rewrite the equation.

$\frac{1 - 4 y}{y} d y = \frac{3 + x^{2}}{x} d x$

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

$\int \frac{1 - 4 y}{y} d y = \int \frac{3 + x^{2}}{x} d x$

Step 2.2

Integrate the left side.

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Step 2.2.1

Split the fraction into multiple fractions.

$\int \frac{1}{y} + \frac{- 4 y}{y} d y = \int \frac{3 + x^{2}}{x} d x$

Step 2.2.2

Split the single integral into multiple integrals.

$\int \frac{1}{y} d y + \int \frac{- 4 y}{y} d y = \int \frac{3 + x^{2}}{x} d x$

Step 2.2.3

Cancel the common factor of $y$ .

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Step 2.2.3.1

Cancel the common factor.

$\int \frac{1}{y} d y + \int \frac{- 4 y}{y} d y = \int \frac{3 + x^{2}}{x} d x$

Step 2.2.3.2

Divide $- 4$ by $1$ .

$\int \frac{1}{y} d y + \int - 4 d y = \int \frac{3 + x^{2}}{x} d x$

Step 2.2.4

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

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

Step 2.2.5

Apply the constant rule.

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

Step 2.2.6

Simplify.

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

Step 2.2.7

Reorder terms.

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

Step 2.3

Integrate the right side.

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Step 2.3.1

Split the fraction into multiple fractions.

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

Step 2.3.2

Split the single integral into multiple integrals.

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

Step 2.3.3

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

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Step 2.3.3.1

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

$- 4 y + \ln (| y |) + C_{3} = \int \frac{3}{x} d x + \int \frac{x \cdot x}{x} d x$

Step 2.3.3.2

Cancel the common factors.

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Step 2.3.3.2.1

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

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

Step 2.3.3.2.2

Factor $x$ out of $x^{1}$ .

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

Step 2.3.3.2.3

Cancel the common factor.

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

Step 2.3.3.2.4

Rewrite the expression.

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

Step 2.3.3.2.5

Divide $x$ by $1$ .

$- 4 y + \ln (| y |) + C_{3} = \int \frac{3}{x} d x + \int x d x$

Step 2.3.4

Since $3$ is constant with respect to $x$ , move $3$ out of the integral.

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

Step 2.3.5

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

$- 4 y + \ln (| y |) + C_{3} = 3 (\ln (| x |) + C_{4}) + \int x d x$

Step 2.3.6

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

$- 4 y + \ln (| y |) + C_{3} = 3 (\ln (| x |) + C_{4}) + \frac{1}{2} x^{2} + C_{5}$

Step 2.3.7

Simplify.

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

Step 2.3.8

Reorder terms.

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

Step 2.4

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

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