Calculus Examples

$(2 x y^{2} + \frac{x}{y^{2}}) d x + 4 x^{2} (y d) y = 0$

Step 1

Find $\frac{\partial M}{\partial y}$ where $M (x, y) = 2 x y^{2} + \frac{x}{y^{2}}$ .

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Step 1.1

Differentiate $M$ with respect to $y$ .

$\frac{\partial M}{\partial y} = \frac{d}{d y} [2 x y^{2} + \frac{x}{y^{2}}]$

Step 1.2

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

$\frac{\partial M}{\partial y} = \frac{d}{d y} [2 x y^{2}] + \frac{d}{d y} [\frac{x}{y^{2}}]$

Step 1.3

Evaluate $\frac{d}{d y} [2 x y^{2}]$ .

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Step 1.3.1

Since $2 x$ is constant with respect to $y$ , the derivative of $2 x y^{2}$ with respect to $y$ is $2 x \frac{d}{d y} [y^{2}]$ .

$\frac{\partial M}{\partial y} = 2 x \frac{d}{d y} [y^{2}] + \frac{d}{d y} [\frac{x}{y^{2}}]$

Step 1.3.2

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

$\frac{\partial M}{\partial y} = 2 x (2 y) + \frac{d}{d y} [\frac{x}{y^{2}}]$

Step 1.3.3

Multiply $2$ by $2$ .

$\frac{\partial M}{\partial y} = 4 x y + \frac{d}{d y} [\frac{x}{y^{2}}]$

Step 1.4

Evaluate $\frac{d}{d y} [\frac{x}{y^{2}}]$ .

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Step 1.4.1

Since $x$ is constant with respect to $y$ , the derivative of $\frac{x}{y^{2}}$ with respect to $y$ is $x \frac{d}{d y} [\frac{1}{y^{2}}]$ .

$\frac{\partial M}{\partial y} = 4 x y + x \frac{d}{d y} [\frac{1}{y^{2}}]$

Step 1.4.2

Rewrite $\frac{1}{y^{2}}$ as ${(y^{2})}^{- 1}$ .

$\frac{\partial M}{\partial y} = 4 x y + x \frac{d}{d y} [{(y^{2})}^{- 1}]$

Step 1.4.3

Differentiate using the chain rule, which states that $\frac{d}{d y} [f (g (y))]$ is $f' (g (y)) g' (y)$ where $f (y) = y^{- 1}$ and $g (y) = y^{2}$ .

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Step 1.4.3.1

To apply the Chain Rule, set $u$ as $y^{2}$ .

$\frac{\partial M}{\partial y} = 4 x y + x (\frac{d}{d u} [u^{- 1}] \frac{d}{d y} [y^{2}])$

Step 1.4.3.2

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

$\frac{\partial M}{\partial y} = 4 x y + x (- u^{- 2} \frac{d}{d y} [y^{2}])$

Step 1.4.3.3

Replace all occurrences of $u$ with $y^{2}$ .

$\frac{\partial M}{\partial y} = 4 x y + x (- {(y^{2})}^{- 2} \frac{d}{d y} [y^{2}])$

Step 1.4.4

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

$\frac{\partial M}{\partial y} = 4 x y + x (- {(y^{2})}^{- 2} (2 y))$

Step 1.4.5

Multiply the exponents in ${(y^{2})}^{- 2}$ .

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Step 1.4.5.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

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

Step 1.4.5.2

Multiply $2$ by $- 2$ .

$\frac{\partial M}{\partial y} = 4 x y + x (- y^{- 4} (2 y))$

Step 1.4.6

Multiply $2$ by $- 1$ .

$\frac{\partial M}{\partial y} = 4 x y + x (- 2 y^{- 4} y)$

Step 1.4.7

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

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

Step 1.4.8

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

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

Step 1.4.9

Subtract $4$ from $1$ .

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

Step 1.5

Simplify.

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Step 1.5.1

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

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

Step 1.5.2

Combine terms.

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Step 1.5.2.1

Combine $- 2$ and $\frac{1}{y^{3}}$ .

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

Step 1.5.2.2

Move the negative in front of the fraction.

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

Step 1.5.2.3

Combine $x$ and $\frac{2}{y^{3}}$ .

$\frac{\partial M}{\partial y} = 4 x y - \frac{x \cdot 2}{y^{3}}$

Step 1.5.2.4

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

$\frac{\partial M}{\partial y} = 4 x y - \frac{2 x}{y^{3}}$

Step 2

Find $\frac{\partial N}{\partial x}$ where $N (x, y) = 4 x^{2} y$ .

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Step 2.1

Differentiate $N$ with respect to $x$ .

$\frac{\partial N}{\partial x} = \frac{d}{d x} [4 x^{2} y]$

Step 2.2

Since $4 y$ is constant with respect to $x$ , the derivative of $4 x^{2} y$ with respect to $x$ is $4 y \frac{d}{d x} [x^{2}]$ .

$\frac{\partial N}{\partial x} = 4 y \frac{d}{d x} [x^{2}]$

Step 2.3

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

$\frac{\partial N}{\partial x} = 4 y (2 x)$

Step 2.4

Simplify the expression.

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Step 2.4.1

Multiply $2$ by $4$ .

$\frac{\partial N}{\partial x} = 8 y x$

Step 2.4.2

Reorder the factors of $8 y x$ .

$\frac{\partial N}{\partial x} = 8 x y$

Step 3

Check that $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$ .

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Step 3.1

Substitute $4 x y - \frac{2 x}{y^{3}}$ for $\frac{\partial M}{\partial y}$ and $8 x y$ for $\frac{\partial N}{\partial x}$ .

$4 x y - \frac{2 x}{y^{3}} = 8 x y$

Step 3.2

Since the left side does not equal the right side, the equation is not an identity.

$4 x y - \frac{2 x}{y^{3}} = 8 x y$ is not an identity.

Step 4

Find the integration factor $μ (x, y) = e^{\int \frac{\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y}}{M} d y}$ .

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Step 4.1

Substitute $8 x y$ for $\frac{\partial N}{\partial x}$ .

$\frac{8 x y - \frac{\partial M}{\partial y}}{M}$

Step 4.2

Substitute $4 x y - \frac{2 x}{y^{3}}$ for $\frac{\partial M}{\partial y}$ .

$\frac{8 x y - (4 x y - \frac{2 x}{y^{3}})}{M}$

Step 4.3

Substitute $2 x y^{2} + \frac{x}{y^{2}}$ for $M$ .

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Step 4.3.1

Substitute $2 x y^{2} + \frac{x}{y^{2}}$ for $M$ .

$\frac{8 x y - (4 x y - \frac{2 x}{y^{3}})}{2 x y^{2} + \frac{x}{y^{2}}}$

Step 4.3.2

Multiply the numerator and denominator of the fraction by $y^{2}$ .

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Step 4.3.2.1

Multiply $\frac{8 x y - (4 x y - \frac{2 x}{y^{3}})}{2 x y^{2} + \frac{x}{y^{2}}}$ by $\frac{y^{2}}{y^{2}}$ .

$\frac{y^{2}}{y^{2}} \cdot \frac{8 x y - (4 x y - \frac{2 x}{y^{3}})}{2 x y^{2} + \frac{x}{y^{2}}}$

Step 4.3.2.2

Combine.

$\frac{y^{2} (8 x y - (4 x y - \frac{2 x}{y^{3}}))}{y^{2} (2 x y^{2} + \frac{x}{y^{2}})}$

Step 4.3.3

Apply the distributive property.

$\frac{y^{2} (8 x y) + y^{2} (- (4 x y - \frac{2 x}{y^{3}}))}{y^{2} (2 x y^{2}) + y^{2} \frac{x}{y^{2}}}$

Step 4.3.4

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

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Step 4.3.4.1

Cancel the common factor.

$\frac{y^{2} (8 x y) + y^{2} (- (4 x y - \frac{2 x}{y^{3}}))}{y^{2} (2 x y^{2}) + y^{2} \frac{x}{y^{2}}}$

Step 4.3.4.2

Rewrite the expression.

$\frac{y^{2} (8 x y) + y^{2} (- (4 x y - \frac{2 x}{y^{3}}))}{y^{2} (2 x y^{2}) + x}$

Step 4.3.5

Simplify the numerator.

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Step 4.3.5.1

Factor $y^{3}$ out of $y^{2} (8 x y) + y^{2} (- (4 x y - \frac{2 x}{y^{3}}))$ .

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Step 4.3.5.1.1

Factor $y^{3}$ out of $y^{2} (8 x y)$ .

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

Step 4.3.5.1.2

Factor $y^{3}$ out of $y^{2} (- (4 x y - \frac{2 x}{y^{3}}))$ .

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

Step 4.3.5.1.3

Factor $y^{3}$ out of $y^{3} (y^{- 1} (8 x y)) + y^{3} (y^{- 1} (- (4 x y - \frac{2 x}{y^{3}})))$ .

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

Step 4.3.5.2

Rewrite using the commutative property of multiplication.

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

Step 4.3.5.3

Multiply $y^{- 1}$ by $y$ by adding the exponents.

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Step 4.3.5.3.1

Move $y$ .

$\frac{y^{3} (8 (y \cdot y^{- 1}) x + y^{- 1} (- (4 x y - \frac{2 x}{y^{3}})))}{y^{2} (2 x y^{2}) + x}$

Step 4.3.5.3.2

Multiply $y$ by $y^{- 1}$ .

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Step 4.3.5.3.2.1

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

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

Step 4.3.5.3.2.2

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

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

Step 4.3.5.3.3

Subtract $1$ from $1$ .

$\frac{y^{3} (8 y^{0} x + y^{- 1} (- (4 x y - \frac{2 x}{y^{3}})))}{y^{2} (2 x y^{2}) + x}$

Step 4.3.5.4

Simplify $8 y^{0} x$ .

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

Step 4.3.5.5

Rewrite using the commutative property of multiplication.

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

Step 4.3.5.6

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

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

Step 4.3.5.7

Apply the distributive property.

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

Step 4.3.5.8

Cancel the common factor of $y$ .

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Step 4.3.5.8.1

Move the leading negative in $- \frac{1}{y}$ into the numerator.

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

Step 4.3.5.8.2

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

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

Step 4.3.5.8.3

Cancel the common factor.

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

Step 4.3.5.8.4

Rewrite the expression.

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

Step 4.3.5.9

Multiply $4$ by $- 1$ .

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

Step 4.3.5.10

Multiply $- \frac{1}{y} (- \frac{2 x}{y^{3}})$ .

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Step 4.3.5.10.1

Multiply $- 1$ by $- 1$ .

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

Step 4.3.5.10.2

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

$\frac{y^{3} (8 x - 4 x + \frac{1}{y} \cdot \frac{2 x}{y^{3}})}{y^{2} (2 x y^{2}) + x}$

Step 4.3.5.10.3

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

$\frac{y^{3} (8 x - 4 x + \frac{2 x}{y \cdot y^{3}})}{y^{2} (2 x y^{2}) + x}$

Step 4.3.5.10.4

Multiply $y$ by $y^{3}$ by adding the exponents.

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Step 4.3.5.10.4.1

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

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Step 4.3.5.10.4.1.1

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

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

Step 4.3.5.10.4.1.2

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

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

Step 4.3.5.10.4.2

Add $1$ and $3$ .

$\frac{y^{3} (8 x - 4 x + \frac{2 x}{y^{4}})}{y^{2} (2 x y^{2}) + x}$

Step 4.3.5.11

Subtract $4 x$ from $8 x$ .

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

Step 4.3.5.12

Factor $2 x$ out of $4 x + \frac{2 x}{y^{4}}$ .

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Step 4.3.5.12.1

Factor $2 x$ out of $4 x$ .

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

Step 4.3.5.12.2

Factor $2 x$ out of $\frac{2 x}{y^{4}}$ .

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

Step 4.3.5.12.3

Factor $2 x$ out of $2 x (2) + 2 x \frac{1}{y^{4}}$ .

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

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

Step 4.3.5.13

To write $2$ as a fraction with a common denominator, multiply by $\frac{y^{4}}{y^{4}}$ .

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

Step 4.3.5.14

Combine the numerators over the common denominator.

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

Step 4.3.5.15

Combine exponents.

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Step 4.3.5.15.1

Combine $y^{3}$ and $\frac{2 y^{4} + 1}{y^{4}}$ .

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

Step 4.3.5.15.2

Combine $2$ and $\frac{y^{3} (2 y^{4} + 1)}{y^{4}}$ .

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

Step 4.3.5.15.3

Combine $x$ and $\frac{2 (y^{3} (2 y^{4} + 1))}{y^{4}}$ .

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

Step 4.3.5.16

Remove unnecessary parentheses.

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

Step 4.3.5.17

Reduce the expression $\frac{x \cdot 2 y^{3} (2 y^{4} + 1)}{y^{4}}$ by cancelling the common factors.

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Step 4.3.5.17.1

Factor $y^{3}$ out of $x \cdot 2 y^{3} (2 y^{4} + 1)$ .

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

Step 4.3.5.17.2

Factor $y^{3}$ out of $y^{4}$ .

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

Step 4.3.5.17.3

Cancel the common factor.

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

Step 4.3.5.17.4

Rewrite the expression.

$\frac{\frac{x \cdot 2 (2 y^{4} + 1)}{y}}{y^{2} (2 x y^{2}) + x}$

Step 4.3.5.18

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

$\frac{\frac{2 x (2 y^{4} + 1)}{y}}{y^{2} (2 x y^{2}) + x}$

Step 4.3.6

Simplify the denominator.

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Step 4.3.6.1

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

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Step 4.3.6.1.1

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

$\frac{\frac{2 x (2 y^{4} + 1)}{y}}{x (y^{2} (2 y^{2})) + x}$

Step 4.3.6.1.2

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

$\frac{\frac{2 x (2 y^{4} + 1)}{y}}{x (y^{2} (2 y^{2})) + x^{1}}$

Step 4.3.6.1.3

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

$\frac{\frac{2 x (2 y^{4} + 1)}{y}}{x (y^{2} (2 y^{2})) + x \cdot 1}$

Step 4.3.6.1.4

Factor $x$ out of $x (y^{2} (2 y^{2})) + x \cdot 1$ .

$\frac{\frac{2 x (2 y^{4} + 1)}{y}}{x (y^{2} (2 y^{2}) + 1)}$

Step 4.3.6.2

Rewrite using the commutative property of multiplication.

$\frac{\frac{2 x (2 y^{4} + 1)}{y}}{x (2 y^{2} y^{2} + 1)}$

Step 4.3.6.3

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

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Step 4.3.6.3.1

Move $y^{2}$ .

$\frac{\frac{2 x (2 y^{4} + 1)}{y}}{x (2 (y^{2} y^{2}) + 1)}$

Step 4.3.6.3.2

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

$\frac{\frac{2 x (2 y^{4} + 1)}{y}}{x (2 y^{2 + 2} + 1)}$

Step 4.3.6.3.3

Add $2$ and $2$ .

$\frac{\frac{2 x (2 y^{4} + 1)}{y}}{x (2 y^{4} + 1)}$

Step 4.3.7

Multiply the numerator by the reciprocal of the denominator.

$\frac{2 x (2 y^{4} + 1)}{y} \cdot \frac{1}{x (2 y^{4} + 1)}$

Step 4.3.8

Combine.

$\frac{2 x (2 y^{4} + 1) \cdot 1}{y (x (2 y^{4} + 1))}$

Step 4.3.9

Cancel the common factor of $x$ .

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Step 4.3.9.1

Cancel the common factor.

$\frac{2 x (2 y^{4} + 1) \cdot 1}{y (x (2 y^{4} + 1))}$

Step 4.3.9.2

Rewrite the expression.

$\frac{(2 (2 y^{4} + 1)) \cdot 1}{y (2 y^{4} + 1)}$

Step 4.3.10

Cancel the common factor of $2 y^{4} + 1$ .

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Step 4.3.10.1

Cancel the common factor.

$\frac{2 (2 y^{4} + 1) \cdot 1}{y (2 y^{4} + 1)}$

Step 4.3.10.2

Rewrite the expression.

$\frac{(2) \cdot 1}{y}$

Step 4.3.11

Substitute $2 x y^{2} + \frac{x}{y^{2}}$ for $M$ .

$\frac{2}{y}$

Step 4.4

Find the integration factor $μ (x, y) = e^{\int \frac{\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y}}{M} d y}$ .

$μ (x, y) = e^{\int \frac{2}{y} d y}$

Step 5

Evaluate the integral $e^{\int \frac{2}{y} d y}$ .

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Step 5.1

Since $2$ is constant with respect to $y$ , move $2$ out of the integral.

$μ (x, y) = e^{2 \int \frac{1}{y} d y}$

Step 5.2

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

$μ (x, y) = e^{2 (\ln (| y |) + C)}$

Step 5.3

Simplify.

$μ (x, y) = e^{2 \ln (| y |)} + C$

Step 5.4

Simplify each term.

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Step 5.4.1

Simplify $2 \ln (| y |)$ by moving $2$ inside the logarithm.

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

Step 5.4.2

Exponentiation and log are inverse functions.

$μ (x, y) = {| y |}^{2} + C$

Step 5.4.3

Remove the absolute value in ${| y |}^{2}$ because exponentiations with even powers are always positive.

$μ (x, y) = y^{2} + C$

Step 6

Multiply both sides of $(2 x y^{2} + \frac{x}{y^{2}}) d x + 4 x^{2} y d y = 0$ by the integration factor $y^{2}$ .

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Step 6.1

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

$(2 x y^{2} + \frac{x}{y^{2}}) y^{2} d x + 4 x^{2} y d y = 0$

Step 6.2

Apply the distributive property.

$(2 x y^{2} y^{2} + \frac{x}{y^{2}} y^{2}) d x + 4 x^{2} y d y = 0$

Step 6.3

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

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Step 6.3.1

Move $y^{2}$ .

$(2 x (y^{2} y^{2}) + \frac{x}{y^{2}} y^{2}) d x + 4 x^{2} y d y = 0$

Step 6.3.2

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

$(2 x y^{2 + 2} + \frac{x}{y^{2}} y^{2}) d x + 4 x^{2} y d y = 0$

Step 6.3.3

Add $2$ and $2$ .

$(2 x y^{4} + \frac{x}{y^{2}} y^{2}) d x + 4 x^{2} y d y = 0$

Step 6.4

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

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Step 6.4.1

Cancel the common factor.

$(2 x y^{4} + \frac{x}{y^{2}} y^{2}) d x + 4 x^{2} y d y = 0$

Step 6.4.2

Rewrite the expression.

$(2 x y^{4} + x) d x + 4 x^{2} y d y = 0$

Step 6.5

Multiply $4 x^{2} y$ by $y^{2}$ .

$(2 x y^{4} + x) d x + 4 x^{2} y \cdot y^{2} d y = 0$

Step 6.6

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

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Step 6.6.1

Move $y^{2}$ .

$(2 x y^{4} + x) d x + 4 x^{2} (y^{2} y) d y = 0$

Step 6.6.2

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

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Step 6.6.2.1

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

$(2 x y^{4} + x) d x + 4 x^{2} (y^{2} y^{1}) d y = 0$

Step 6.6.2.2

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

$(2 x y^{4} + x) d x + 4 x^{2} y^{2 + 1} d y = 0$

Step 6.6.3

Add $2$ and $1$ .

$(2 x y^{4} + x) d x + 4 x^{2} y^{3} d y = 0$

Step 7

Set $f (x, y)$ equal to the integral of $N (x, y)$ .

$f (x, y) = \int 4 x^{2} y^{3} d y$

Step 8

Integrate $N (x, y) = 4 x^{2} y^{3}$ to find $f (x, y)$ .

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Step 8.1

Since $4 x^{2}$ is constant with respect to $y$ , move $4 x^{2}$ out of the integral.

$f (x, y) = 4 x^{2} \int y^{3} d y$

Step 8.2

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

$f (x, y) = 4 x^{2} (\frac{1}{4} y^{4} + C)$

Step 8.3

Simplify the answer.

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Step 8.3.1

Rewrite $4 x^{2} (\frac{1}{4} y^{4} + C)$ as $4 x^{2} \frac{1}{4} y^{4} + C$ .

$f (x, y) = 4 x^{2} \frac{1}{4} y^{4} + C$

Step 8.3.2

Simplify.

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Step 8.3.2.1

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

$f (x, y) = x^{2} \frac{4}{4} y^{4} + C$

Step 8.3.2.2

Combine $x^{2}$ and $\frac{4}{4}$ .

$f (x, y) = \frac{x^{2} \cdot 4}{4} y^{4} + C$

Step 8.3.2.3

Move $4$ to the left of $x^{2}$ .

$f (x, y) = \frac{4 \cdot x^{2}}{4} y^{4} + C$

Step 8.3.2.4

Multiply $4$ by $x^{2}$ .

$f (x, y) = \frac{4 x^{2}}{4} y^{4} + C$

Step 8.3.2.5

Cancel the common factor of $4$ .

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Step 8.3.2.5.1

Cancel the common factor.

$f (x, y) = \frac{4 x^{2}}{4} y^{4} + C$

Step 8.3.2.5.2

Divide $x^{2}$ by $1$ .

$f (x, y) = x^{2} y^{4} + C$

Step 9

Since the integral of $g (x)$ will contain an integration constant, we can replace $C$ with $g (x)$ .

$f (x, y) = x^{2} y^{4} + g (x)$

Step 10

Set $\frac{\partial f}{\partial x} = M (x, y)$ .

$\frac{\partial f}{\partial x} = 2 x y^{4} + x$

Step 11

Find $\frac{\partial f}{\partial x}$ .

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Step 11.1

Differentiate $f$ with respect to $x$ .

$\frac{d}{d x} [x^{2} y^{4} + g (x)] = 2 x y^{4} + x$

Step 11.2

By the Sum Rule, the derivative of $x^{2} y^{4} + g (x)$ with respect to $x$ is $\frac{d}{d x} [x^{2} y^{4}] + \frac{d}{d x} [g (x)]$ .

$\frac{d}{d x} [x^{2} y^{4}] + \frac{d}{d x} [g (x)] = 2 x y^{4} + x$

Step 11.3

Evaluate $\frac{d}{d x} [x^{2} y^{4}]$ .

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Step 11.3.1

Since $y^{4}$ is constant with respect to $x$ , the derivative of $x^{2} y^{4}$ with respect to $x$ is $y^{4} \frac{d}{d x} [x^{2}]$ .

$y^{4} \frac{d}{d x} [x^{2}] + \frac{d}{d x} [g (x)] = 2 x y^{4} + x$

Step 11.3.2

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

$y^{4} (2 x) + \frac{d}{d x} [g (x)] = 2 x y^{4} + x$

Step 11.3.3

Move $2$ to the left of $y^{4}$ .

$2 y^{4} x + \frac{d}{d x} [g (x)] = 2 x y^{4} + x$

Step 11.4

Differentiate using the function rule which states that the derivative of $g (x)$ is $\frac{d g}{d x}$ .

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

Step 11.5

Reorder terms.

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

Step 12

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

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Step 12.1

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

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Step 12.1.1

Subtract $2 y^{4} x$ from both sides of the equation.

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

Step 12.1.2

Combine the opposite terms in $2 x y^{4} + x - 2 y^{4} x$ .

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Step 12.1.2.1

Reorder the factors in the terms $2 x y^{4}$ and $- 2 y^{4} x$ .

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

Step 12.1.2.2

Subtract $2 y^{4} x$ from $2 y^{4} x$ .

$\frac{d g}{d x} = x + 0$

Step 12.1.2.3

Add $x$ and $0$ .

$\frac{d g}{d x} = x$

Step 13

Find the antiderivative of $x$ to find $g (x)$ .

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Step 13.1

Integrate both sides of $\frac{d g}{d x} = x$ .

$\int \frac{d g}{d x} d x = \int x d x$

Step 13.2

Evaluate $\int \frac{d g}{d x} d x$ .

$g (x) = \int x d x$

Step 13.3

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

$g (x) = \frac{1}{2} x^{2} + C$

Step 14

Substitute for $g (x)$ in $f (x, y) = x^{2} y^{4} + g (x)$ .

$f (x, y) = x^{2} y^{4} + \frac{1}{2} x^{2} + C$

Step 15

Simplify $f (x, y) = x^{2} y^{4} + \frac{1}{2} x^{2} + C$ .

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Step 15.1

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

$f (x, y) = x^{2} y^{4} + \frac{x^{2}}{2} + C$

Step 15.2

Reorder the factors of $x^{2} y^{4}$ .

$f (x, y) = y^{4} x^{2} + \frac{x^{2}}{2} + C$

Step 15.3

To write $y^{4} x^{2}$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$f (x, y) = y^{4} x^{2} \cdot \frac{2}{2} + \frac{x^{2}}{2} + C$

Step 15.4

Combine $y^{4} x^{2}$ and $\frac{2}{2}$ .

$f (x, y) = \frac{y^{4} x^{2} \cdot 2}{2} + \frac{x^{2}}{2} + C$

Step 15.5

Combine the numerators over the common denominator.

$f (x, y) = \frac{y^{4} x^{2} \cdot 2 + x^{2}}{2} + C$

Step 15.6

Simplify the numerator.

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Step 15.6.1

Factor $x^{2}$ out of $y^{4} x^{2} \cdot 2 + x^{2}$ .

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Step 15.6.1.1

Factor $x^{2}$ out of $y^{4} x^{2} \cdot 2$ .

$f (x, y) = \frac{x^{2} (y^{4} \cdot 2) + x^{2}}{2} + C$

Step 15.6.1.2

Multiply by $1$ .

$f (x, y) = \frac{x^{2} (y^{4} \cdot 2) + x^{2} \cdot 1}{2} + C$

Step 15.6.1.3

Factor $x^{2}$ out of $x^{2} (y^{4} \cdot 2) + x^{2} \cdot 1$ .

$f (x, y) = \frac{x^{2} (y^{4} \cdot 2 + 1)}{2} + C$

Step 15.6.2

Move $2$ to the left of $y^{4}$ .

$f (x, y) = \frac{x^{2} (2 y^{4} + 1)}{2} + C$