Calculus Examples

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

Step 1

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

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Step 1.1

Differentiate $M$ with respect to $y$ .

$\frac{\partial M}{\partial y} = \frac{d}{d y} [y \ln (y) - 2 x y e^{y}]$

Step 1.2

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

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

Step 1.3

Evaluate $\frac{d}{d y} [y \ln (y)]$ .

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Step 1.3.1

Differentiate using the Product Rule which states that $\frac{d}{d y} [f (y) g (y)]$ is $f (y) \frac{d}{d y} [g (y)] + g (y) \frac{d}{d y} [f (y)]$ where $f (y) = y$ and $g (y) = \ln (y)$ .

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

Step 1.3.2

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

$\frac{\partial M}{\partial y} = y \frac{1}{y} + \ln (y) \frac{d}{d y} [y] + \frac{d}{d y} [- 2 x y e^{y}]$

Step 1.3.3

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

$\frac{\partial M}{\partial y} = y \frac{1}{y} + \ln (y) \cdot 1 + \frac{d}{d y} [- 2 x y e^{y}]$

Step 1.3.4

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

$\frac{\partial M}{\partial y} = \frac{y}{y} + \ln (y) \cdot 1 + \frac{d}{d y} [- 2 x y e^{y}]$

Step 1.3.5

Cancel the common factor of $y$ .

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Step 1.3.5.1

Cancel the common factor.

$\frac{\partial M}{\partial y} = \frac{y}{y} + \ln (y) \cdot 1 + \frac{d}{d y} [- 2 x y e^{y}]$

Step 1.3.5.2

Rewrite the expression.

$\frac{\partial M}{\partial y} = 1 + \ln (y) \cdot 1 + \frac{d}{d y} [- 2 x y e^{y}]$

Step 1.3.6

Multiply $\ln (y)$ by $1$ .

$\frac{\partial M}{\partial y} = 1 + \ln (y) + \frac{d}{d y} [- 2 x y e^{y}]$

Step 1.4

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

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Step 1.4.1

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

$\frac{\partial M}{\partial y} = 1 + \ln (y) - 2 x \frac{d}{d y} [y e^{y}]$

Step 1.4.2

Differentiate using the Product Rule which states that $\frac{d}{d y} [f (y) g (y)]$ is $f (y) \frac{d}{d y} [g (y)] + g (y) \frac{d}{d y} [f (y)]$ where $f (y) = y$ and $g (y) = e^{y}$ .

$\frac{\partial M}{\partial y} = 1 + \ln (y) - 2 x (y \frac{d}{d y} [e^{y}] + e^{y} \frac{d}{d y} [y])$

Step 1.4.3

Differentiate using the Exponential Rule which states that $\frac{d}{d y} [a^{y}]$ is $a^{y} \ln (a)$ where $a$ = $e$ .

$\frac{\partial M}{\partial y} = 1 + \ln (y) - 2 x (y e^{y} + e^{y} \frac{d}{d y} [y])$

Step 1.4.4

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

$\frac{\partial M}{\partial y} = 1 + \ln (y) - 2 x (y e^{y} + e^{y} \cdot 1)$

Step 1.4.5

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

$\frac{\partial M}{\partial y} = 1 + \ln (y) - 2 x (y e^{y} + e^{y})$

Step 1.5

Simplify.

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Step 1.5.1

Apply the distributive property.

$\frac{\partial M}{\partial y} = 1 + \ln (y) - 2 x (y e^{y}) - 2 x e^{y}$

Step 1.5.2

Remove unnecessary parentheses.

$\frac{\partial M}{\partial y} = 1 + \ln (y) - 2 x y e^{y} - 2 x e^{y}$

Step 1.5.3

Reorder terms.

$\frac{\partial M}{\partial y} = - 2 e^{y} x y - 2 e^{y} x + 1 + \ln (y)$

Step 2

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

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Step 2.1

Differentiate $N$ with respect to $x$ .

$\frac{\partial N}{\partial x} = \frac{d}{d x} [x (1 - x y e^{y})]$

Step 2.2

Differentiate using the Product Rule which states that $\frac{d}{d x} [f (x) g (x)]$ is $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ where $f (x) = x$ and $g (x) = 1 - x y e^{y}$ .

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

Step 2.3

Differentiate.

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Step 2.3.1

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

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

Step 2.3.2

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

$\frac{\partial N}{\partial x} = x (0 + \frac{d}{d x} [- x y e^{y}]) + (1 - x y e^{y}) \frac{d}{d x} [x]$

Step 2.3.3

Add $0$ and $\frac{d}{d x} [- x y e^{y}]$ .

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

Step 2.3.4

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

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

Step 2.3.5

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

$\frac{\partial N}{\partial x} = x (- y e^{y} \cdot 1) + (1 - x y e^{y}) \frac{d}{d x} [x]$

Step 2.3.6

Multiply $- 1$ by $1$ .

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

Step 2.3.7

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

$\frac{\partial N}{\partial x} = x (- y e^{y}) + (1 - x y e^{y}) \cdot 1$

Step 2.3.8

Multiply $1 - x y e^{y}$ by $1$ .

$\frac{\partial N}{\partial x} = x (- y e^{y}) + 1 - x y e^{y}$

Step 2.4

Subtract $x y e^{y}$ from $x (- y e^{y})$ .

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Step 2.4.1

Reorder $x$ and $- 1$ .

$\frac{\partial N}{\partial x} = - 1 \cdot x y e^{y} - x y e^{y} + 1$

Step 2.4.2

Subtract $x y e^{y}$ from $- 1 \cdot x y e^{y}$ .

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

Step 3

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

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Step 3.1

Substitute $- 2 e^{y} x y - 2 e^{y} x + 1 + \ln (y)$ for $\frac{\partial M}{\partial y}$ and $- 2 x y e^{y} + 1$ for $\frac{\partial N}{\partial x}$ .

$- 2 e^{y} x y - 2 e^{y} x + 1 + \ln (y) = - 2 x y e^{y} + 1$

Step 3.2

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

$- 2 e^{y} x y - 2 e^{y} x + 1 + \ln (y) = - 2 x y e^{y} + 1$ 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 $- 2 x y e^{y} + 1$ for $\frac{\partial N}{\partial x}$ .

$\frac{- 2 x y e^{y} + 1 - \frac{\partial M}{\partial y}}{M}$

Step 4.2

Substitute $- 2 e^{y} x y - 2 e^{y} x + 1 + \ln (y)$ for $\frac{\partial M}{\partial y}$ .

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

Step 4.3

Substitute $y \ln (y) - 2 x y e^{y}$ for $M$ .

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Step 4.3.1

Substitute $y \ln (y) - 2 x y e^{y}$ for $M$ .

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

Step 4.3.2

Simplify the numerator.

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Step 4.3.2.1

Apply the distributive property.

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

Step 4.3.2.2

Simplify.

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Step 4.3.2.2.1

Multiply $- 2$ by $- 1$ .

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

Step 4.3.2.2.2

Multiply $- 2$ by $- 1$ .

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

Step 4.3.2.2.3

Multiply $- 1$ by $1$ .

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

Step 4.3.2.3

Remove parentheses.

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

Step 4.3.2.4

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

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Step 4.3.2.4.1

Move $e^{y}$ .

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

Step 4.3.2.4.2

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

$\frac{0 + 1 + 2 e^{y} x - 1 - \ln (y)}{y \ln (y) - 2 x y e^{y}}$

Step 4.3.2.5

Add $0$ and $1$ .

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

Step 4.3.2.6

Subtract $1$ from $1$ .

$\frac{2 e^{y} x + 0 - \ln (y)}{y \ln (y) - 2 x y e^{y}}$

Step 4.3.2.7

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

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

Step 4.3.3

Factor $y$ out of $y \ln (y) - 2 x y e^{y}$ .

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Step 4.3.3.1

Factor $y$ out of $y \ln (y)$ .

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

Step 4.3.3.2

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

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

Step 4.3.3.3

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

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

Step 4.3.4

Cancel the common factor of $2 e^{y} x - \ln (y)$ and $\ln (y) - 2 x e^{y}$ .

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Step 4.3.4.1

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

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

Step 4.3.4.2

Factor $- 1$ out of $- \ln (y)$ .

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

Step 4.3.4.3

Factor $- 1$ out of $- (- 2 e^{y} x) - (\ln (y))$ .

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

Step 4.3.4.4

Rewrite $- (- 2 e^{y} x + \ln (y))$ as $- 1 (- 2 e^{y} x + \ln (y))$ .

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

Step 4.3.4.5

Reorder terms.

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

Step 4.3.4.6

Cancel the common factor.

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

Step 4.3.4.7

Rewrite the expression.

$\frac{- 1}{y}$

Step 4.3.5

Substitute $y \ln (y) - 2 x y e^{y}$ for $M$ .

$- \frac{1}{y}$

Step 4.4

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

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

Step 5

Evaluate the integral $e^{\int - \frac{1}{y} d y}$ .

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Step 5.1

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

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

Step 5.2

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

$μ (x, y) = e^{- (\ln (| y |) + C)}$

Step 5.3

Simplify.

$μ (x, y) = e^{- \ln (| y |)} + C$

Step 5.4

Simplify each term.

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Step 5.4.1

Simplify $- \ln (| y |)$ by moving $- 1$ inside the logarithm.

$μ (x, y) = e^{\ln ({| y |}^{- 1})} + C$

Step 5.4.2

Exponentiation and log are inverse functions.

$μ (x, y) = {| y |}^{- 1} + C$

Step 5.4.3

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

$μ (x, y) = \frac{1}{| y |} + C$

Step 6

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

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Step 6.1

Multiply $y \ln (y) - 2 x y e^{y}$ by $\frac{1}{y}$ .

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

Step 6.2

Multiply $y \ln (y) - 2 x y e^{y}$ by $\frac{1}{y}$ .

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

Step 6.3

Factor $y$ out of $y \ln (y) - 2 x y e^{y}$ .

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Step 6.3.1

Factor $y$ out of $y \ln (y)$ .

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

Step 6.3.2

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

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

Step 6.3.3

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

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

Step 6.4

Cancel the common factor of $y$ .

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Step 6.4.1

Cancel the common factor.

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

Step 6.4.2

Divide $\ln (y) - 2 x e^{y}$ by $1$ .

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

Step 6.5

Multiply $x (1 - x y e^{y})$ by $\frac{1}{y}$ .

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

Step 6.6

Apply the distributive property.

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

Step 6.7

Multiply $x$ by $1$ .

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

Step 6.8

Rewrite using the commutative property of multiplication.

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

Step 6.9

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

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Step 6.9.1

Move $x$ .

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

Step 6.9.2

Multiply $x$ by $x$ .

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

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

Step 6.10

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

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

Step 6.11

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

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Step 6.11.1

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

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

Step 6.11.2

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

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

Step 6.11.3

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

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

Step 6.11.4

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

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

Step 7

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

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

Step 8

Integrate $M (x, y) = \ln (y) - 2 x e^{y}$ to find $f (x, y)$ .

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Step 8.1

Split the single integral into multiple integrals.

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

Step 8.2

Apply the constant rule.

$f (x, y) = \ln (y) x + C + \int - 2 x e^{y} d x$

Step 8.3

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

$f (x, y) = \ln (y) x + C - 2 e^{y} \int x d x$

Step 8.4

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

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

Step 8.5

Simplify.

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

Step 8.6

Simplify.

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Step 8.6.1

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

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

Step 8.6.2

Cancel the common factor of $- 2$ and $2$ .

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Step 8.6.2.1

Factor $2$ out of $- 2$ .

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

Step 8.6.2.2

Cancel the common factors.

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Step 8.6.2.2.1

Factor $2$ out of $2$ .

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

Step 8.6.2.2.2

Cancel the common factor.

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

Step 8.6.2.2.3

Rewrite the expression.

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

Step 8.6.2.2.4

Divide $- 1$ by $1$ .

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

Step 9

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

$f (x, y) = \ln (y) x - e^{y} x^{2} + g (y)$

Step 10

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

$\frac{\partial f}{\partial y} = \frac{x (1 - x y e^{y})}{y}$

Step 11

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

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Step 11.1

Differentiate $f$ with respect to $y$ .

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

Step 11.2

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

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

Step 11.3

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

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Step 11.3.1

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

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

Step 11.3.2

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

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

Step 11.3.3

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

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

Step 11.4

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

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Step 11.4.1

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

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

Step 11.4.2

Differentiate using the Exponential Rule which states that $\frac{d}{d y} [a^{y}]$ is $a^{y} \ln (a)$ where $a$ = $e$ .

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

Step 11.5

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

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

Step 11.6

Simplify.

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Step 11.6.1

Reorder terms.

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

Step 11.6.2

Reorder factors in $\frac{d g}{d y} - e^{y} x^{2} + \frac{x}{y}$ .

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

Step 12

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

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Step 12.1

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

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Step 12.1.1

Subtract $\frac{x (1 - x y e^{y})}{y}$ from both sides of the equation.

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

Step 12.1.2

Combine the numerators over the common denominator.

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

Step 12.1.3

Simplify each term.

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Step 12.1.3.1

Apply the distributive property.

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

Step 12.1.3.2

Multiply $- 1$ by $1$ .

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

Step 12.1.3.3

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

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Step 12.1.3.3.1

Move $x$ .

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

Step 12.1.3.3.2

Multiply $x$ by $x$ .

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

Step 12.1.3.4

Multiply $- x^{2} (- y e^{y})$ .

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Step 12.1.3.4.1

Multiply $- 1$ by $- 1$ .

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

Step 12.1.3.4.2

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

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

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

Step 12.1.4

Subtract $x$ from $x$ .

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

Step 12.1.5

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

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

Step 12.1.6

Cancel the common factor of $y$ .

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Step 12.1.6.1

Cancel the common factor.

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

Step 12.1.6.2

Divide $x^{2} e^{y}$ by $1$ .

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

Step 12.1.7

Combine the opposite terms in $\frac{d g}{d y} - x^{2} e^{y} + x^{2} e^{y}$ .

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Step 12.1.7.1

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

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

Step 12.1.7.2

Add $\frac{d g}{d y}$ and $0$ .

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

Step 13

Find the antiderivative of $0$ to find $g (y)$ .

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Step 13.1

Integrate both sides of $\frac{d g}{d y} = 0$ .

$\int \frac{d g}{d y} d y = \int 0 d y$

Step 13.2

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

$g (y) = \int 0 d y$

Step 13.3

The integral of $0$ with respect to $y$ is $0$ .

$g (y) = 0 + C$

Step 13.4

Add $0$ and $C$ .

$g (y) = C$

Step 14

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

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

Step 15

Reorder factors in $f (x, y) = \ln (y) x - e^{y} x^{2} + C$ .

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