Calculus Examples

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

Step 1

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

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Step 1.1

Differentiate $M$ with respect to $y$ .

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

Step 1.2

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

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

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

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

Step 1.3.2

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

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

Step 1.3.3

Multiply $2$ by $2$ .

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

Step 1.4

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

$\frac{\partial M}{\partial y} = 4 x y + 1$

Step 2

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

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Step 2.1

Differentiate $N$ with respect to $x$ .

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

Step 2.2

Differentiate.

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Step 2.2.1

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

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

Step 2.2.2

Since $2 y^{3}$ is constant with respect to $x$ , the derivative of $2 y^{3}$ with respect to $x$ is $0$ .

$\frac{\partial N}{\partial x} = 0 + \frac{d}{d x} [- x]$

Step 2.3

Evaluate $\frac{d}{d x} [- x]$ .

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Step 2.3.1

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

$\frac{\partial N}{\partial x} = 0 - \frac{d}{d x} [x]$

Step 2.3.2

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

$\frac{\partial N}{\partial x} = 0 - 1 \cdot 1$

Step 2.3.3

Multiply $- 1$ by $1$ .

$\frac{\partial N}{\partial x} = 0 - 1$

Step 2.4

Subtract $1$ from $0$ .

$\frac{\partial N}{\partial x} = - 1$

Step 3

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

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Step 3.1

Substitute $4 x y + 1$ for $\frac{\partial M}{\partial y}$ and $- 1$ for $\frac{\partial N}{\partial x}$ .

$4 x y + 1 = - 1$

Step 3.2

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

$4 x y + 1 = - 1$ 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 $- 1$ for $\frac{\partial N}{\partial x}$ .

$\frac{- 1 - \frac{\partial M}{\partial y}}{M}$

Step 4.2

Substitute $4 x y + 1$ for $\frac{\partial M}{\partial y}$ .

$\frac{- 1 - (4 x y + 1)}{M}$

Step 4.3

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

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Step 4.3.1

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

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

Step 4.3.2

Simplify the numerator.

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Step 4.3.2.1

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

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Step 4.3.2.1.1

Reorder $- 1$ and $- (4 x y + 1)$ .

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

Step 4.3.2.1.2

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

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

Step 4.3.2.1.3

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

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

Step 4.3.2.2

Add $1$ and $1$ .

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

Step 4.3.2.3

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

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Step 4.3.2.3.1

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

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

Step 4.3.2.3.2

Factor $2$ out of $2$ .

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

Step 4.3.2.3.3

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

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

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

Step 4.3.2.4

Multiply $- 1$ by $2$ .

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

Step 4.3.3

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

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Step 4.3.3.1

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

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

Step 4.3.3.2

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

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

Step 4.3.3.3

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

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

Step 4.3.3.4

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

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

Step 4.3.4

Cancel the common factor of $2 x y + 1$ .

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Step 4.3.4.1

Cancel the common factor.

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

Step 4.3.4.2

Rewrite the expression.

$\frac{- 2}{y}$

Step 4.3.5

Substitute $2 x y^{2} + y$ 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 $- 1$ is constant with respect to $y$ , move $- 1$ out of the integral.

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

Step 5.2

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 5.3

Multiply $2$ by $- 1$ .

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

Step 5.4

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

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

Step 5.5

Simplify.

$μ (x, y) = e^{- 2 \ln (| y |)} + C$

Step 5.6

Simplify each term.

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Step 5.6.1

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

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

Step 5.6.2

Exponentiation and log are inverse functions.

$μ (x, y) = {| y |}^{- 2} + C$

Step 5.6.3

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

$μ (x, y) = y^{- 2} + C$

Step 5.6.4

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

$μ (x, y) = \frac{1}{y^{2}} + C$

Step 6

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

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Step 6.1

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

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

Step 6.2

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

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

Step 6.3

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

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Step 6.3.1

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

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

Step 6.3.2

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

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

Step 6.3.3

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

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

Step 6.3.4

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

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

Step 6.4

Cancel the common factors.

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Step 6.4.1

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

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

Step 6.4.2

Cancel the common factor.

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

Step 6.4.3

Rewrite the expression.

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

Step 6.5

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

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

Step 6.6

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

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

Step 7

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

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

Step 8

Integrate $M (x, y) = \frac{2 x y + 1}{y}$ to find $f (x, y)$ .

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Step 8.1

Split the fraction into multiple fractions.

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

Step 8.2

Split the single integral into multiple integrals.

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

Step 8.3

Cancel the common factor of $y$ .

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Step 8.3.1

Cancel the common factor.

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

Step 8.3.2

Divide $2 x$ by $1$ .

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

Step 8.4

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

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

Step 8.5

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

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

Step 8.6

Apply the constant rule.

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

Step 8.7

Simplify.

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Step 8.7.1

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

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

Step 8.7.2

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

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

Step 8.8

Simplify.

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

Step 9

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

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

Step 10

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

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

Step 11

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

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Step 11.1

Differentiate $f$ with respect to $y$ .

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

Step 11.2

Differentiate.

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Step 11.2.1

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

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

Step 11.2.2

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

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

Step 11.3

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

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Step 11.3.1

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

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

Step 11.3.2

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

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

Step 11.3.3

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

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

Step 11.4

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

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

Step 11.5

Simplify.

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Step 11.5.1

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

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

Step 11.5.2

Combine terms.

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Step 11.5.2.1

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

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

Step 11.5.2.2

Subtract $\frac{x}{y^{2}}$ from $0$ .

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

Step 11.5.3

Reorder terms.

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

Step 12

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

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Step 12.1

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

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Step 12.1.1

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

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Step 12.1.1.1

Subtract $\frac{2 y^{3} - x}{y^{2}}$ from both sides of the equation.

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

Step 12.1.1.2

Combine the numerators over the common denominator.

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

Step 12.1.1.3

Simplify each term.

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Step 12.1.1.3.1

Apply the distributive property.

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

Step 12.1.1.3.2

Multiply $2$ by $- 1$ .

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

Step 12.1.1.3.3

Multiply $- - x$ .

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Step 12.1.1.3.3.1

Multiply $- 1$ by $- 1$ .

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

Step 12.1.1.3.3.2

Multiply $x$ by $1$ .

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

Step 12.1.1.4

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

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Step 12.1.1.4.1

Add $- x$ and $x$ .

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

Step 12.1.1.4.2

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

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

Step 12.1.1.5

Cancel the common factor of $y^{3}$ and $y^{2}$ .

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Step 12.1.1.5.1

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

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

Step 12.1.1.5.2

Cancel the common factors.

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Step 12.1.1.5.2.1

Multiply by $1$ .

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

Step 12.1.1.5.2.2

Cancel the common factor.

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

Step 12.1.1.5.2.3

Rewrite the expression.

$\frac{d g}{d y} + \frac{- 2 y}{1} = 0$

Step 12.1.1.5.2.4

Divide $- 2 y$ by $1$ .

$\frac{d g}{d y} - 2 y = 0$

Step 12.1.2

Add $2 y$ to both sides of the equation.

$\frac{d g}{d y} = 2 y$

Step 13

Find the antiderivative of $2 y$ to find $g (y)$ .

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Step 13.1

Integrate both sides of $\frac{d g}{d y} = 2 y$ .

$\int \frac{d g}{d y} d y = \int 2 y d y$

Step 13.2

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

$g (y) = \int 2 y d y$

Step 13.3

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

$g (y) = 2 \int y d y$

Step 13.4

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

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

Step 13.5

Simplify the answer.

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Step 13.5.1

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

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

Step 13.5.2

Simplify.

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Step 13.5.2.1

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

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

Step 13.5.2.2

Cancel the common factor of $2$ .

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Step 13.5.2.2.1

Cancel the common factor.

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

Step 13.5.2.2.2

Rewrite the expression.

$g (y) = 1 y^{2} + C$

Step 13.5.2.3

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

$g (y) = y^{2} + C$

Step 14

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

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