Calculus Examples

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

Step 1

Find $\frac{\partial M}{\partial y}$ where $M (x, y) = y^{3} + y + 1$ .

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Step 1.1

Differentiate $M$ with respect to $y$ .

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

Step 1.2

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

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

Step 1.3

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

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

Step 1.4

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

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

Step 1.5

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

$\frac{\partial M}{\partial y} = 3 y^{2} + 1 + 0$

Step 1.6

Add $3 y^{2} + 1$ and $0$ .

$\frac{\partial M}{\partial y} = 3 y^{2} + 1$

Step 2

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

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Step 2.1

Differentiate $N$ with respect to $x$ .

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

Step 2.2

Differentiate.

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Step 2.2.1

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

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

Step 2.2.2

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

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

Step 2.3

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

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Step 2.3.1

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

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

Step 2.3.2

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

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

Step 2.3.3

Multiply $- 3$ by $1$ .

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

Step 2.4

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

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Step 2.4.1

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

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

Step 2.4.2

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

$\frac{\partial N}{\partial x} = 2 x - 3 y^{2} - 1 \cdot 1$

Step 2.4.3

Multiply $- 1$ by $1$ .

$\frac{\partial N}{\partial x} = 2 x - 3 y^{2} - 1$

Step 2.5

Reorder terms.

$\frac{\partial N}{\partial x} = - 3 y^{2} + 2 x - 1$

Step 3

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

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Step 3.1

Substitute $3 y^{2} + 1$ for $\frac{\partial M}{\partial y}$ and $- 3 y^{2} + 2 x - 1$ for $\frac{\partial N}{\partial x}$ .

$3 y^{2} + 1 = - 3 y^{2} + 2 x - 1$

Step 3.2

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

$3 y^{2} + 1 = - 3 y^{2} + 2 x - 1$ 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 $3 y^{2} + 1$ for $\frac{\partial M}{\partial y}$ .

$\frac{3 y^{2} + 1 - \frac{\partial N}{\partial x}}{N}$

Step 4.2

Substitute $- 3 y^{2} + 2 x - 1$ for $\frac{\partial N}{\partial x}$ .

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

Step 4.3

Substitute $x^{2} - 3 x y^{2} - x$ for $N$ .

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Step 4.3.1

Substitute $x^{2} - 3 x y^{2} - x$ for $N$ .

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

Step 4.3.2

Simplify the numerator.

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Step 4.3.2.1

Apply the distributive property.

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

Step 4.3.2.2

Simplify.

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Step 4.3.2.2.1

Multiply $- 3$ by $- 1$ .

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

Step 4.3.2.2.2

Multiply $2$ by $- 1$ .

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

Step 4.3.2.2.3

Multiply $- 1$ by $- 1$ .

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

Step 4.3.2.3

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

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

Step 4.3.2.4

Add $1$ and $1$ .

$\frac{6 y^{2} - 2 x + 2}{x^{2} - 3 x y^{2} - x}$

Step 4.3.2.5

Factor $2$ out of $6 y^{2} - 2 x + 2$ .

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Step 4.3.2.5.1

Factor $2$ out of $6 y^{2}$ .

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

Step 4.3.2.5.2

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

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

Step 4.3.2.5.3

Factor $2$ out of $2$ .

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

Step 4.3.2.5.4

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

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

Step 4.3.2.5.5

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

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

Step 4.3.3

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

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Step 4.3.3.1

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

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

Step 4.3.3.2

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

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

Step 4.3.3.3

Factor $x$ out of $- x$ .

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

Step 4.3.3.4

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

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

Step 4.3.3.5

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

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

Step 4.3.4

Cancel the common factor of $3 y^{2} - x + 1$ and $x - 3 y^{2} - 1$ .

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Step 4.3.4.1

Factor $- 1$ out of $3 y^{2}$ .

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

Step 4.3.4.2

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

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

Step 4.3.4.3

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

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

Step 4.3.4.4

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

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

Step 4.3.4.5

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

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

Step 4.3.4.6

Rewrite $- (- 3 y^{2} + x - 1)$ as $- 1 (- 3 y^{2} + x - 1)$ .

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

Step 4.3.4.7

Reorder terms.

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

Step 4.3.4.8

Cancel the common factor.

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

Step 4.3.4.9

Rewrite the expression.

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

Step 4.3.5

Multiply $2$ by $- 1$ .

$\frac{- 2}{x}$

Step 4.3.6

Move the negative in front of the fraction.

$- \frac{2}{x}$

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

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

Step 5.2

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 5.3

Multiply $2$ by $- 1$ .

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

Step 5.4

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

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

Step 5.5

Simplify.

$μ (x, y) = e^{- 2 \ln (| x |)} + C$

Step 5.6

Simplify each term.

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Step 5.6.1

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

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

Step 5.6.2

Exponentiation and log are inverse functions.

$μ (x, y) = {| x |}^{- 2} + C$

Step 5.6.3

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

$μ (x, y) = x^{- 2} + C$

Step 5.6.4

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

$μ (x, y) = \frac{1}{x^{2}} + C$

Step 6

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

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Step 6.1

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

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

Step 6.2

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

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

Step 6.3

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

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

Step 6.4

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

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

Step 6.5

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

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Step 6.5.1

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

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

Step 6.5.2

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

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

Step 6.5.3

Factor $x$ out of $- x$ .

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

Step 6.5.4

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

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

Step 6.5.5

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

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

Step 6.6

Cancel the common factors.

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Step 6.6.1

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

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

Step 6.6.2

Cancel the common factor.

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

Step 6.6.3

Rewrite the expression.

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

Step 7

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

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

Step 8

Integrate $M (x, y) = \frac{y^{3} + y + 1}{x^{2}}$ to find $f (x, y)$ .

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Step 8.1

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

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

Step 8.2

Move $x^{2}$ out of the denominator by raising it to the $- 1$ power.

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

Step 8.3

Multiply the exponents in ${(x^{2})}^{- 1}$ .

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Step 8.3.1

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

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

Step 8.3.2

Multiply $2$ by $- 1$ .

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

Step 8.4

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

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

Step 8.5

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

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

Step 9

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

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

Step 10

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

$\frac{\partial f}{\partial y} = \frac{x - 3 y^{2} - 1}{x}$

Step 11

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

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Step 11.1

Differentiate $f$ with respect to $y$ .

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

Step 11.2

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

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

Step 11.3

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

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Step 11.3.1

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

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

Step 11.3.2

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

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

Step 11.3.3

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

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

Step 11.3.4

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

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

Step 11.3.5

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

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

Step 11.3.6

Add $3 y^{2} + 1$ and $0$ .

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

Step 11.4

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

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

Step 11.5

Simplify.

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Step 11.5.1

Apply the distributive property.

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

Step 11.5.2

Combine terms.

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Step 11.5.2.1

Multiply $3$ by $- 1$ .

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

Step 11.5.2.2

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

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

Step 11.5.2.3

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

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

Step 11.5.2.4

Move the negative in front of the fraction.

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

Step 11.5.2.5

Multiply $- 1$ by $1$ .

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

Step 11.5.3

Reorder terms.

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

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 $\frac{x - 3 y^{2} - 1}{x}$ from both sides of the equation.

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

Step 12.1.1.2

Combine the numerators over the common denominator.

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

Step 12.1.1.3

Simplify each term.

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Step 12.1.1.3.1

Apply the distributive property.

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

Step 12.1.1.3.2

Simplify.

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Step 12.1.1.3.2.1

Multiply $- 3$ by $- 1$ .

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

Step 12.1.1.3.2.2

Multiply $- 1$ by $- 1$ .

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

Step 12.1.1.4

Combine the opposite terms in $- 3 y^{2} - 1 - x + 3 y^{2} + 1$ .

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Step 12.1.1.4.1

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

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

Step 12.1.1.4.2

Add $- 1 - x$ and $0$ .

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

Step 12.1.1.4.3

Add $- 1$ and $1$ .

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

Step 12.1.1.4.4

Add $- x$ and $0$ .

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

Step 12.1.1.5

Cancel the common factor of $x$ .

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Step 12.1.1.5.1

Cancel the common factor.

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

Step 12.1.1.5.2

Divide $- 1$ by $1$ .

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

Step 12.1.2

Add $1$ to both sides of the equation.

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

Step 13

Find the antiderivative of $1$ to find $g (y)$ .

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Step 13.1

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

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

Step 13.2

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

$g (y) = \int d y$

Step 13.3

Apply the constant rule.

$g (y) = y + C$

Step 14

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

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

Step 15

Simplify each term.

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Step 15.1

Apply the distributive property.

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

Step 15.2

Simplify.

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Step 15.2.1

Rewrite using the commutative property of multiplication.

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

Step 15.2.2

Rewrite using the commutative property of multiplication.

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

Step 15.2.3

Multiply $- \frac{1}{x}$ by $1$ .

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

Step 15.3

Simplify each term.

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Step 15.3.1

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

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

Step 15.3.2

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

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