Calculus Examples

$y (8 x - 9 y) d x + 2 x (x - 3 y) d y = 0$

Step 1

Find $\frac{\partial M}{\partial y}$ where $M (x, y) = y (8 x - 9 y)$ .

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Step 1.1

Differentiate $M$ with respect to $y$ .

$\frac{\partial M}{\partial y} = \frac{d}{d y} [y (8 x - 9 y)]$

Step 1.2

Differentiate using the Product Rule which states that $\frac{d}{d y} [f (y) g (y)]$ is $f (y) \frac{d}{d y} [g (y)] + g (y) \frac{d}{d y} [f (y)]$ where $f (y) = y$ and $g (y) = 8 x - 9 y$ .

$\frac{\partial M}{\partial y} = y \frac{d}{d y} [8 x - 9 y] + (8 x - 9 y) \frac{d}{d y} [y]$

Step 1.3

Differentiate.

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Step 1.3.1

By the Sum Rule, the derivative of $8 x - 9 y$ with respect to $y$ is $\frac{d}{d y} [8 x] + \frac{d}{d y} [- 9 y]$ .

$\frac{\partial M}{\partial y} = y (\frac{d}{d y} [8 x] + \frac{d}{d y} [- 9 y]) + (8 x - 9 y) \frac{d}{d y} [y]$

Step 1.3.2

Since $8 x$ is constant with respect to $y$ , the derivative of $8 x$ with respect to $y$ is $0$ .

$\frac{\partial M}{\partial y} = y (0 + \frac{d}{d y} [- 9 y]) + (8 x - 9 y) \frac{d}{d y} [y]$

Step 1.3.3

Add $0$ and $\frac{d}{d y} [- 9 y]$ .

$\frac{\partial M}{\partial y} = y \frac{d}{d y} [- 9 y] + (8 x - 9 y) \frac{d}{d y} [y]$

Step 1.3.4

Since $- 9$ is constant with respect to $y$ , the derivative of $- 9 y$ with respect to $y$ is $- 9 \frac{d}{d y} [y]$ .

$\frac{\partial M}{\partial y} = y (- 9 \frac{d}{d y} [y]) + (8 x - 9 y) \frac{d}{d y} [y]$

Step 1.3.5

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

$\frac{\partial M}{\partial y} = y (- 9 \cdot 1) + (8 x - 9 y) \frac{d}{d y} [y]$

Step 1.3.6

Simplify the expression.

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Step 1.3.6.1

Multiply $- 9$ by $1$ .

$\frac{\partial M}{\partial y} = y \cdot - 9 + (8 x - 9 y) \frac{d}{d y} [y]$

Step 1.3.6.2

Move $- 9$ to the left of $y$ .

$\frac{\partial M}{\partial y} = - 9 \cdot y + (8 x - 9 y) \frac{d}{d y} [y]$

Step 1.3.7

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

$\frac{\partial M}{\partial y} = - 9 y + (8 x - 9 y) \cdot 1$

Step 1.3.8

Simplify by adding terms.

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Step 1.3.8.1

Multiply $8 x - 9 y$ by $1$ .

$\frac{\partial M}{\partial y} = - 9 y + 8 x - 9 y$

Step 1.3.8.2

Subtract $9 y$ from $- 9 y$ .

$\frac{\partial M}{\partial y} = 8 x - 18 y$

Step 2

Find $\frac{\partial N}{\partial x}$ where $N (x, y) = 2 x (x - 3 y)$ .

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Step 2.1

Differentiate $N$ with respect to $x$ .

$\frac{\partial N}{\partial x} = \frac{d}{d x} [2 x (x - 3 y)]$

Step 2.2

Since $2$ is constant with respect to $x$ , the derivative of $2 x (x - 3 y)$ with respect to $x$ is $2 \frac{d}{d x} [x (x - 3 y)]$ .

$\frac{\partial N}{\partial x} = 2 \frac{d}{d x} [x (x - 3 y)]$

Step 2.3

Differentiate using the Product Rule which states that $\frac{d}{d x} [f (x) g (x)]$ is $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ where $f (x) = x$ and $g (x) = x - 3 y$ .

$\frac{\partial N}{\partial x} = 2 (x \frac{d}{d x} [x - 3 y] + (x - 3 y) \frac{d}{d x} [x])$

Step 2.4

Differentiate.

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Step 2.4.1

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

$\frac{\partial N}{\partial x} = 2 (x (\frac{d}{d x} [x] + \frac{d}{d x} [- 3 y]) + (x - 3 y) \frac{d}{d x} [x])$

Step 2.4.2

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

$\frac{\partial N}{\partial x} = 2 (x (1 + \frac{d}{d x} [- 3 y]) + (x - 3 y) \frac{d}{d x} [x])$

Step 2.4.3

Since $- 3 y$ is constant with respect to $x$ , the derivative of $- 3 y$ with respect to $x$ is $0$ .

$\frac{\partial N}{\partial x} = 2 (x (1 + 0) + (x - 3 y) \frac{d}{d x} [x])$

Step 2.4.4

Simplify the expression.

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Step 2.4.4.1

Add $1$ and $0$ .

$\frac{\partial N}{\partial x} = 2 (x \cdot 1 + (x - 3 y) \frac{d}{d x} [x])$

Step 2.4.4.2

Multiply $x$ by $1$ .

$\frac{\partial N}{\partial x} = 2 (x + (x - 3 y) \frac{d}{d x} [x])$

Step 2.4.5

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

$\frac{\partial N}{\partial x} = 2 (x + (x - 3 y) \cdot 1)$

Step 2.4.6

Simplify by adding terms.

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Step 2.4.6.1

Multiply $x - 3 y$ by $1$ .

$\frac{\partial N}{\partial x} = 2 (x + x - 3 y)$

Step 2.4.6.2

Add $x$ and $x$ .

$\frac{\partial N}{\partial x} = 2 (2 x - 3 y)$

Step 2.5

Simplify.

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Step 2.5.1

Apply the distributive property.

$\frac{\partial N}{\partial x} = 2 (2 x) + 2 (- 3 y)$

Step 2.5.2

Combine terms.

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Step 2.5.2.1

Multiply $2$ by $2$ .

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

Step 2.5.2.2

Multiply $- 3$ by $2$ .

$\frac{\partial N}{\partial x} = 4 x - 6 y$

Step 3

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

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Step 3.1

Substitute $8 x - 18 y$ for $\frac{\partial M}{\partial y}$ and $4 x - 6 y$ for $\frac{\partial N}{\partial x}$ .

$8 x - 18 y = 4 x - 6 y$

Step 3.2

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

$8 x - 18 y = 4 x - 6 y$ is not an identity.

Step 4

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

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Step 4.1

Substitute $8 x - 18 y$ for $\frac{\partial M}{\partial y}$ .

$\frac{8 x - 18 y - \frac{\partial N}{\partial x}}{N}$

Step 4.2

Substitute $4 x - 6 y$ for $\frac{\partial N}{\partial x}$ .

$\frac{8 x - 18 y - (4 x - 6 y)}{N}$

Step 4.3

Substitute $2 x (x - 3 y)$ for $N$ .

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Step 4.3.1

Substitute $2 x (x - 3 y)$ for $N$ .

$\frac{8 x - 18 y - (4 x - 6 y)}{2 x (x - 3 y)}$

Step 4.3.2

Cancel the common factor of $8 x - 18 y - (4 x - 6 y)$ and $2$ .

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Step 4.3.2.1

Factor $- 1$ out of $8 x$ .

$\frac{- (- 8 x) - 18 y - (4 x - 6 y)}{2 x (x - 3 y)}$

Step 4.3.2.2

Factor $- 1$ out of $- 18 y$ .

$\frac{- (- 8 x) - (18 y) - (4 x - 6 y)}{2 x (x - 3 y)}$

Step 4.3.2.3

Factor $- 1$ out of $- (- 8 x) - (18 y)$ .

$\frac{- (- 8 x + 18 y) - (4 x - 6 y)}{2 x (x - 3 y)}$

Step 4.3.2.4

Factor $- 1$ out of $- (- 8 x + 18 y) - (4 x - 6 y)$ .

$\frac{- (- 8 x + 18 y + 4 x - 6 y)}{2 x (x - 3 y)}$

Step 4.3.2.5

Rewrite $- (- 8 x + 18 y + 4 x - 6 y)$ as $- 1 (- 8 x + 18 y + 4 x - 6 y)$ .

$\frac{- 1 (- 8 x + 18 y + 4 x - 6 y)}{2 x (x - 3 y)}$

Step 4.3.2.6

Factor $2$ out of $- 1 (- 8 x + 18 y + 4 x - 6 y)$ .

$\frac{2 (- 1 (- 4 x + 9 y + 2 x - 3 y))}{2 x (x - 3 y)}$

Step 4.3.2.7

Cancel the common factors.

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Step 4.3.2.7.1

Factor $2$ out of $2 x (x - 3 y)$ .

$\frac{2 (- 1 (- 4 x + 9 y + 2 x - 3 y))}{2 (x (x - 3 y))}$

Step 4.3.2.7.2

Cancel the common factor.

$\frac{2 (- 1 (- 4 x + 9 y + 2 x - 3 y))}{2 (x (x - 3 y))}$

Step 4.3.2.7.3

Rewrite the expression.

$\frac{- 1 (- 4 x + 9 y + 2 x - 3 y)}{x (x - 3 y)}$

Step 4.3.3

Simplify the numerator.

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Step 4.3.3.1

Add $- 4 x$ and $2 x$ .

$\frac{- 1 (- 2 x + 9 y - 3 y)}{x (x - 3 y)}$

Step 4.3.3.2

Subtract $3 y$ from $9 y$ .

$\frac{- 1 (- 2 x + 6 y)}{x (x - 3 y)}$

Step 4.3.3.3

Factor $2$ out of $- 2 x + 6 y$ .

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Step 4.3.3.3.1

Factor $2$ out of $- 2 x$ .

$\frac{- 1 (2 (- x) + 6 y)}{x (x - 3 y)}$

Step 4.3.3.3.2

Factor $2$ out of $6 y$ .

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

Step 4.3.3.3.3

Factor $2$ out of $2 (- x) + 2 (3 y)$ .

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

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

Step 4.3.3.4

Multiply $- 1$ by $2$ .

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

Step 4.3.4

Cancel the common factor of $- x + 3 y$ and $x - 3 y$ .

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Step 4.3.4.1

Factor $- 1$ out of $- x$ .

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

Step 4.3.4.2

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

$\frac{- 2 (- (x) - (- 3 y))}{x (x - 3 y)}$

Step 4.3.4.3

Factor $- 1$ out of $- (x) - (- 3 y)$ .

$\frac{- 2 (- (x - 3 y))}{x (x - 3 y)}$

Step 4.3.4.4

Rewrite $- (x - 3 y)$ as $- 1 (x - 3 y)$ .

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

Step 4.3.4.5

Cancel the common factor.

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

Step 4.3.4.6

Rewrite the expression.

$\frac{- 2 \cdot (- 1)}{x}$

Step 4.3.5

Multiply $- 2$ by $- 1$ .

$\frac{2}{x}$

Step 4.4

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

$μ (x, y) = e^{\int \frac{2}{x} d x}$

Step 5

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

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Step 5.1

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

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

Step 5.2

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

$μ (x, y) = e^{2 (\ln (| x |) + C)}$

Step 5.3

Simplify.

$μ (x, y) = e^{2 \ln (| x |)} + C$

Step 5.4

Simplify each term.

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Step 5.4.1

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

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

Step 5.4.2

Exponentiation and log are inverse functions.

$μ (x, y) = {| x |}^{2} + C$

Step 5.4.3

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

$μ (x, y) = x^{2} + C$

Step 6

Multiply both sides of $y (8 x - 9 y) d x + 2 x (x - 3 y) d y = 0$ by the integration factor $x^{2}$ .

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Step 6.1

Multiply $y (8 x - 9 y)$ by $x^{2}$ .

$y (8 x - 9 y) x^{2} d x + 2 x (x - 3 y) d y = 0$

Step 6.2

Apply the distributive property.

$(y (8 x) + y (- 9 y)) x^{2} d x + 2 x (x - 3 y) d y = 0$

Step 6.3

Rewrite using the commutative property of multiplication.

$(8 y x + y (- 9 y)) x^{2} d x + 2 x (x - 3 y) d y = 0$

Step 6.4

Rewrite using the commutative property of multiplication.

$(8 y x - 9 y \cdot y) x^{2} d x + 2 x (x - 3 y) d y = 0$

Step 6.5

Multiply $y$ by $y$ by adding the exponents.

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Step 6.5.1

Move $y$ .

$(8 y x - 9 (y \cdot y)) x^{2} d x + 2 x (x - 3 y) d y = 0$

Step 6.5.2

Multiply $y$ by $y$ .

$(8 y x - 9 y^{2}) x^{2} d x + 2 x (x - 3 y) d y = 0$

Step 6.6

Apply the distributive property.

$(8 y x \cdot x^{2} - 9 y^{2} x^{2}) d x + 2 x (x - 3 y) d y = 0$

Step 6.7

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

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Step 6.7.1

Move $x^{2}$ .

$(8 y (x^{2} x) - 9 y^{2} x^{2}) d x + 2 x (x - 3 y) d y = 0$

Step 6.7.2

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

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Step 6.7.2.1

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

$(8 y (x^{2} x^{1}) - 9 y^{2} x^{2}) d x + 2 x (x - 3 y) d y = 0$

Step 6.7.2.2

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

$(8 y x^{2 + 1} - 9 y^{2} x^{2}) d x + 2 x (x - 3 y) d y = 0$

Step 6.7.3

Add $2$ and $1$ .

$(8 y x^{3} - 9 y^{2} x^{2}) d x + 2 x (x - 3 y) d y = 0$

Step 6.8

Multiply $2 x (x - 3 y)$ by $x^{2}$ .

$(8 y x^{3} - 9 y^{2} x^{2}) d x + 2 x (x - 3 y) x^{2} d y = 0$

Step 6.9

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

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Step 6.9.1

Move $x^{2}$ .

$(8 y x^{3} - 9 y^{2} x^{2}) d x + 2 (x^{2} x) (x - 3 y) d y = 0$

Step 6.9.2

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

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Step 6.9.2.1

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

$(8 y x^{3} - 9 y^{2} x^{2}) d x + 2 (x^{2} x^{1}) (x - 3 y) d y = 0$

Step 6.9.2.2

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

$(8 y x^{3} - 9 y^{2} x^{2}) d x + 2 x^{2 + 1} (x - 3 y) d y = 0$

Step 6.9.3

Add $2$ and $1$ .

$(8 y x^{3} - 9 y^{2} x^{2}) d x + 2 x^{3} (x - 3 y) d y = 0$

Step 6.10

Apply the distributive property.

$(8 y x^{3} - 9 y^{2} x^{2}) d x + (2 x^{3} x + 2 x^{3} (- 3 y)) d y = 0$

Step 6.11

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

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Step 6.11.1

Move $x$ .

$(8 y x^{3} - 9 y^{2} x^{2}) d x + (2 (x \cdot x^{3}) + 2 x^{3} (- 3 y)) d y = 0$

Step 6.11.2

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

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Step 6.11.2.1

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

$(8 y x^{3} - 9 y^{2} x^{2}) d x + (2 (x^{1} x^{3}) + 2 x^{3} (- 3 y)) d y = 0$

Step 6.11.2.2

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

$(8 y x^{3} - 9 y^{2} x^{2}) d x + (2 x^{1 + 3} + 2 x^{3} (- 3 y)) d y = 0$

Step 6.11.3

Add $1$ and $3$ .

$(8 y x^{3} - 9 y^{2} x^{2}) d x + (2 x^{4} + 2 x^{3} (- 3 y)) d y = 0$

Step 6.12

Rewrite using the commutative property of multiplication.

$(8 y x^{3} - 9 y^{2} x^{2}) d x + (2 x^{4} + 2 \cdot - 3 x^{3} y) d y = 0$

Step 6.13

Multiply $2$ by $- 3$ .

$(8 y x^{3} - 9 y^{2} x^{2}) d x + (2 x^{4} - 6 x^{3} y) d y = 0$

Step 7

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

$f (x, y) = \int 8 y x^{3} - 9 y^{2} x^{2} d x$

Step 8

Integrate $M (x, y) = 8 y x^{3} - 9 y^{2} x^{2}$ to find $f (x, y)$ .

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Step 8.1

Split the single integral into multiple integrals.

$f (x, y) = \int 8 y x^{3} d x + \int - 9 y^{2} x^{2} d x$

Step 8.2

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

$f (x, y) = 8 y \int x^{3} d x + \int - 9 y^{2} x^{2} d x$

Step 8.3

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

$f (x, y) = 8 y (\frac{1}{4} x^{4} + C) + \int - 9 y^{2} x^{2} d x$

Step 8.4

Since $- 9 y^{2}$ is constant with respect to $x$ , move $- 9 y^{2}$ out of the integral.

$f (x, y) = 8 y (\frac{1}{4} x^{4} + C) - 9 y^{2} \int x^{2} d x$

Step 8.5

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

$f (x, y) = 8 y (\frac{1}{4} x^{4} + C) - 9 y^{2} (\frac{1}{3} x^{3} + C)$

Step 8.6

Simplify.

$f (x, y) = 8 \cdot \frac{1}{4} y x^{4} - 9 y^{2} (\frac{1}{3} x^{3}) + C$

Step 8.7

Simplify.

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Step 8.7.1

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

$f (x, y) = \frac{8}{4} y x^{4} - 9 y^{2} (\frac{1}{3} x^{3}) + C$

Step 8.7.2

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

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Step 8.7.2.1

Factor $4$ out of $8$ .

$f (x, y) = \frac{4 \cdot 2}{4} y x^{4} - 9 y^{2} (\frac{1}{3} x^{3}) + C$

Step 8.7.2.2

Cancel the common factors.

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Step 8.7.2.2.1

Factor $4$ out of $4$ .

$f (x, y) = \frac{4 \cdot 2}{4 (1)} y x^{4} - 9 y^{2} (\frac{1}{3} x^{3}) + C$

Step 8.7.2.2.2

Cancel the common factor.

$f (x, y) = \frac{4 \cdot 2}{4 \cdot 1} y x^{4} - 9 y^{2} (\frac{1}{3} x^{3}) + C$

Step 8.7.2.2.3

Rewrite the expression.

$f (x, y) = \frac{2}{1} y x^{4} - 9 y^{2} (\frac{1}{3} x^{3}) + C$

Step 8.7.2.2.4

Divide $2$ by $1$ .

$f (x, y) = 2 y x^{4} - 9 y^{2} (\frac{1}{3} x^{3}) + C$

Step 8.7.3

Combine $\frac{1}{3}$ and $x^{3}$ .

$f (x, y) = 2 y x^{4} - 9 y^{2} \frac{x^{3}}{3} + C$

Step 8.7.4

Combine $- 9$ and $\frac{x^{3}}{3}$ .

$f (x, y) = 2 y x^{4} + y^{2} \frac{- 9 x^{3}}{3} + C$

Step 8.7.5

Combine $y^{2}$ and $\frac{- 9 x^{3}}{3}$ .

$f (x, y) = 2 y x^{4} + \frac{y^{2} (- 9 x^{3})}{3} + C$

Step 8.7.6

Cancel the common factor of $- 9$ and $3$ .

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Step 8.7.6.1

Factor $3$ out of $y^{2} (- 9 x^{3})$ .

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

Step 8.7.6.2

Cancel the common factors.

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Step 8.7.6.2.1

Factor $3$ out of $3$ .

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

Step 8.7.6.2.2

Cancel the common factor.

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

Step 8.7.6.2.3

Rewrite the expression.

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

Step 8.7.6.2.4

Divide $y^{2} (- 3 x^{3})$ by $1$ .

$f (x, y) = 2 y x^{4} + y^{2} (- 3 x^{3}) + C$

Step 9

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

$f (x, y) = 2 y x^{4} + y^{2} (- 3 x^{3}) + g (y)$

Step 10

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

$\frac{\partial f}{\partial y} = 2 x^{4} - 6 x^{3} y$

Step 11

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

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Step 11.1

Differentiate $f$ with respect to $y$ .

$\frac{d}{d y} [2 y x^{4} + y^{2} (- 3 x^{3}) + g (y)] = 2 x^{4} - 6 x^{3} y$

Step 11.2

By the Sum Rule, the derivative of $2 y x^{4} + y^{2} (- 3 x^{3}) + g (y)$ with respect to $y$ is $\frac{d}{d y} [2 y x^{4}] + \frac{d}{d y} [y^{2} (- 3 x^{3})] + \frac{d}{d y} [g (y)]$ .

$\frac{d}{d y} [2 y x^{4}] + \frac{d}{d y} [y^{2} (- 3 x^{3})] + \frac{d}{d y} [g (y)] = 2 x^{4} - 6 x^{3} y$

Step 11.3

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

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Step 11.3.1

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

$2 x^{4} \frac{d}{d y} [y] + \frac{d}{d y} [y^{2} (- 3 x^{3})] + \frac{d}{d y} [g (y)] = 2 x^{4} - 6 x^{3} y$

Step 11.3.2

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

$2 x^{4} \cdot 1 + \frac{d}{d y} [y^{2} (- 3 x^{3})] + \frac{d}{d y} [g (y)] = 2 x^{4} - 6 x^{3} y$

Step 11.3.3

Multiply $2$ by $1$ .

$2 x^{4} + \frac{d}{d y} [y^{2} (- 3 x^{3})] + \frac{d}{d y} [g (y)] = 2 x^{4} - 6 x^{3} y$

Step 11.4

Evaluate $\frac{d}{d y} [y^{2} (- 3 x^{3})]$ .

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Step 11.4.1

Since $- 3 x^{3}$ is constant with respect to $y$ , the derivative of $y^{2} (- 3 x^{3})$ with respect to $y$ is $- 3 x^{3} \frac{d}{d y} [y^{2}]$ .

$2 x^{4} - 3 x^{3} \frac{d}{d y} [y^{2}] + \frac{d}{d y} [g (y)] = 2 x^{4} - 6 x^{3} y$

Step 11.4.2

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

$2 x^{4} - 3 x^{3} (2 y) + \frac{d}{d y} [g (y)] = 2 x^{4} - 6 x^{3} y$

Step 11.4.3

Multiply $2$ by $- 3$ .

$2 x^{4} - 6 x^{3} y + \frac{d}{d y} [g (y)] = 2 x^{4} - 6 x^{3} y$

Step 11.5

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

$2 x^{4} - 6 x^{3} y + \frac{d g}{d y} = 2 x^{4} - 6 x^{3} y$

Step 11.6

Reorder terms.

$\frac{d g}{d y} + 2 x^{4} - 6 x^{3} y = 2 x^{4} - 6 x^{3} y$

Step 12

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

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Step 12.1

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

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Step 12.1.1

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

$\frac{d g}{d y} - 6 x^{3} y = 2 x^{4} - 6 x^{3} y - 2 x^{4}$

Step 12.1.2

Add $6 x^{3} y$ to both sides of the equation.

$\frac{d g}{d y} = 2 x^{4} - 6 x^{3} y - 2 x^{4} + 6 x^{3} y$

Step 12.1.3

Combine the opposite terms in $2 x^{4} - 6 x^{3} y - 2 x^{4} + 6 x^{3} y$ .

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Step 12.1.3.1

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

$\frac{d g}{d y} = - 6 x^{3} y + 0 + 6 x^{3} y$

Step 12.1.3.2

Add $- 6 x^{3} y$ and $0$ .

$\frac{d g}{d y} = - 6 x^{3} y + 6 x^{3} y$

Step 12.1.3.3

Add $- 6 x^{3} y$ and $6 x^{3} y$ .

$\frac{d g}{d y} = 0$

Step 13

Find the antiderivative of $0$ to find $g (y)$ .

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Step 13.1

Integrate both sides of $\frac{d g}{d y} = 0$ .

$\int \frac{d g}{d y} d y = \int 0 d y$

Step 13.2

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

$g (y) = \int 0 d y$

Step 13.3

The integral of $0$ with respect to $y$ is $0$ .

$g (y) = 0 + C$

Step 13.4

Add $0$ and $C$ .

$g (y) = C$

Step 14

Substitute for $g (y)$ in $f (x, y) = 2 y x^{4} + y^{2} (- 3 x^{3}) + g (y)$ .

$f (x, y) = 2 y x^{4} + y^{2} (- 3 x^{3}) + C$

Step 15

Rewrite using the commutative property of multiplication.

$f (x, y) = 2 y x^{4} - 3 y^{2} x^{3} + C$