Calculus Examples

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

Step 1

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

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Step 1.1

Rewrite.

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

Step 2

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

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Step 2.1

Differentiate $M$ with respect to $y$ .

$\frac{\partial M}{\partial y} = \frac{d}{d y} [- (y + 1)]$

Step 2.2

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

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

Step 2.3

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

$\frac{\partial M}{\partial y} = - (\frac{d}{d y} [y] + \frac{d}{d y} [1])$

Step 2.4

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

$\frac{\partial M}{\partial y} = - (1 + \frac{d}{d y} [1])$

Step 2.5

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

$\frac{\partial M}{\partial y} = - (1 + 0)$

Step 2.6

Simplify the expression.

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Step 2.6.1

Add $1$ and $0$ .

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

Step 2.6.2

Multiply $- 1$ by $1$ .

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

Step 3

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

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Step 3.1

Differentiate $N$ with respect to $x$ .

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

Step 3.2

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

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

Step 3.3

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

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Step 3.3.1

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

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

Step 3.3.2

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

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

Step 3.3.3

Multiply $y$ by $1$ .

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

Step 3.4

Differentiate using the Constant Rule.

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Step 3.4.1

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

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

Step 3.4.2

Add $y$ and $0$ .

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

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 $y$ for $\frac{\partial N}{\partial x}$ .

$- 1 = y$

Step 4.2

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

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

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

Step 5.2

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

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

Step 5.3

Substitute $- (y + 1)$ for $M$ .

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Step 5.3.1

Substitute $- (y + 1)$ for $M$ .

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

Step 5.3.2

Cancel the common factor of $y + 1$ .

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Step 5.3.2.1

Cancel the common factor.

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

Step 5.3.2.2

Rewrite the expression.

$\frac{1}{- 1}$

Step 5.3.2.3

Move the negative one from the denominator of $\frac{1}{- 1}$ .

$- 1 \cdot 1$

Step 5.3.3

Substitute $- (y + 1)$ for $M$ .

$- 1$

Step 5.4

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

$μ (x, y) = e^{\int - 1 d y}$

Step 6

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

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Step 6.1

Apply the constant rule.

$μ (x, y) = e^{- y + C}$

Step 6.2

Simplify.

$μ (x, y) = e^{- y} + C$

Step 7

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

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Step 7.1

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

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

Step 7.2

Apply the distributive property.

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

Step 7.3

Multiply $- 1$ by $1$ .

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

Step 7.4

Apply the distributive property.

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

Step 7.5

Rewrite $- 1 e^{- y}$ as $- e^{- y}$ .

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

Step 7.6

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

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

Step 7.7

Apply the distributive property.

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

Step 8

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

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

Step 9

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

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Step 9.1

Apply the constant rule.

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

Step 10

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

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

Step 11

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

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

Step 12

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

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Step 12.1

Differentiate $f$ with respect to $y$ .

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

Step 12.2

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

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

Step 12.3

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

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Step 12.3.1

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

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

Step 12.3.2

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

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

Step 12.3.3

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

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

Step 12.3.4

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

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

Step 12.3.5

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 $g (y) = - y$ .

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Step 12.3.5.1

To apply the Chain Rule, set $u_{1}$ as $- y$ .

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

Step 12.3.5.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u_{1}} [a^{u_{1}}]$ is $a^{u_{1}} \ln (a)$ where $a$ = $e$ .

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

Step 12.3.5.3

Replace all occurrences of $u_{1}$ with $- y$ .

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

Step 12.3.6

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

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

Step 12.3.7

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

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

Step 12.3.8

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

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

Step 12.3.9

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

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

Step 12.3.10

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 $g (y) = - y$ .

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Step 12.3.10.1

To apply the Chain Rule, set $u_{2}$ as $- y$ .

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

Step 12.3.10.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u_{2}} [a^{u_{2}}]$ is $a^{u_{2}} \ln (a)$ where $a$ = $e$ .

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

Step 12.3.10.3

Replace all occurrences of $u_{2}$ with $- y$ .

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

Step 12.3.11

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

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

Step 12.3.12

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

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

Step 12.3.13

Multiply $- 1$ by $1$ .

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

Step 12.3.14

Move $- 1$ to the left of $e^{- y}$ .

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

Step 12.3.15

Rewrite $- 1 e^{- y}$ as $- e^{- y}$ .

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

Step 12.3.16

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

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

Step 12.3.17

Multiply $- 1$ by $1$ .

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

Step 12.3.18

Move $- 1$ to the left of $e^{- y}$ .

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

Step 12.3.19

Rewrite $- 1 e^{- y}$ as $- e^{- y}$ .

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

Step 12.3.20

Multiply $- 1$ by $- 1$ .

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

Step 12.3.21

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

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

Step 12.4

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

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

Step 12.5

Simplify.

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Step 12.5.1

Apply the distributive property.

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

Step 12.5.2

Apply the distributive property.

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

Step 12.5.3

Combine terms.

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Step 12.5.3.1

Multiply $- 1$ by $- 1$ .

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

Step 12.5.3.2

Multiply $y$ by $1$ .

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

Step 12.5.3.3

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

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Step 12.5.3.3.1

Reorder $x$ and $- 1$ .

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

Step 12.5.3.3.2

Add $- 1 \cdot x e^{- y}$ and $x e^{- y}$ .

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

Step 12.5.3.4

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

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

Step 12.5.4

Reorder terms.

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

Step 12.5.5

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

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

Step 13

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

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Step 13.1

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

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Step 13.1.1

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

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

Step 13.1.2

Combine the opposite terms in $y x e^{- y} - y e^{- y} - x y e^{- y}$ .

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Step 13.1.2.1

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

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

Step 13.1.2.2

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

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

Step 13.1.2.3

Add $- y e^{- y}$ and $0$ .

$\frac{d g}{d y} = - y e^{- y}$

Step 14

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

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Step 14.1

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

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

Step 14.2

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

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

Step 14.3

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

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

Step 14.4

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 14.5

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 14.6

Simplify.

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Step 14.6.1

Multiply $- 1$ by $- 1$ .

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

Step 14.6.2

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

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

Step 14.7

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

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Step 14.7.1

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

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Step 14.7.1.1

Differentiate $- y$ .

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

Step 14.7.1.2

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 14.7.1.3

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 14.7.1.4

Multiply $- 1$ by $1$ .

$- 1$

Step 14.7.2

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

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

Step 14.8

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 14.9

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

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

Step 14.10

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

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

Step 14.11

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

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

Step 14.12

Simplify.

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Step 14.12.1

Apply the distributive property.

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

Step 14.12.2

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

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Step 14.12.2.1

Multiply $- 1$ by $- 1$ .

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

Step 14.12.2.2

Multiply $y$ by $1$ .

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

Step 14.12.3

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

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Step 14.12.3.1

Multiply $- 1$ by $- 1$ .

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

Step 14.12.3.2

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

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

Step 15

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

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

Step 16

Simplify $f (x, y) = (- y e^{- y} - e^{- y}) x + y e^{- y} + e^{- y} + C$ .

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Step 16.1

Apply the distributive property.

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

Step 16.2

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

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