Calculus Examples

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

Step 1

Find $\frac{\partial M}{\partial y}$ where $M (x, y) = y^{4} + 2 y$ .

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Step 1.1

Differentiate $M$ with respect to $y$ .

$\frac{\partial M}{\partial y} = \frac{d}{d y} [y^{4} + 2 y]$

Step 1.2

Differentiate.

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Step 1.2.1

By the Sum Rule, the derivative of $y^{4} + 2 y$ with respect to $y$ is $\frac{d}{d y} [y^{4}] + \frac{d}{d y} [2 y]$ .

$\frac{\partial M}{\partial y} = \frac{d}{d y} [y^{4}] + \frac{d}{d y} [2 y]$

Step 1.2.2

Differentiate using the Power Rule which states that $\frac{d}{d y} [y^{n}]$ is $n y^{n - 1}$ where $n = 4$ .

$\frac{\partial M}{\partial y} = 4 y^{3} + \frac{d}{d y} [2 y]$

Step 1.3

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

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Step 1.3.1

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

$\frac{\partial M}{\partial y} = 4 y^{3} + 2 \frac{d}{d y} [y]$

Step 1.3.2

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

$\frac{\partial M}{\partial y} = 4 y^{3} + 2 \cdot 1$

Step 1.3.3

Multiply $2$ by $1$ .

$\frac{\partial M}{\partial y} = 4 y^{3} + 2$

Step 2

Find $\frac{\partial N}{\partial x}$ where $N (x, y) = x y^{3} + 2 y^{4} - 4 x$ .

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Step 2.1

Differentiate $N$ with respect to $x$ .

$\frac{\partial N}{\partial x} = \frac{d}{d x} [x y^{3} + 2 y^{4} - 4 x]$

Step 2.2

By the Sum Rule, the derivative of $x y^{3} + 2 y^{4} - 4 x$ with respect to $x$ is $\frac{d}{d x} [x y^{3}] + \frac{d}{d x} [2 y^{4}] + \frac{d}{d x} [- 4 x]$ .

$\frac{\partial N}{\partial x} = \frac{d}{d x} [x y^{3}] + \frac{d}{d x} [2 y^{4}] + \frac{d}{d x} [- 4 x]$

Step 2.3

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

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Step 2.3.1

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

$\frac{\partial N}{\partial x} = y^{3} \frac{d}{d x} [x] + \frac{d}{d x} [2 y^{4}] + \frac{d}{d x} [- 4 x]$

Step 2.3.2

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

$\frac{\partial N}{\partial x} = y^{3} \cdot 1 + \frac{d}{d x} [2 y^{4}] + \frac{d}{d x} [- 4 x]$

Step 2.3.3

Multiply $y^{3}$ by $1$ .

$\frac{\partial N}{\partial x} = y^{3} + \frac{d}{d x} [2 y^{4}] + \frac{d}{d x} [- 4 x]$

Step 2.4

Since $2 y^{4}$ is constant with respect to $x$ , the derivative of $2 y^{4}$ with respect to $x$ is $0$ .

$\frac{\partial N}{\partial x} = y^{3} + 0 + \frac{d}{d x} [- 4 x]$

Step 2.5

Evaluate $\frac{d}{d x} [- 4 x]$ .

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Step 2.5.1

Since $- 4$ is constant with respect to $x$ , the derivative of $- 4 x$ with respect to $x$ is $- 4 \frac{d}{d x} [x]$ .

$\frac{\partial N}{\partial x} = y^{3} + 0 - 4 \frac{d}{d x} [x]$

Step 2.5.2

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

$\frac{\partial N}{\partial x} = y^{3} + 0 - 4 \cdot 1$

Step 2.5.3

Multiply $- 4$ by $1$ .

$\frac{\partial N}{\partial x} = y^{3} + 0 - 4$

Step 2.6

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

$\frac{\partial N}{\partial x} = y^{3} - 4$

Step 3

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

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Step 3.1

Substitute $4 y^{3} + 2$ for $\frac{\partial M}{\partial y}$ and $y^{3} - 4$ for $\frac{\partial N}{\partial x}$ .

$4 y^{3} + 2 = y^{3} - 4$

Step 3.2

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

$4 y^{3} + 2 = y^{3} - 4$ 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 $y^{3} - 4$ for $\frac{\partial N}{\partial x}$ .

$\frac{y^{3} - 4 - \frac{\partial M}{\partial y}}{M}$

Step 4.2

Substitute $4 y^{3} + 2$ for $\frac{\partial M}{\partial y}$ .

$\frac{y^{3} - 4 - (4 y^{3} + 2)}{M}$

Step 4.3

Substitute $y^{4} + 2 y$ for $M$ .

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Step 4.3.1

Substitute $y^{4} + 2 y$ for $M$ .

$\frac{y^{3} - 4 - (4 y^{3} + 2)}{y^{4} + 2 y}$

Step 4.3.2

Simplify the numerator.

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Step 4.3.2.1

Apply the distributive property.

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

Step 4.3.2.2

Multiply $4$ by $- 1$ .

$\frac{y^{3} - 4 - 4 y^{3} - 1 \cdot 2}{y^{4} + 2 y}$

Step 4.3.2.3

Multiply $- 1$ by $2$ .

$\frac{y^{3} - 4 - 4 y^{3} - 2}{y^{4} + 2 y}$

Step 4.3.2.4

Subtract $4 y^{3}$ from $y^{3}$ .

$\frac{- 3 y^{3} - 4 - 2}{y^{4} + 2 y}$

Step 4.3.2.5

Subtract $2$ from $- 4$ .

$\frac{- 3 y^{3} - 6}{y^{4} + 2 y}$

Step 4.3.2.6

Factor $3$ out of $- 3 y^{3} - 6$ .

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Step 4.3.2.6.1

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

$\frac{3 (- y^{3}) - 6}{y^{4} + 2 y}$

Step 4.3.2.6.2

Factor $3$ out of $- 6$ .

$\frac{3 (- y^{3}) + 3 (- 2)}{y^{4} + 2 y}$

Step 4.3.2.6.3

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

$\frac{3 (- y^{3} - 2)}{y^{4} + 2 y}$

Step 4.3.3

Factor $y$ out of $y^{4} + 2 y$ .

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Step 4.3.3.1

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

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

Step 4.3.3.2

Factor $y$ out of $2 y$ .

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

Step 4.3.3.3

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

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

Step 4.3.4

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

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Step 4.3.4.1

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

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

Step 4.3.4.2

Rewrite $- 2$ as $- 1 (2)$ .

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

Step 4.3.4.3

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

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

Step 4.3.4.4

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

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

Step 4.3.4.5

Cancel the common factor.

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

Step 4.3.4.6

Rewrite the expression.

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

Step 4.3.5

Multiply $3$ by $- 1$ .

$\frac{- 3}{y}$

Step 4.3.6

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

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Step 6.1

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

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

Step 6.2

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

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

Step 6.3

Factor $y$ out of $y^{4} + 2 y$ .

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Step 6.3.1

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

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

Step 6.3.2

Factor $y$ out of $2 y$ .

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

Step 6.3.3

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

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

Step 6.4

Cancel the common factors.

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Step 6.4.1

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

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

Step 6.4.2

Cancel the common factor.

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

Step 6.4.3

Rewrite the expression.

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

Step 6.5

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

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

Step 6.6

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

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

Step 7

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

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

Step 8

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

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Step 8.1

Apply the constant rule.

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

Step 8.2

Simplify the answer.

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Step 8.2.1

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

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

Step 8.2.2

Rewrite $\frac{(y^{3} + 2) x}{y^{2}} + C$ as $\frac{y^{3} x + 2 x}{y^{2}} + C$ .

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

Step 9

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

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

Step 10

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

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

Step 11

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

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Step 11.1

Differentiate $f$ with respect to $y$ .

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

Step 11.2

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

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

Step 11.3

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

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Step 11.3.1

Differentiate using the Quotient Rule which states that $\frac{d}{d y} [\frac{f (y)}{g (y)}]$ is $\frac{g (y) \frac{d}{d y} [f (y)] - f (y) \frac{d}{d y} [g (y)]}{{g (y)}^{2}}$ where $f (y) = y^{3} x + 2 x$ and $g (y) = y^{2}$ .

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

Step 11.3.2

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

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

Step 11.3.3

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

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

Step 11.3.4

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

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

Step 11.3.5

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

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

Step 11.3.6

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

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

Step 11.3.7

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

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

Step 11.3.8

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

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

Step 11.3.9

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

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Step 11.3.9.1

Move $y^{2}$ .

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

Step 11.3.9.2

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

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

Step 11.3.9.3

Add $2$ and $2$ .

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

Step 11.3.10

Move $3$ to the left of $y^{4}$ .

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

Step 11.3.11

Multiply $2$ by $- 1$ .

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

Step 11.3.12

Multiply the exponents in ${(y^{2})}^{2}$ .

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Step 11.3.12.1

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

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

Step 11.3.12.2

Multiply $2$ by $2$ .

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

Step 11.3.13

Factor $y$ out of $3 y^{4} x - 2 (y^{3} x + 2 x) y$ .

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Step 11.3.13.1

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

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

Step 11.3.13.2

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

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

Step 11.3.13.3

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

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

Step 11.3.14

Cancel the common factors.

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Step 11.3.14.1

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

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

Step 11.3.14.2

Cancel the common factor.

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

Step 11.3.14.3

Rewrite the expression.

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

Step 11.4

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

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

Step 11.5

Simplify.

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Step 11.5.1

Apply the distributive property.

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

Step 11.5.2

Combine terms.

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Step 11.5.2.1

Multiply $2$ by $- 2$ .

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

Step 11.5.2.2

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

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

Step 11.5.2.3

Multiply $y^{3}$ by $1$ .

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

Step 11.5.2.4

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

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

Step 11.5.2.5

Combine the numerators over the common denominator.

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

Step 11.5.3

Reorder terms.

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

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

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

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

Step 12.1.2

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

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Step 12.1.2.1

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

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

Step 12.1.2.2

Add $4 x$ to both sides of the equation.

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

Step 12.1.2.3

Combine the opposite terms in $x y^{3} + 2 y^{4} - 4 x - y^{3} x + 4 x$ .

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Step 12.1.2.3.1

Reorder the factors in the terms $x y^{3}$ and $- y^{3} x$ .

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

Step 12.1.2.3.2

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

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

Step 12.1.2.3.3

Add $2 y^{4} - 4 x$ and $0$ .

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

Step 12.1.2.3.4

Add $- 4 x$ and $4 x$ .

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

Step 12.1.2.3.5

Add $2 y^{4}$ and $0$ .

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

Step 12.1.3

Divide each term in $y^{3} \frac{d g}{d y} = 2 y^{4}$ by $y^{3}$ and simplify.

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Step 12.1.3.1

Divide each term in $y^{3} \frac{d g}{d y} = 2 y^{4}$ by $y^{3}$ .

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

Step 12.1.3.2

Simplify the left side.

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Step 12.1.3.2.1

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

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Step 12.1.3.2.1.1

Cancel the common factor.

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

Step 12.1.3.2.1.2

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

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

Step 12.1.3.3

Simplify the right side.

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Step 12.1.3.3.1

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

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Step 12.1.3.3.1.1

Factor $y^{3}$ out of $2 y^{4}$ .

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

Step 12.1.3.3.1.2

Cancel the common factors.

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Step 12.1.3.3.1.2.1

Multiply by $1$ .

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

Step 12.1.3.3.1.2.2

Cancel the common factor.

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

Step 12.1.3.3.1.2.3

Rewrite the expression.

$\frac{d g}{d y} = \frac{2 y}{1}$

Step 12.1.3.3.1.2.4

Divide $2 y$ by $1$ .

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

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

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

Step 15

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

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Step 15.1

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

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Step 15.1.1

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

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

Step 15.1.2

Factor $x$ out of $2 x$ .

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

Step 15.1.3

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

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

Step 15.2

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

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

Step 15.3

Combine the numerators over the common denominator.

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

Step 15.4

Simplify the numerator.

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Step 15.4.1

Apply the distributive property.

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

Step 15.4.2

Move $2$ to the left of $x$ .

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

Step 15.4.3

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

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Step 15.4.3.1

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

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

Step 15.4.3.2

Add $2$ and $2$ .

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