Calculus Examples

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

Step 1

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

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Step 1.1

Differentiate $M$ with respect to $y$ .

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

Step 1.2

Differentiate.

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Step 1.2.1

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

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

Step 1.2.2

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

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

Step 1.3

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

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Step 1.3.1

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

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

Step 1.3.2

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

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

Step 1.3.3

Multiply $3$ by $3$ .

$\frac{\partial M}{\partial y} = 2 y + 9 x y^{2}$

Step 1.4

Reorder terms.

$\frac{\partial M}{\partial y} = 9 y^{2} x + 2 y$

Step 2

Find $\frac{\partial N}{\partial x}$ where $N (x, y) = 1 - x y$ .

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Step 2.1

Differentiate $N$ with respect to $x$ .

$\frac{\partial N}{\partial x} = \frac{d}{d x} [1 - x y]$

Step 2.2

Differentiate.

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Step 2.2.1

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

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

Step 2.2.2

Since $1$ is constant with respect to $x$ , the derivative of $1$ with respect to $x$ is $0$ .

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

Step 2.3

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

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Step 2.3.1

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

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

Step 2.3.2

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

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

Step 2.3.3

Multiply $- 1$ by $1$ .

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

Step 2.4

Subtract $y$ from $0$ .

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

Step 3

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

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Step 3.1

Substitute $9 y^{2} x + 2 y$ for $\frac{\partial M}{\partial y}$ and $- y$ for $\frac{\partial N}{\partial x}$ .

$9 y^{2} x + 2 y = - y$

Step 3.2

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

$9 y^{2} x + 2 y = - y$ 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 $- y$ for $\frac{\partial N}{\partial x}$ .

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

Step 4.2

Substitute $9 y^{2} x + 2 y$ for $\frac{\partial M}{\partial y}$ .

$\frac{- y - (9 y^{2} x + 2 y)}{M}$

Step 4.3

Substitute $y^{2} + 3 x y^{3}$ for $M$ .

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Step 4.3.1

Substitute $y^{2} + 3 x y^{3}$ for $M$ .

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

Step 4.3.2

Simplify the numerator.

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Step 4.3.2.1

Apply the distributive property.

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

Step 4.3.2.2

Multiply $9$ by $- 1$ .

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

Step 4.3.2.3

Multiply $2$ by $- 1$ .

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

Step 4.3.2.4

Subtract $2 y$ from $- y$ .

$\frac{- 9 y^{2} x - 3 y}{y^{2} + 3 x y^{3}}$

Step 4.3.2.5

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

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Step 4.3.2.5.1

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

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

Step 4.3.2.5.2

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

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

Step 4.3.2.5.3

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

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

Step 4.3.3

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

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Step 4.3.3.1

Multiply by $1$ .

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

Step 4.3.3.2

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

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

Step 4.3.3.3

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

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

Step 4.3.4

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

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Step 4.3.4.1

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

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

Step 4.3.4.2

Cancel the common factors.

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Step 4.3.4.2.1

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

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

Step 4.3.4.2.2

Cancel the common factor.

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

Step 4.3.4.2.3

Rewrite the expression.

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

Step 4.3.5

Reorder the terms.

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

Step 4.3.6

Cancel the common factor of $- 3 x y - 1$ and $3 x y + 1$ .

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Step 4.3.6.1

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

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

Step 4.3.6.2

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

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

Step 4.3.6.3

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

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

Step 4.3.6.4

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

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

Step 4.3.6.5

Cancel the common factor.

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

Step 4.3.6.6

Rewrite the expression.

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

Step 4.3.7

Multiply $3$ by $- 1$ .

$\frac{- 3}{y}$

Step 4.3.8

Substitute $y^{2} + 3 x y^{3}$ for $M$ .

$- \frac{3}{y}$

Step 4.4

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

$μ (x, y) = e^{\int - \frac{3}{y} d y}$

Step 5

Evaluate the integral $e^{\int - \frac{3}{y} d y}$ .

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Step 5.1

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

$μ (x, y) = e^{- \int \frac{3}{y} d y}$

Step 5.2

Since $3$ is constant with respect to $y$ , move $3$ out of the integral.

$μ (x, y) = e^{- (3 \int \frac{1}{y} d y)}$

Step 5.3

Multiply $3$ by $- 1$ .

$μ (x, y) = e^{- 3 \int \frac{1}{y} d y}$

Step 5.4

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

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

Step 5.5

Simplify.

$μ (x, y) = e^{- 3 \ln (| y |)} + C$

Step 5.6

Simplify each term.

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Step 5.6.1

Simplify $- 3 \ln (| y |)$ by moving $- 3$ inside the logarithm.

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

Step 5.6.2

Exponentiation and log are inverse functions.

$μ (x, y) = {| y |}^{- 3} + C$

Step 5.6.3

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

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

Step 6

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

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Step 6.1

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

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

Step 6.2

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

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

Step 6.3

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

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Step 6.3.1

Multiply by $1$ .

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

Step 6.3.2

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

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

Step 6.3.3

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

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

Step 6.4

Cancel the common factors.

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Step 6.4.1

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

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

Step 6.4.2

Cancel the common factor.

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

Step 6.4.3

Rewrite the expression.

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

Step 6.5

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

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

Step 6.6

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

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

Step 7

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

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

Step 8

Integrate $M (x, y) = \frac{1 + 3 x y}{y}$ to find $f (x, y)$ .

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Step 8.1

Split the fraction into multiple fractions.

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

Step 8.2

Split the single integral into multiple integrals.

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

Step 8.3

Cancel the common factor of $y$ .

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Step 8.3.1

Cancel the common factor.

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

Step 8.3.2

Divide $3 x$ by $1$ .

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

Step 8.4

Apply the constant rule.

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

Step 8.5

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

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

Step 8.6

Since $3$ is constant with respect to $x$ , move $3$ out of the integral.

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

Step 8.7

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

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

Step 8.8

Simplify.

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

Step 8.9

Combine $3$ and $\frac{1}{2}$ .

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

Step 9

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

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

Step 10

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

$\frac{\partial f}{\partial y} = \frac{1 - x y}{y^{3}}$

Step 11

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

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Step 11.1

Differentiate $f$ with respect to $y$ .

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

Step 11.2

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

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

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 $x \frac{d}{d y} [\frac{1}{y}]$ .

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

Step 11.3.2

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

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

Step 11.3.3

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

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

Step 11.4

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

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

Step 11.5

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

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

Step 11.6

Simplify.

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Step 11.6.1

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

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

Step 11.6.2

Combine terms.

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Step 11.6.2.1

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

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

Step 11.6.2.2

Add $- \frac{x}{y^{2}}$ and $0$ .

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

Step 11.6.3

Reorder terms.

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

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 $\frac{1 - x y}{y^{3}}$ from both sides of the equation.

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

Step 12.1.1.2

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

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

Step 12.1.1.3

Combine the numerators over the common denominator.

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

Step 12.1.1.4

Simplify the numerator.

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Step 12.1.1.4.1

Apply the distributive property.

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

Step 12.1.1.4.2

Multiply $- 1$ by $1$ .

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

Step 12.1.1.4.3

Multiply $- (- x y)$ .

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Step 12.1.1.4.3.1

Multiply $- 1$ by $- 1$ .

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

Step 12.1.1.4.3.2

Multiply $x$ by $1$ .

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

Step 12.1.1.5

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

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

Step 12.1.1.6

Write each expression with a common denominator of $y^{3}$ , by multiplying each by an appropriate factor of $1$ .

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Step 12.1.1.6.1

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

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

Step 12.1.1.6.2

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

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Step 12.1.1.6.2.1

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

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Step 12.1.1.6.2.1.1

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

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

Step 12.1.1.6.2.1.2

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

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

Step 12.1.1.6.2.2

Add $2$ and $1$ .

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

Step 12.1.1.7

Combine the numerators over the common denominator.

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

Step 12.1.1.8

Simplify the numerator.

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Step 12.1.1.8.1

Add $- x y$ and $x y$ .

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

Step 12.1.1.8.2

Add $\frac{d g}{d y} y^{3} - 1$ and $0$ .

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

Step 12.1.2

Set the numerator equal to zero.

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

Step 12.1.3

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

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Step 12.1.3.1

Add $1$ to both sides of the equation.

$\frac{d g}{d y} y^{3} = 1$

Step 12.1.3.2

Divide each term in $\frac{d g}{d y} y^{3} = 1$ by $y^{3}$ and simplify.

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Step 12.1.3.2.1

Divide each term in $\frac{d g}{d y} y^{3} = 1$ by $y^{3}$ .

$\frac{\frac{d g}{d y} y^{3}}{y^{3}} = \frac{1}{y^{3}}$

Step 12.1.3.2.2

Simplify the left side.

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Step 12.1.3.2.2.1

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

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Step 12.1.3.2.2.1.1

Cancel the common factor.

$\frac{\frac{d g}{d y} y^{3}}{y^{3}} = \frac{1}{y^{3}}$

Step 12.1.3.2.2.1.2

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

$\frac{d g}{d y} = \frac{1}{y^{3}}$

Step 13

Find the antiderivative of $\frac{1}{y^{3}}$ to find $g (y)$ .

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Step 13.1

Integrate both sides of $\frac{d g}{d y} = \frac{1}{y^{3}}$ .

$\int \frac{d g}{d y} d y = \int \frac{1}{y^{3}} d y$

Step 13.2

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

$g (y) = \int \frac{1}{y^{3}} d y$

Step 13.3

Move $y^{3}$ out of the denominator by raising it to the $- 1$ power.

$g (y) = \int {(y^{3})}^{- 1} d y$

Step 13.4

Multiply the exponents in ${(y^{3})}^{- 1}$ .

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Step 13.4.1

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

$g (y) = \int y^{3 \cdot - 1} d y$

Step 13.4.2

Multiply $3$ by $- 1$ .

$g (y) = \int y^{- 3} d y$

Step 13.5

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

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

Step 13.6

Simplify the answer.

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Step 13.6.1

Rewrite $- \frac{1}{2} y^{- 2} + C$ as $- \frac{1}{2} \cdot \frac{1}{y^{2}} + C$ .

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

Step 13.6.2

Simplify.

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Step 13.6.2.1

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

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

Step 13.6.2.2

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

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

Step 13.6.2.3

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

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

Step 14

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

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

Step 15

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

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