Calculus Examples

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

Step 1

Rewrite the differential equation to fit the Exact differential equation technique.

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Step 1.1

Subtract $(2 x y + x^{2}) d y$ from both sides of the equation.

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

Step 2

Find $\frac{\partial M}{\partial y}$ where $M (x, y) = 2 x y + 3 y^{2}$ .

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Step 2.1

Differentiate $M$ with respect to $y$ .

$\frac{\partial M}{\partial y} = \frac{d}{d y} [2 x y + 3 y^{2}]$

Step 2.2

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

$\frac{\partial M}{\partial y} = \frac{d}{d y} [2 x y] + \frac{d}{d y} [3 y^{2}]$

Step 2.3

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

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Step 2.3.1

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

$\frac{\partial M}{\partial y} = 2 x \frac{d}{d y} [y] + \frac{d}{d y} [3 y^{2}]$

Step 2.3.2

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

$\frac{\partial M}{\partial y} = 2 x \cdot 1 + \frac{d}{d y} [3 y^{2}]$

Step 2.3.3

Multiply $2$ by $1$ .

$\frac{\partial M}{\partial y} = 2 x + \frac{d}{d y} [3 y^{2}]$

Step 2.4

Evaluate $\frac{d}{d y} [3 y^{2}]$ .

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Step 2.4.1

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

$\frac{\partial M}{\partial y} = 2 x + 3 \frac{d}{d y} [y^{2}]$

Step 2.4.2

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

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

Step 2.4.3

Multiply $2$ by $3$ .

$\frac{\partial M}{\partial y} = 2 x + 6 y$

Step 3

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

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Step 3.1

Differentiate $N$ with respect to $x$ .

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

Step 3.2

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

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

Step 3.3

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

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

Step 3.4

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

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

Step 3.5

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

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

Step 3.6

Multiply $2$ by $1$ .

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

Step 3.7

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

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

Step 3.8

Simplify.

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Step 3.8.1

Apply the distributive property.

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

Step 3.8.2

Combine terms.

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Step 3.8.2.1

Multiply $2$ by $- 1$ .

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

Step 3.8.2.2

Multiply $2$ by $- 1$ .

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

Step 3.8.3

Reorder terms.

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

Step 4

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

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Step 4.1

Substitute $2 x + 6 y$ for $\frac{\partial M}{\partial y}$ and $- 2 x - 2 y$ for $\frac{\partial N}{\partial x}$ .

$2 x + 6 y = - 2 x - 2 y$

Step 4.2

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

$2 x + 6 y = - 2 x - 2 y$ is not an identity.

Step 5

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 5.1

Substitute $2 x + 6 y$ for $\frac{\partial M}{\partial y}$ .

$\frac{2 x + 6 y - \frac{\partial N}{\partial x}}{N}$

Step 5.2

Substitute $- 2 x - 2 y$ for $\frac{\partial N}{\partial x}$ .

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

Step 5.3

Substitute $- (2 x y + x^{2})$ for $N$ .

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Step 5.3.1

Substitute $- (2 x y + x^{2})$ for $N$ .

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

Step 5.3.2

Simplify the numerator.

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Step 5.3.2.1

Apply the distributive property.

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

Step 5.3.2.2

Multiply $- 2$ by $- 1$ .

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

Step 5.3.2.3

Multiply $- 2$ by $- 1$ .

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

Step 5.3.2.4

Add $2 x$ and $2 x$ .

$\frac{4 x + 6 y + 2 y}{- (2 x y + x^{2})}$

Step 5.3.2.5

Add $6 y$ and $2 y$ .

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

Step 5.3.2.6

Factor $4$ out of $4 x + 8 y$ .

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Step 5.3.2.6.1

Factor $4$ out of $4 x$ .

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

Step 5.3.2.6.2

Factor $4$ out of $8 y$ .

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

Step 5.3.2.6.3

Factor $4$ out of $4 (x) + 4 (2 y)$ .

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

Step 5.3.3

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

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Step 5.3.3.1

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

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

Step 5.3.3.2

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

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

Step 5.3.3.3

Factor $x$ out of $x (2 y) + x \cdot x$ .

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

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

Step 5.3.4

Cancel the common factor of $x + 2 y$ and $2 y + x$ .

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Step 5.3.4.1

Reorder terms.

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

Step 5.3.4.2

Cancel the common factor.

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

Step 5.3.4.3

Rewrite the expression.

$\frac{4}{- x}$

Step 5.3.5

Move the negative in front of the fraction.

$- \frac{4}{x}$

Step 5.4

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

$μ (x, y) = e^{\int - \frac{4}{x} d x}$

Step 6

Evaluate the integral $e^{\int - \frac{4}{x} d x}$ .

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Step 6.1

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

$μ (x, y) = e^{- \int \frac{4}{x} d x}$

Step 6.2

Since $4$ is constant with respect to $x$ , move $4$ out of the integral.

$μ (x, y) = e^{- (4 \int \frac{1}{x} d x)}$

Step 6.3

Multiply $4$ by $- 1$ .

$μ (x, y) = e^{- 4 \int \frac{1}{x} d x}$

Step 6.4

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

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

Step 6.5

Simplify.

$μ (x, y) = e^{- 4 \ln (| x |)} + C$

Step 6.6

Simplify each term.

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Step 6.6.1

Simplify $- 4 \ln (| x |)$ by moving $- 4$ inside the logarithm.

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

Step 6.6.2

Exponentiation and log are inverse functions.

$μ (x, y) = {| x |}^{- 4} + C$

Step 6.6.3

Remove the absolute value in ${| x |}^{- 4}$ because exponentiations with even powers are always positive.

$μ (x, y) = x^{- 4} + C$

Step 6.6.4

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

$μ (x, y) = \frac{1}{x^{4}} + C$

Step 7

Multiply both sides of $(2 x y + 3 y^{2}) d x - (2 x y + x^{2}) d y = 0$ by the integration factor $\frac{1}{x^{4}}$ .

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Step 7.1

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

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

Step 7.2

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

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

Step 7.3

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

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Step 7.3.1

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

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

Step 7.3.2

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

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

Step 7.3.3

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

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

Step 7.4

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

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

Step 7.5

Apply the distributive property.

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

Step 7.6

Multiply $2$ by $- 1$ .

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

Step 7.7

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

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

Step 7.8

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

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Step 7.8.1

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

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

Step 7.8.2

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

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

Step 7.8.3

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

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

Step 7.9

Cancel the common factors.

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Step 7.9.1

Factor $x$ out of $x^{4}$ .

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

Step 7.9.2

Cancel the common factor.

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

Step 7.9.3

Rewrite the expression.

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

Step 7.10

Factor $- 1$ out of $- 2 y$ .

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

Step 7.11

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

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

Step 7.12

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

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

Step 7.13

Rewrite $- (2 y + x)$ as $- 1 (2 y + x)$ .

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

Step 7.14

Move the negative in front of the fraction.

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

Step 8

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

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

Step 9

Integrate $N (x, y) = - \frac{2 y + x}{x^{3}}$ to find $f (x, y)$ .

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Step 9.1

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

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

Step 9.2

Since $\frac{1}{x^{3}}$ is constant with respect to $y$ , move $\frac{1}{x^{3}}$ out of the integral.

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

Step 9.3

Remove parentheses.

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

Step 9.4

Split the single integral into multiple integrals.

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

Step 9.5

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

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

Step 9.6

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

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

Step 9.7

Apply the constant rule.

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

Step 9.8

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

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

Step 9.9

Simplify.

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

Step 10

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

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

Step 11

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

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

Step 12

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

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Step 12.1

Differentiate $f$ with respect to $x$ .

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

Step 12.2

By the Sum Rule, the derivative of $- \frac{1}{x^{3}} (y^{2} + x y) + g (x)$ with respect to $x$ is $\frac{d}{d x} [- \frac{1}{x^{3}} (y^{2} + x y)] + \frac{d}{d x} [g (x)]$ .

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

Step 12.3

Evaluate $\frac{d}{d x} [- \frac{1}{x^{3}} (y^{2} + x y)]$ .

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Step 12.3.1

Since $- 1$ is constant with respect to $x$ , the derivative of $- \frac{1}{x^{3}} (y^{2} + x y)$ with respect to $x$ is $- \frac{d}{d x} [\frac{1}{x^{3}} (y^{2} + x y)]$ .

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

Step 12.3.2

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 $f (x) = \frac{1}{x^{3}}$ and $g (x) = y^{2} + x y$ .

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

Step 12.3.3

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

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

Step 12.3.4

Since $y^{2}$ is constant with respect to $x$ , the derivative of $y^{2}$ with respect to $x$ is $0$ .

$- (\frac{1}{x^{3}} (0 + \frac{d}{d x} [x y]) + (y^{2} + x y) \frac{d}{d x} [\frac{1}{x^{3}}]) + \frac{d}{d x} [g (x)] = \frac{y (2 x + 3 y)}{x^{4}}$

Step 12.3.5

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

$- (\frac{1}{x^{3}} (0 + y \frac{d}{d x} [x]) + (y^{2} + x y) \frac{d}{d x} [\frac{1}{x^{3}}]) + \frac{d}{d x} [g (x)] = \frac{y (2 x + 3 y)}{x^{4}}$

Step 12.3.6

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

$- (\frac{1}{x^{3}} (0 + y \cdot 1) + (y^{2} + x y) \frac{d}{d x} [\frac{1}{x^{3}}]) + \frac{d}{d x} [g (x)] = \frac{y (2 x + 3 y)}{x^{4}}$

Step 12.3.7

Rewrite $\frac{1}{x^{3}}$ as ${(x^{3})}^{- 1}$ .

$- (\frac{1}{x^{3}} (0 + y \cdot 1) + (y^{2} + x y) \frac{d}{d x} [{(x^{3})}^{- 1}]) + \frac{d}{d x} [g (x)] = \frac{y (2 x + 3 y)}{x^{4}}$

Step 12.3.8

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{- 1}$ and $g (x) = x^{3}$ .

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Step 12.3.8.1

To apply the Chain Rule, set $u$ as $x^{3}$ .

$- (\frac{1}{x^{3}} (0 + y \cdot 1) + (y^{2} + x y) (\frac{d}{d u} [u^{- 1}] \frac{d}{d x} [x^{3}])) + \frac{d}{d x} [g (x)] = \frac{y (2 x + 3 y)}{x^{4}}$

Step 12.3.8.2

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

$- (\frac{1}{x^{3}} (0 + y \cdot 1) + (y^{2} + x y) (- u^{- 2} \frac{d}{d x} [x^{3}])) + \frac{d}{d x} [g (x)] = \frac{y (2 x + 3 y)}{x^{4}}$

Step 12.3.8.3

Replace all occurrences of $u$ with $x^{3}$ .

$- (\frac{1}{x^{3}} (0 + y \cdot 1) + (y^{2} + x y) (- {(x^{3})}^{- 2} \frac{d}{d x} [x^{3}])) + \frac{d}{d x} [g (x)] = \frac{y (2 x + 3 y)}{x^{4}}$

Step 12.3.9

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 3$ .

$- (\frac{1}{x^{3}} (0 + y \cdot 1) + (y^{2} + x y) (- {(x^{3})}^{- 2} (3 x^{2}))) + \frac{d}{d x} [g (x)] = \frac{y (2 x + 3 y)}{x^{4}}$

Step 12.3.10

Multiply $y$ by $1$ .

$- (\frac{1}{x^{3}} (0 + y) + (y^{2} + x y) (- {(x^{3})}^{- 2} (3 x^{2}))) + \frac{d}{d x} [g (x)] = \frac{y (2 x + 3 y)}{x^{4}}$

Step 12.3.11

Add $0$ and $y$ .

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

Step 12.3.12

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

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

Step 12.3.13

Multiply the exponents in ${(x^{3})}^{- 2}$ .

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Step 12.3.13.1

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

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

Step 12.3.13.2

Multiply $3$ by $- 2$ .

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

Step 12.3.14

Multiply $3$ by $- 1$ .

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

Step 12.3.15

Multiply $x^{- 6}$ by $x^{2}$ by adding the exponents.

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Step 12.3.15.1

Move $x^{2}$ .

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

Step 12.3.15.2

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

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

Step 12.3.15.3

Subtract $6$ from $2$ .

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

Step 12.3.16

To write $(y^{2} + x y) (- 3 x^{- 4})$ as a fraction with a common denominator, multiply by $\frac{x^{3}}{x^{3}}$ .

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

Step 12.3.17

Combine the numerators over the common denominator.

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

Step 12.3.18

Multiply $x^{- 4}$ by $x^{3}$ by adding the exponents.

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Step 12.3.18.1

Move $x^{3}$ .

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

Step 12.3.18.2

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

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

Step 12.3.18.3

Subtract $4$ from $3$ .

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

Step 12.4

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

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

Step 12.5

Simplify.

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Step 12.5.1

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

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

Step 12.5.2

Apply the distributive property.

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

Step 12.5.3

Combine terms.

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Step 12.5.3.1

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

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

Step 12.5.3.2

Move the negative in front of the fraction.

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

Step 12.5.3.3

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

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

Step 12.5.3.4

Move $3$ to the left of $y^{2}$ .

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

Step 12.5.3.5

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

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

Step 12.5.3.6

Move the negative in front of the fraction.

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

Step 12.5.3.7

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

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

Step 12.5.3.8

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

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

Step 12.5.3.9

Move $3$ to the left of $x$ .

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

Step 12.5.3.10

Move $3$ to the left of $y$ .

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

Step 12.5.3.11

Cancel the common factor of $x$ .

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Step 12.5.3.11.1

Cancel the common factor.

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

Step 12.5.3.11.2

Divide $3 y$ by $1$ .

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

Step 12.5.3.12

Multiply $3$ by $- 1$ .

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

Step 12.5.3.13

Subtract $3 y$ from $y$ .

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

Step 12.5.3.14

To write $\frac{d g}{d x}$ as a fraction with a common denominator, multiply by $\frac{x^{3}}{x^{3}}$ .

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

Step 12.5.3.15

Combine the numerators over the common denominator.

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

Step 12.5.4

Reorder terms.

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

Step 12.5.5

Simplify the numerator.

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Step 12.5.5.1

Apply the distributive property.

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

Step 12.5.5.2

Multiply $- 2$ by $- 1$ .

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

Step 12.5.5.3

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

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Step 12.5.5.3.1

Multiply $- 1$ by $- 1$ .

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

Step 12.5.5.3.2

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

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

Step 12.5.5.4

To write $x^{3} \frac{d g}{d x}$ as a fraction with a common denominator, multiply by $\frac{x}{x}$ .

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

Step 12.5.5.5

Combine the numerators over the common denominator.

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

Step 12.5.5.6

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

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Step 12.5.5.6.1

Move $x$ .

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

Step 12.5.5.6.2

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

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Step 12.5.5.6.2.1

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

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

Step 12.5.5.6.2.2

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

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

Step 12.5.5.6.3

Add $1$ and $3$ .

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

Step 12.5.5.7

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

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

Step 12.5.5.8

Combine the numerators over the common denominator.

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

Step 12.5.6

Multiply the numerator by the reciprocal of the denominator.

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

Step 12.5.7

Multiply $\frac{2 y x + x^{4} \frac{d g}{d x} + 3 y^{2}}{x} \cdot \frac{1}{x^{3}}$ .

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Step 12.5.7.1

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

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

Step 12.5.7.2

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

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Step 12.5.7.2.1

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

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Step 12.5.7.2.1.1

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

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

Step 12.5.7.2.1.2

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

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

Step 12.5.7.2.2

Add $1$ and $3$ .

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

Step 13

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

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Step 13.1

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

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Step 13.1.1

Since the expression on each side of the equation has the same denominator, the numerators must be equal.

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

Step 13.1.2

Simplify $y (2 x + 3 y)$ .

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Step 13.1.2.1

Rewrite.

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

Step 13.1.2.2

Simplify by adding zeros.

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

Step 13.1.2.3

Apply the distributive property.

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

Step 13.1.2.4

Reorder.

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Step 13.1.2.4.1

Rewrite using the commutative property of multiplication.

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

Step 13.1.2.4.2

Rewrite using the commutative property of multiplication.

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

Step 13.1.2.5

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

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Step 13.1.2.5.1

Move $y$ .

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

Step 13.1.2.5.2

Multiply $y$ by $y$ .

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

Step 13.1.3

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

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Step 13.1.3.1

Subtract $2 y x$ from both sides of the equation.

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

Step 13.1.3.2

Subtract $3 y^{2}$ from both sides of the equation.

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

Step 13.1.3.3

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

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Step 13.1.3.3.1

Subtract $2 y x$ from $2 y x$ .

$x^{4} \frac{d g}{d x} = 3 y^{2} + 0 - 3 y^{2}$

Step 13.1.3.3.2

Add $3 y^{2}$ and $0$ .

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

Step 13.1.3.3.3

Subtract $3 y^{2}$ from $3 y^{2}$ .

$x^{4} \frac{d g}{d x} = 0$

Step 13.1.4

Divide each term in $x^{4} \frac{d g}{d x} = 0$ by $x^{4}$ and simplify.

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Step 13.1.4.1

Divide each term in $x^{4} \frac{d g}{d x} = 0$ by $x^{4}$ .

$\frac{x^{4} \frac{d g}{d x}}{x^{4}} = \frac{0}{x^{4}}$

Step 13.1.4.2

Simplify the left side.

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Step 13.1.4.2.1

Cancel the common factor of $x^{4}$ .

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Step 13.1.4.2.1.1

Cancel the common factor.

$\frac{x^{4} \frac{d g}{d x}}{x^{4}} = \frac{0}{x^{4}}$

Step 13.1.4.2.1.2

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

$\frac{d g}{d x} = \frac{0}{x^{4}}$

Step 13.1.4.3

Simplify the right side.

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Step 13.1.4.3.1

Divide $0$ by $x^{4}$ .

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

Step 14

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

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Step 14.1

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

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

Step 14.2

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

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

Step 14.3

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

$g (x) = 0 + C$

Step 14.4

Add $0$ and $C$ .

$g (x) = C$

Step 15

Substitute for $g (x)$ in $f (x, y) = - \frac{1}{x^{3}} (y^{2} + x y) + g (x)$ .

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

Step 16

Simplify each term.

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Step 16.1

Apply the distributive property.

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

Step 16.2

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

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

Step 16.3

Cancel the common factor of $x$ .

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Step 16.3.1

Move the leading negative in $- \frac{1}{x^{3}}$ into the numerator.

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

Step 16.3.2

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

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

Step 16.3.3

Factor $x$ out of $x y$ .

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

Step 16.3.4

Cancel the common factor.

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

Step 16.3.5

Rewrite the expression.

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

Step 16.4

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

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

Step 16.5

Move the negative in front of the fraction.

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