Calculus Examples

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

Step 1

Rewrite the differential equation as a function of $\frac{y}{x}$ .

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Step 1.1

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

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

Step 1.2

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

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

Step 1.3

Apply the distributive property.

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

Step 1.4

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

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Step 1.4.1

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

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

Step 1.4.2

Cancel the common factor.

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

Step 1.4.3

Rewrite the expression.

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

Step 1.5

Combine $y^{2}$ and $\frac{1}{x^{2}}$ .

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

Step 1.6

Apply the distributive property.

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

Step 1.7

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

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Step 1.7.1

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

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

Step 1.7.2

Cancel the common factor.

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

Step 1.7.3

Rewrite the expression.

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

Step 1.8

Cancel the common factor of $x$ .

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Step 1.8.1

Factor $x$ out of $2 x y$ .

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

Step 1.8.2

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

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

Step 1.8.3

Cancel the common factor.

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

Step 1.8.4

Rewrite the expression.

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

Step 1.9

Combine $2$ and $\frac{1}{x}$ .

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

Step 1.10

Combine $y$ and $\frac{2}{x}$ .

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

Step 1.11

Move $2$ to the left of $y$ .

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

Step 1.12

Use the power of quotient rule $\frac{a^{m}}{b^{m}} = {(\frac{a}{b})}^{m}$ .

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

Step 1.13

Factor out $\frac{y}{x}$ from $\frac{2 y}{x}$ .

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Step 1.13.1

Factor $\frac{y}{x}$ out of $\frac{2 y}{x}$ .

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

Step 1.13.2

Reorder $\frac{y}{x}$ and $2$ .

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

Step 2

Let $V = \frac{y}{x}$ . Substitute $V$ for $\frac{y}{x}$ .

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

Step 3

Solve $V = \frac{y}{x}$ for $y$ .

$y = V x$

Step 4

Use the product rule to find the derivative of $y = V x$ with respect to $x$ .

$\frac{d y}{d x} = x \frac{d V}{d x} + V$

Step 5

Substitute $x \frac{d V}{d x} + V$ for $\frac{d y}{d x}$ .

$x \frac{d V}{d x} + V = \frac{2 + V^{2}}{3 + 2 V}$

Step 6

Solve the substituted differential equation.

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Step 6.1

Separate the variables.

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Step 6.1.1

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

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Step 6.1.1.1

Move all terms not containing $\frac{d V}{d x}$ to the right side of the equation.

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Step 6.1.1.1.1

Subtract $V$ from both sides of the equation.

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

Step 6.1.1.1.2

Split the fraction $\frac{2 + V^{2}}{3 + 2 V}$ into two fractions.

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

Step 6.1.1.1.3

Find the common denominator.

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Step 6.1.1.1.3.1

Write $- V$ as a fraction with denominator $1$ .

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

Step 6.1.1.1.3.2

Multiply $\frac{- V}{1}$ by $\frac{3 + 2 V}{3 + 2 V}$ .

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

Step 6.1.1.1.3.3

Multiply $\frac{- V}{1}$ by $\frac{3 + 2 V}{3 + 2 V}$ .

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

Step 6.1.1.1.4

Combine the numerators over the common denominator.

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

Step 6.1.1.1.5

Simplify each term.

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Step 6.1.1.1.5.1

Apply the distributive property.

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

Step 6.1.1.1.5.2

Multiply $3$ by $- 1$ .

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

Step 6.1.1.1.5.3

Rewrite using the commutative property of multiplication.

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

Step 6.1.1.1.5.4

Simplify each term.

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Step 6.1.1.1.5.4.1

Multiply $V$ by $V$ by adding the exponents.

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Step 6.1.1.1.5.4.1.1

Move $V$ .

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

Step 6.1.1.1.5.4.1.2

Multiply $V$ by $V$ .

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

Step 6.1.1.1.5.4.2

Multiply $- 1$ by $2$ .

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

Step 6.1.1.1.6

Subtract $2 V^{2}$ from $V^{2}$ .

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

Step 6.1.1.1.7

Reorder terms.

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

Step 6.1.1.1.8

Split the fraction $\frac{- V^{2} - 3 V + 2}{2 V + 3}$ into two fractions.

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

Step 6.1.1.1.9

Factor $V$ out of $- V^{2} - 3 V$ .

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Step 6.1.1.1.9.1

Factor $V$ out of $- V^{2}$ .

$x \frac{d V}{d x} = \frac{V (- V) - 3 V}{2 V + 3} + \frac{2}{2 V + 3}$

Step 6.1.1.1.9.2

Factor $V$ out of $- 3 V$ .

$x \frac{d V}{d x} = \frac{V (- V) + V \cdot - 3}{2 V + 3} + \frac{2}{2 V + 3}$

Step 6.1.1.1.9.3

Factor $V$ out of $V (- V) + V \cdot - 3$ .

$x \frac{d V}{d x} = \frac{V (- V - 3)}{2 V + 3} + \frac{2}{2 V + 3}$

Step 6.1.1.2

Divide each term in $x \frac{d V}{d x} = \frac{V (- V - 3)}{2 V + 3} + \frac{2}{2 V + 3}$ by $x$ and simplify.

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Step 6.1.1.2.1

Divide each term in $x \frac{d V}{d x} = \frac{V (- V - 3)}{2 V + 3} + \frac{2}{2 V + 3}$ by $x$ .

$\frac{x \frac{d V}{d x}}{x} = \frac{\frac{V (- V - 3)}{2 V + 3}}{x} + \frac{\frac{2}{2 V + 3}}{x}$

Step 6.1.1.2.2

Simplify the left side.

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Step 6.1.1.2.2.1

Cancel the common factor of $x$ .

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Step 6.1.1.2.2.1.1

Cancel the common factor.

$\frac{x \frac{d V}{d x}}{x} = \frac{\frac{V (- V - 3)}{2 V + 3}}{x} + \frac{\frac{2}{2 V + 3}}{x}$

Step 6.1.1.2.2.1.2

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

$\frac{d V}{d x} = \frac{\frac{V (- V - 3)}{2 V + 3}}{x} + \frac{\frac{2}{2 V + 3}}{x}$

Step 6.1.1.2.3

Simplify the right side.

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Step 6.1.1.2.3.1

Simplify terms.

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Step 6.1.1.2.3.1.1

Combine the numerators over the common denominator.

$\frac{d V}{d x} = \frac{\frac{V (- V - 3)}{2 V + 3} + \frac{2}{2 V + 3}}{x}$

Step 6.1.1.2.3.1.2

Combine the numerators over the common denominator.

$\frac{d V}{d x} = \frac{\frac{V (- V - 3) + 2}{2 V + 3}}{x}$

Step 6.1.1.2.3.2

Simplify the numerator.

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Step 6.1.1.2.3.2.1

Apply the distributive property.

$\frac{d V}{d x} = \frac{\frac{V (- V) + V \cdot - 3 + 2}{2 V + 3}}{x}$

Step 6.1.1.2.3.2.2

Rewrite using the commutative property of multiplication.

$\frac{d V}{d x} = \frac{\frac{- V \cdot V + V \cdot - 3 + 2}{2 V + 3}}{x}$

Step 6.1.1.2.3.2.3

Move $- 3$ to the left of $V$ .

$\frac{d V}{d x} = \frac{\frac{- V \cdot V - 3 \cdot V + 2}{2 V + 3}}{x}$

Step 6.1.1.2.3.2.4

Multiply $V$ by $V$ by adding the exponents.

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Step 6.1.1.2.3.2.4.1

Move $V$ .

$\frac{d V}{d x} = \frac{\frac{- (V \cdot V) - 3 \cdot V + 2}{2 V + 3}}{x}$

Step 6.1.1.2.3.2.4.2

Multiply $V$ by $V$ .

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

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

Step 6.1.1.2.3.3

Simplify with factoring out.

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Step 6.1.1.2.3.3.1

Factor $- 1$ out of $- V^{2}$ .

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

Step 6.1.1.2.3.3.2

Factor $- 1$ out of $- 3 V$ .

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

Step 6.1.1.2.3.3.3

Factor $- 1$ out of $- (V^{2}) - (3 V)$ .

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

Step 6.1.1.2.3.3.4

Rewrite $2$ as $- 1 (- 2)$ .

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

Step 6.1.1.2.3.3.5

Factor $- 1$ out of $- (V^{2} + 3 V) - 1 (- 2)$ .

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

Step 6.1.1.2.3.3.6

Simplify the expression.

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Step 6.1.1.2.3.3.6.1

Rewrite $- (V^{2} + 3 V - 2)$ as $- 1 (V^{2} + 3 V - 2)$ .

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

Step 6.1.1.2.3.3.6.2

Move the negative in front of the fraction.

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

Step 6.1.1.2.3.4

Multiply the numerator by the reciprocal of the denominator.

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

Step 6.1.1.2.3.5

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

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

Step 6.1.2

Regroup factors.

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

Step 6.1.3

Multiply both sides by $\frac{2 V + 3}{V^{2} + 3 V - 2}$ .

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

Step 6.1.4

Simplify.

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Step 6.1.4.1

Rewrite using the commutative property of multiplication.

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

Step 6.1.4.2

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

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

Step 6.1.4.3

Cancel the common factor of $2 V + 3$ .

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Step 6.1.4.3.1

Move the leading negative in $- \frac{2 V + 3}{V^{2} + 3 V - 2}$ into the numerator.

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

Step 6.1.4.3.2

Factor $2 V + 3$ out of $- (2 V + 3)$ .

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

Step 6.1.4.3.3

Factor $2 V + 3$ out of $(2 V + 3) x$ .

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

Step 6.1.4.3.4

Cancel the common factor.

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

Step 6.1.4.3.5

Rewrite the expression.

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

Step 6.1.4.4

Cancel the common factor of $V^{2} + 3 V - 2$ .

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Step 6.1.4.4.1

Cancel the common factor.

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

Step 6.1.4.4.2

Rewrite the expression.

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

Step 6.1.5

Rewrite the equation.

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

Step 6.2

Integrate both sides.

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Step 6.2.1

Set up an integral on each side.

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

Step 6.2.2

Integrate the left side.

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Step 6.2.2.1

Let $u = V^{2} + 3 V - 2$ . Then $d u = (2 V + 3) d V$ , so $\frac{1}{2 V + 3} d u = d V$ . Rewrite using $u$ and $d$ $u$ .

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Step 6.2.2.1.1

Let $u = V^{2} + 3 V - 2$ . Find $\frac{d u}{d V}$ .

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Step 6.2.2.1.1.1

Differentiate $V^{2} + 3 V - 2$ .

$\frac{d}{d V} [V^{2} + 3 V - 2]$

Step 6.2.2.1.1.2

Differentiate.

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Step 6.2.2.1.1.2.1

By the Sum Rule, the derivative of $V^{2} + 3 V - 2$ with respect to $V$ is $\frac{d}{d V} [V^{2}] + \frac{d}{d V} [3 V] + \frac{d}{d V} [- 2]$ .

$\frac{d}{d V} [V^{2}] + \frac{d}{d V} [3 V] + \frac{d}{d V} [- 2]$

Step 6.2.2.1.1.2.2

Differentiate using the Power Rule which states that $\frac{d}{d V} [V^{n}]$ is $n V^{n - 1}$ where $n = 2$ .

$2 V + \frac{d}{d V} [3 V] + \frac{d}{d V} [- 2]$

Step 6.2.2.1.1.3

Evaluate $\frac{d}{d V} [3 V]$ .

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Step 6.2.2.1.1.3.1

Since $3$ is constant with respect to $V$ , the derivative of $3 V$ with respect to $V$ is $3 \frac{d}{d V} [V]$ .

$2 V + 3 \frac{d}{d V} [V] + \frac{d}{d V} [- 2]$

Step 6.2.2.1.1.3.2

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

$2 V + 3 \cdot 1 + \frac{d}{d V} [- 2]$

Step 6.2.2.1.1.3.3

Multiply $3$ by $1$ .

$2 V + 3 + \frac{d}{d V} [- 2]$

Step 6.2.2.1.1.4

Differentiate using the Constant Rule.

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Step 6.2.2.1.1.4.1

Since $- 2$ is constant with respect to $V$ , the derivative of $- 2$ with respect to $V$ is $0$ .

$2 V + 3 + 0$

Step 6.2.2.1.1.4.2

Add $2 V + 3$ and $0$ .

$2 V + 3$

Step 6.2.2.1.2

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

$\int \frac{1}{u} d u = \int - \frac{1}{x} d x$

Step 6.2.2.2

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

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

Step 6.2.2.3

Replace all occurrences of $u$ with $V^{2} + 3 V - 2$ .

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

Step 6.2.3

Integrate the right side.

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Step 6.2.3.1

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

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

Step 6.2.3.2

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

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

Step 6.2.3.3

Simplify.

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

Step 6.2.4

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

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

Step 7

Substitute $\frac{y}{x}$ for $V$ .

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

Step 8

Solve $\ln (| {(\frac{y}{x})}^{2} + 3 \frac{y}{x} - 2 |) = - \ln (| x |) + C$ for $y$ .

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Step 8.1

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

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

Step 8.2

Use the product property of logarithms, $\log_{b} (x) + \log_{b} (y) = \log_{b} (x y)$ .

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

Step 8.3

Simplify each term.

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Step 8.3.1

Apply the product rule to $\frac{y}{x}$ .

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

Step 8.3.2

Combine $3$ and $\frac{y}{x}$ .

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

Step 8.4

To multiply absolute values, multiply the terms inside each absolute value.

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

Step 8.5

Apply the distributive property.

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

Step 8.6

Simplify.

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Step 8.6.1

Cancel the common factor of $x$ .

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Step 8.6.1.1

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

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

Step 8.6.1.2

Cancel the common factor.

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

Step 8.6.1.3

Rewrite the expression.

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

Step 8.6.2

Cancel the common factor of $x$ .

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Step 8.6.2.1

Cancel the common factor.

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

Step 8.6.2.2

Rewrite the expression.

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