Calculus Examples

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

Step 1

Rewrite the differential equation to fit the Exact differential equation technique.

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Step 1.1

Subtract $(e^{y} + 2 x y - 2 x) d y$ from both sides of the equation.

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

Step 2

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

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Step 2.1

Differentiate $M$ with respect to $y$ .

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

Step 2.2

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 3

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

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Step 3.1

Differentiate $N$ with respect to $x$ .

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

Step 3.2

Since $- 1$ is constant with respect to $x$ , the derivative of $- (e^{y} + 2 x y - 2 x)$ with respect to $x$ is $- \frac{d}{d x} [e^{y} + 2 x y - 2 x]$ .

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

Step 3.3

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

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

Step 3.4

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

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

Step 3.5

Add $0$ and $\frac{d}{d x} [2 x y]$ .

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

Step 3.6

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

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

Step 3.7

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

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

Step 3.8

Multiply $2$ by $1$ .

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

Step 3.9

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

$\frac{\partial N}{\partial x} = - (2 y - 2 \frac{d}{d x} [x])$

Step 3.10

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

$\frac{\partial N}{\partial x} = - (2 y - 2 \cdot 1)$

Step 3.11

Multiply $- 2$ by $1$ .

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

Step 3.12

Simplify.

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Step 3.12.1

Apply the distributive property.

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

Step 3.12.2

Combine terms.

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Step 3.12.2.1

Multiply $2$ by $- 1$ .

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

Step 3.12.2.2

Multiply $- 1$ by $- 2$ .

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

Step 4

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

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Step 4.1

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

$1 = - 2 y + 2$

Step 4.2

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

$1 = - 2 y + 2$ is not an identity.

Step 5

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 5.1

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

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

Step 5.2

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

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

Step 5.3

Substitute $y$ for $M$ .

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Step 5.3.1

Substitute $y$ for $M$ .

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

Step 5.3.2

Subtract $1$ from $2$ .

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

Step 5.3.3

Factor $- 1$ out of $- 2 y$ .

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

Step 5.3.4

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

$\frac{- (2 y) - 1 (- 1)}{y}$

Step 5.3.5

Factor $- 1$ out of $- (2 y) - 1 (- 1)$ .

$\frac{- (2 y - 1)}{y}$

Step 5.3.6

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

$\frac{- 1 (2 y - 1)}{y}$

Step 5.3.7

Substitute $y$ for $M$ .

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

Step 5.4

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

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

Step 6

Evaluate the integral $e^{\int - \frac{2 y - 1}{y} d y}$ .

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Step 6.1

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

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

Step 6.2

Divide $2 y - 1$ by $y$ .

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Step 6.2.1

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

$y$

$0$

$2 y$

$1$

Step 6.2.2

Divide the highest order term in the dividend $2 y$ by the highest order term in divisor $y$ .

					$2$
	$y$	+	$0$		$2 y$	-	$1$

Step 6.2.3

Multiply the new quotient term by the divisor.

				$2$
$y$	+	$0$		$2 y$	-	$1$
			+	$2 y$	+	$0$

Step 6.2.4

The expression needs to be subtracted from the dividend, so change all the signs in $2 y + 0$

				$2$
$y$	+	$0$		$2 y$	-	$1$
			-	$2 y$	-	$0$

Step 6.2.5

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

				$2$
$y$	+	$0$		$2 y$	-	$1$
			-	$2 y$	-	$0$
					-	$1$

Step 6.2.6

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

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

Step 6.3

Split the single integral into multiple integrals.

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

Step 6.4

Apply the constant rule.

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

Step 6.5

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

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

Step 6.6

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

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

Step 6.7

Simplify.

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

Step 7

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

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Step 7.1

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

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

Step 7.2

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

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

Step 7.3

Apply the distributive property.

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

Step 7.4

Simplify.

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Step 7.4.1

Multiply $2$ by $- 1$ .

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

Step 7.4.2

Multiply $- 2$ by $- 1$ .

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

Step 7.5

Apply the distributive property.

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

Step 7.6

Multiply $e^{y}$ by $e^{- 2 y + \ln (y)}$ by adding the exponents.

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Step 7.6.1

Move $e^{- 2 y + \ln (y)}$ .

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

Step 7.6.2

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

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

Step 7.6.3

Add $- 2 y$ and $y$ .

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

Step 8

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

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

Step 9

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

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Step 9.1

Apply the constant rule.

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

Step 10

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

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

Step 11

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

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

Step 12

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

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Step 12.1

Differentiate $f$ with respect to $y$ .

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

Step 12.2

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

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

Step 12.3

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

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Step 12.3.1

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

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

Step 12.3.2

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 $g (y) = e^{- 2 y + \ln (y)}$ .

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

Step 12.3.3

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 $g (y) = - 2 y + \ln (y)$ .

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Step 12.3.3.1

To apply the Chain Rule, set $u$ as $- 2 y + \ln (y)$ .

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

Step 12.3.3.2

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

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

Step 12.3.3.3

Replace all occurrences of $u$ with $- 2 y + \ln (y)$ .

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

Step 12.3.4

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

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

Step 12.3.5

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

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

Step 12.3.6

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

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

Step 12.3.7

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

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

Step 12.3.8

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

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

Step 12.3.9

Multiply $- 2$ by $1$ .

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

Step 12.3.10

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

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

Step 12.4

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

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

Step 12.5

Simplify.

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Step 12.5.1

Apply the distributive property.

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

Step 12.5.2

Reorder terms.

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

Step 12.5.3

Simplify each term.

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Step 12.5.3.1

Apply the distributive property.

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

Step 12.5.3.2

Move $- 2$ to the left of $e^{- 2 y + \ln (y)} x y$ .

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

Step 12.5.3.3

Cancel the common factor of $y$ .

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Step 12.5.3.3.1

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

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

Step 12.5.3.3.2

Cancel the common factor.

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

Step 12.5.3.3.3

Rewrite the expression.

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

Step 12.5.3.4

Remove parentheses.

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

Step 12.5.4

Add $e^{- 2 y + \ln (y)} x$ and $e^{- 2 y + \ln (y)} x$ .

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

Step 12.5.5

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

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

Step 13

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

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Step 13.1

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

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Step 13.1.1

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

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

Step 13.1.2

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

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Step 13.1.2.1

Add $- 2 x y e^{- 2 y + \ln (y)}$ and $2 x y e^{- 2 y + \ln (y)}$ .

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

Step 13.1.2.2

Add $2 x e^{- 2 y + \ln (y)} + e^{- y + \ln (y)}$ and $0$ .

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

Step 13.1.2.3

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

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

Step 13.1.2.4

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

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

Step 13.1.3

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 13.1.4

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

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Step 13.1.4.1

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 13.1.4.2

Simplify the left side.

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Step 13.1.4.2.1

Dividing two negative values results in a positive value.

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

Step 13.1.4.2.2

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

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

Step 13.1.4.3

Simplify the right side.

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Step 13.1.4.3.1

Move the negative one from the denominator of $\frac{e^{- y + \ln (y)}}{- 1}$ .

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

Step 13.1.4.3.2

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

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

Step 14

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

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Step 14.1

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 14.2

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

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

Step 14.3

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

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

Step 14.4

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 14.5

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

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

Step 14.6

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 14.7

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 14.8

Simplify.

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Step 14.8.1

Multiply $- 1$ by $- 1$ .

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

Step 14.8.2

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

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

Step 14.9

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

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Step 14.9.1

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

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Step 14.9.1.1

Differentiate $- y$ .

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

Step 14.9.1.2

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 14.9.1.3

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 14.9.1.4

Multiply $- 1$ by $1$ .

$- 1$

Step 14.9.2

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

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

Step 14.10

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 14.11

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

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

Step 14.12

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

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

Step 14.13

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

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

Step 14.14

Simplify.

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Step 14.14.1

Apply the distributive property.

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

Step 14.14.2

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

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Step 14.14.2.1

Multiply $- 1$ by $- 1$ .

$g (y) = 1 (y e^{- y}) - - e^{- y} + C$

Step 14.14.2.2

Multiply $y$ by $1$ .

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

Step 14.14.3

Multiply $- - e^{- y}$ .

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Step 14.14.3.1

Multiply $- 1$ by $- 1$ .

$g (y) = y e^{- y} + 1 e^{- y} + C$

Step 14.14.3.2

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

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

Step 15

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

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

Step 16

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

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