Calculus Examples

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

Step 1

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

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Step 1.1

Subtract $(x^{3} + y^{3}) d x$ from both sides of the equation.

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

Step 1.2

Rewrite.

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

Step 2

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

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Step 2.1

Differentiate $M$ with respect to $y$ .

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

Step 2.2

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

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

Step 2.3

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

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

Step 2.4

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

$\frac{\partial M}{\partial y} = - (0 + \frac{d}{d y} [y^{3}])$

Step 2.5

Add $0$ and $\frac{d}{d y} [y^{3}]$ .

$\frac{\partial M}{\partial y} = - \frac{d}{d y} [y^{3}]$

Step 2.6

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

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

Step 2.7

Multiply $3$ by $- 1$ .

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

Step 3

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

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Step 3.1

Differentiate $N$ with respect to $x$ .

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

Step 3.2

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

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

Step 3.3

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

$\frac{\partial N}{\partial x} = y^{2} \cdot 1$

Step 3.4

Multiply $y^{2}$ by $1$ .

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

Step 4

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

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Step 4.1

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

$- 3 y^{2} = y^{2}$

Step 4.2

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

$- 3 y^{2} = y^{2}$ 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 $- 3 y^{2}$ for $\frac{\partial M}{\partial y}$ .

$\frac{- 3 y^{2} - \frac{\partial N}{\partial x}}{N}$

Step 5.2

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

$\frac{- 3 y^{2} - y^{2}}{N}$

Step 5.3

Substitute $x y^{2}$ for $N$ .

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Step 5.3.1

Substitute $x y^{2}$ for $N$ .

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

Step 5.3.2

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

$\frac{- 4 y^{2}}{x y^{2}}$

Step 5.3.3

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

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Step 5.3.3.1

Cancel the common factor.

$\frac{- 4 y^{2}}{x y^{2}}$

Step 5.3.3.2

Rewrite the expression.

$\frac{- 4}{x}$

Step 5.3.4

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

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Step 7.1

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

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

Step 7.2

Apply the distributive property.

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

Step 7.3

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

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

Step 7.4

Simplify the numerator.

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Step 7.4.1

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

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

Step 7.4.2

Since both terms are perfect cubes, factor using the difference of cubes formula, $a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2})$ where $a = - x$ and $b = y$ .

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

Step 7.4.3

Simplify.

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Step 7.4.3.1

Apply the product rule to $- x$ .

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

Step 7.4.3.2

Raise $- 1$ to the power of $2$ .

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

Step 7.4.3.3

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

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

Step 7.5

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

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

Step 7.6

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

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

Step 7.7

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

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

Step 7.8

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

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

Step 7.9

Move the negative in front of the fraction.

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

Step 7.10

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

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

Step 7.11

Cancel the common factor of $x$ .

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Step 7.11.1

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

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

Step 7.11.2

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

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

Step 7.11.3

Cancel the common factor.

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

Step 7.11.4

Rewrite the expression.

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

Step 7.12

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

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

Step 8

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

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

Step 9

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

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Step 9.1

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

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

Step 9.2

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

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

Step 9.3

Simplify the answer.

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Step 9.3.1

Rewrite $\frac{1}{x^{3}} (\frac{1}{3} y^{3} + C)$ as $\frac{1}{x^{3}} \cdot \frac{1}{3} y^{3} + C$ .

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

Step 9.3.2

Simplify.

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Step 9.3.2.1

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

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

Step 9.3.2.2

Move $3$ to the left of $x^{3}$ .

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

Step 9.3.2.3

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

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

Step 9.3.2.4

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

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

Step 10

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

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

Step 11

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

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

Step 12

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

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Step 12.1

Differentiate $f$ with respect to $x$ .

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

Step 12.2

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

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

Step 12.3

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

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Step 12.3.1

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

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

Step 12.3.2

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

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

Step 12.3.3

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 12.3.3.1

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

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

Step 12.3.3.2

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

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

Step 12.3.3.3

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

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

Step 12.3.4

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

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

Step 12.3.5

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

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Step 12.3.5.1

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

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

Step 12.3.5.2

Multiply $3$ by $- 2$ .

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

Step 12.3.6

Multiply $3$ by $- 1$ .

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

Step 12.3.7

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

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Step 12.3.7.1

Move $x^{2}$ .

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

Step 12.3.7.2

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

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

Step 12.3.7.3

Subtract $6$ from $2$ .

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

Step 12.3.8

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

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

Step 12.3.9

Combine $\frac{- 3 y^{3}}{3}$ and $x^{- 4}$ .

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

Step 12.3.10

Move $x^{- 4}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

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

Step 12.3.11

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

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Step 12.3.11.1

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

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

Step 12.3.11.2

Cancel the common factors.

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Step 12.3.11.2.1

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

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

Step 12.3.11.2.2

Cancel the common factor.

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

Step 12.3.11.2.3

Rewrite the expression.

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

Step 12.3.12

Move the negative in front of the fraction.

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

Step 12.4

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

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

Step 12.5

Reorder terms.

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

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

Move all terms containing variables to the left side of the equation.

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Step 13.1.1.1

Add $\frac{(x + y) (x^{2} - x y + y^{2})}{x^{4}}$ to both sides of the equation.

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

Step 13.1.1.2

Combine the numerators over the common denominator.

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

Step 13.1.1.3

Simplify each term.

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Step 13.1.1.3.1

Expand $(x + y) (x^{2} - x y + y^{2})$ by multiplying each term in the first expression by each term in the second expression.

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

Step 13.1.1.3.2

Simplify each term.

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Step 13.1.1.3.2.1

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

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Step 13.1.1.3.2.1.1

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

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Step 13.1.1.3.2.1.1.1

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

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

Step 13.1.1.3.2.1.1.2

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

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

Step 13.1.1.3.2.1.2

Add $1$ and $2$ .

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

Step 13.1.1.3.2.2

Rewrite using the commutative property of multiplication.

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

Step 13.1.1.3.2.3

Multiply $x$ by $x$ by adding the exponents.

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Step 13.1.1.3.2.3.1

Move $x$ .

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

Step 13.1.1.3.2.3.2

Multiply $x$ by $x$ .

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

Step 13.1.1.3.2.4

Rewrite using the commutative property of multiplication.

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

Step 13.1.1.3.2.5

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

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Step 13.1.1.3.2.5.1

Move $y$ .

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

Step 13.1.1.3.2.5.2

Multiply $y$ by $y$ .

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

Step 13.1.1.3.2.6

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

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Step 13.1.1.3.2.6.1

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

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Step 13.1.1.3.2.6.1.1

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

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

Step 13.1.1.3.2.6.1.2

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

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

Step 13.1.1.3.2.6.2

Add $1$ and $2$ .

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

Step 13.1.1.3.3

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

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Step 13.1.1.3.3.1

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

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

Step 13.1.1.3.3.2

Add $- x^{2} y$ and $x^{2} y$ .

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

Step 13.1.1.3.3.3

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

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

Step 13.1.1.3.3.4

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

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

Step 13.1.1.3.3.5

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

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

Step 13.1.1.3.3.6

Add $x^{3}$ and $0$ .

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

Step 13.1.1.4

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

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Step 13.1.1.4.1

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

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

Step 13.1.1.4.2

Add $x^{3}$ and $0$ .

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

Step 13.1.1.5

Cancel the common factor of $x^{3}$ and $x^{4}$ .

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Step 13.1.1.5.1

Multiply by $1$ .

$\frac{d g}{d x} + \frac{x^{3} \cdot 1}{x^{4}} = 0$

Step 13.1.1.5.2

Cancel the common factors.

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Step 13.1.1.5.2.1

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

$\frac{d g}{d x} + \frac{x^{3} \cdot 1}{x^{3} x} = 0$

Step 13.1.1.5.2.2

Cancel the common factor.

$\frac{d g}{d x} + \frac{x^{3} \cdot 1}{x^{3} x} = 0$

Step 13.1.1.5.2.3

Rewrite the expression.

$\frac{d g}{d x} + \frac{1}{x} = 0$

Step 13.1.2

Subtract $\frac{1}{x}$ from both sides of the equation.

$\frac{d g}{d x} = - \frac{1}{x}$

Step 14

Find the antiderivative of $- \frac{1}{x}$ to find $g (x)$ .

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Step 14.1

Integrate both sides of $\frac{d g}{d x} = - \frac{1}{x}$ .

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

Step 14.2

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

$g (x) = \int - \frac{1}{x} d x$

Step 14.3

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

$g (x) = - \int \frac{1}{x} d x$

Step 14.4

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

$g (x) = - (\ln (| x |) + C)$

Step 14.5

Simplify.

$g (x) = - \ln (| x |) + C$

Step 15

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

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