Calculus Examples

$y d x + (2 x + 1 - x y) d y = 0$

Step 1

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

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Step 1.1

Differentiate $M$ with respect to $y$ .

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

Step 1.2

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

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

Step 2

Find $\frac{\partial N}{\partial x}$ where $N (x, y) = 2 x + 1 - x y$ .

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Step 2.1

Differentiate $N$ with respect to $x$ .

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

Step 2.2

By the Sum Rule, the derivative of $2 x + 1 - x y$ with respect to $x$ is $\frac{d}{d x} [2 x] + \frac{d}{d x} [1] + \frac{d}{d x} [- x y]$ .

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

Step 2.3

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

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Step 2.3.1

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

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

Step 2.3.2

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

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

Step 2.3.3

Multiply $2$ by $1$ .

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

Step 2.4

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

$\frac{\partial N}{\partial x} = 2 + 0 + \frac{d}{d x} [- x y]$

Step 2.5

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

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Step 2.5.1

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

$\frac{\partial N}{\partial x} = 2 + 0 - y \frac{d}{d x} [x]$

Step 2.5.2

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

$\frac{\partial N}{\partial x} = 2 + 0 - y \cdot 1$

Step 2.5.3

Multiply $- 1$ by $1$ .

$\frac{\partial N}{\partial x} = 2 + 0 - y$

Step 2.6

Simplify.

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Step 2.6.1

Add $2$ and $0$ .

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

Step 2.6.2

Reorder terms.

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

Step 3

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

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Step 3.1

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

$1 = - y + 2$

Step 3.2

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

$1 = - y + 2$ 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 $- y + 2$ for $\frac{\partial N}{\partial x}$ .

$\frac{- y + 2 - \frac{\partial M}{\partial y}}{M}$

Step 4.2

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

$\frac{- y + 2 - 1}{M}$

Step 4.3

Substitute $y$ for $M$ .

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Step 4.3.1

Substitute $y$ for $M$ .

$\frac{- y + 2 - 1}{y}$

Step 4.3.2

Subtract $1$ from $2$ .

$\frac{- y + 1}{y}$

Step 4.3.3

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

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

Step 4.3.4

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

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

Step 4.3.5

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

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

Step 4.3.6

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

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

Step 4.3.7

Substitute $y$ for $M$ .

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

Step 4.4

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

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

Step 5

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

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Step 5.1

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

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

Step 5.2

Divide $y - 1$ by $y$ .

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Step 5.2.1

Set up the polynomials to be divided. If there is not a term for every exponent, insert one with a value of $0$ .

$y$

$0$

$y$

$1$

Step 5.2.2

Divide the highest order term in the dividend $y$ by the highest order term in divisor $y$ .

					$1$
	$y$	+	$0$		$y$	-	$1$

Step 5.2.3

Multiply the new quotient term by the divisor.

				$1$
$y$	+	$0$		$y$	-	$1$
			+	$y$	+	$0$

Step 5.2.4

The expression needs to be subtracted from the dividend, so change all the signs in $y + 0$

				$1$
$y$	+	$0$		$y$	-	$1$
			-	$y$	-	$0$

Step 5.2.5

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

				$1$
$y$	+	$0$		$y$	-	$1$
			-	$y$	-	$0$
					-	$1$

Step 5.2.6

The final answer is the quotient plus the remainder over the divisor.

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

Step 5.3

Split the single integral into multiple integrals.

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

Step 5.4

Apply the constant rule.

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

Step 5.5

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

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

Step 5.6

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

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

Step 5.7

Simplify.

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

Step 6

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

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Step 6.1

Multiply $y$ by $e^{- y + \ln (y)}$ .

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

Step 6.2

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

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

Step 6.3

Apply the distributive property.

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

Step 6.4

Multiply $e^{- y + \ln (y)}$ by $1$ .

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

Step 7

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

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

Step 8

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

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Step 8.1

Apply the constant rule.

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

Step 9

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

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

Step 10

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

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

Step 11

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

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Step 11.1

Differentiate $f$ with respect to $y$ .

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

Step 11.2

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

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

Step 11.3

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

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Step 11.3.1

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

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

Step 11.3.2

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 $g (y) = e^{- y + \ln (y)}$ .

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

Step 11.3.3

Differentiate using the chain rule, which states that $\frac{d}{d y} [f (g (y))]$ is $f' (g (y)) g' (y)$ where $f (y) = e^{y}$ and $g (y) = - y + \ln (y)$ .

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Step 11.3.3.1

To apply the Chain Rule, set $u$ as $- y + \ln (y)$ .

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

Step 11.3.3.2

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

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

Step 11.3.3.3

Replace all occurrences of $u$ with $- y + \ln (y)$ .

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

Step 11.3.4

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

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

Step 11.3.5

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

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

Step 11.3.6

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

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

Step 11.3.7

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

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

Step 11.3.8

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

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

Step 11.3.9

Multiply $- 1$ by $1$ .

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

Step 11.3.10

Multiply $e^{- y + \ln (y)}$ by $1$ .

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

Step 11.4

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

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

Step 11.5

Simplify.

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Step 11.5.1

Apply the distributive property.

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

Step 11.5.2

Reorder terms.

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

Step 11.5.3

Simplify each term.

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Step 11.5.3.1

Apply the distributive property.

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

Step 11.5.3.2

Move $- 1$ to the left of $e^{- y + \ln (y)} x y$ .

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

Step 11.5.3.3

Cancel the common factor of $y$ .

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Step 11.5.3.3.1

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

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

Step 11.5.3.3.2

Cancel the common factor.

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

Step 11.5.3.3.3

Rewrite the expression.

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

Step 11.5.3.4

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

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

Step 11.5.4

Add $e^{- y + \ln (y)} x$ and $e^{- y + \ln (y)} x$ .

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

Step 11.5.5

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

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

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 the terms containing a logarithm to the left side of the equation.

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

Step 12.1.2

Combine the opposite terms in $- x y e^{- y + \ln (y)} + 2 x e^{- y + \ln (y)} - 2 x e^{- y + \ln (y)} - e^{- y + \ln (y)} + x y e^{- y + \ln (y)}$ .

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Step 12.1.2.1

Subtract $2 x e^{- y + \ln (y)}$ from $2 x e^{- y + \ln (y)}$ .

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

Step 12.1.2.2

Add $- x y e^{- y + \ln (y)}$ and $0$ .

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

Step 12.1.2.3

Add $- x y e^{- y + \ln (y)}$ and $x y e^{- y + \ln (y)}$ .

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

Step 12.1.2.4

Subtract $e^{- y + \ln (y)}$ from $0$ .

$- e^{- y + \ln (y)} = - \frac{d g}{d y}$

Step 12.1.3

Since $\frac{d g}{d y}$ is on the right side of the equation, switch the sides so it is on the left side of the equation.

$- \frac{d g}{d y} = - e^{- y + \ln (y)}$

Step 12.1.4

Divide each term in $- \frac{d g}{d y} = - e^{- y + \ln (y)}$ by $- 1$ and simplify.

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Step 12.1.4.1

Divide each term in $- \frac{d g}{d y} = - e^{- y + \ln (y)}$ by $- 1$ .

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

Step 12.1.4.2

Simplify the left side.

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Step 12.1.4.2.1

Dividing two negative values results in a positive value.

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

Step 12.1.4.2.2

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

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

Step 12.1.4.3

Simplify the right side.

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Step 12.1.4.3.1

Dividing two negative values results in a positive value.

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

Step 12.1.4.3.2

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

$\frac{d g}{d y} = e^{- y + \ln (y)}$

Step 13

Find the antiderivative of $e^{- y + \ln (y)}$ to find $g (y)$ .

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Step 13.1

Integrate both sides of $\frac{d g}{d y} = e^{- y + \ln (y)}$ .

$\int \frac{d g}{d y} d y = \int e^{- y + \ln (y)} d y$

Step 13.2

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

$g (y) = \int e^{- y + \ln (y)} d y$

Step 13.3

Rewrite $\int e^{- y + \ln (y)} d y$ as $\int e^{- y} e^{\ln (y)} d y$ .

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

Step 13.4

Rewrite $e^{\ln (y)}$ as $y$ .

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

Step 13.5

Integrate by parts using the formula $\int u d v = u v - \int v d u$ , where $u = y$ and $d v = e^{- y}$ .

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

Step 13.6

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

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

Step 13.7

Simplify.

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Step 13.7.1

Multiply $- 1$ by $- 1$ .

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

Step 13.7.2

Multiply $\int e^{- y} d y$ by $1$ .

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

Step 13.8

Let $u = - y$ . Then $d u = - d y$ , so $- d u = d y$ . Rewrite using $u$ and $d$ $u$ .

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Step 13.8.1

Let $u = - y$ . Find $\frac{d u}{d y}$ .

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Step 13.8.1.1

Differentiate $- y$ .

$\frac{d}{d y} [- y]$

Step 13.8.1.2

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

$- \frac{d}{d y} [y]$

Step 13.8.1.3

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

$- 1 \cdot 1$

Step 13.8.1.4

Multiply $- 1$ by $1$ .

$- 1$

Step 13.8.2

Rewrite the problem using $u$ and $d u$ .

$g (y) = y (- e^{- y}) + \int - e^{u} d u$

Step 13.9

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

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

Step 13.10

The integral of $e^{u}$ with respect to $u$ is $e^{u}$ .

$g (y) = y (- e^{- y}) - (e^{u} + C)$

Step 13.11

Rewrite $y (- e^{- y}) - (e^{u} + C)$ as $- y e^{- y} - e^{u} + C$ .

$g (y) = - y e^{- y} - e^{u} + C$

Step 13.12

Replace all occurrences of $u$ with $- y$ .

$g (y) = - y e^{- y} - e^{- y} + C$

Step 14

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

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

Step 15

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

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