Calculus Examples

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

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 $N (x, y) = x - x y + 2$ .

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Step 2.1

Differentiate $N$ with respect to $x$ .

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

Step 2.2

Differentiate.

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Step 2.2.1

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

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

Step 2.2.2

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

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

Step 2.3

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

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Step 2.3.1

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

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

Step 2.3.2

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

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

Step 2.3.3

Multiply $- 1$ by $1$ .

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

Step 2.4

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

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

Step 2.5

Simplify.

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Step 2.5.1

Add $1 - y$ and $0$ .

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

Step 2.5.2

Reorder terms.

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

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

$1 = - y + 1$

Step 3.2

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

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

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

Step 4.2

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

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

Step 4.3

Substitute $y$ for $M$ .

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Step 4.3.1

Substitute $y$ for $M$ .

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

Step 4.3.2

Simplify the numerator.

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Step 4.3.2.1

Subtract $1$ from $1$ .

$\frac{- y + 0}{y}$

Step 4.3.2.2

Add $- y$ and $0$ .

$\frac{- y}{y}$

Step 4.3.3

Substitute $y$ for $M$ .

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Step 4.3.3.1

Cancel the common factor.

$\frac{- y}{y}$

Step 4.3.3.2

Divide $- 1$ by $1$ .

$- 1$

Step 4.4

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

$μ (x, y) = e^{\int - 1 d y}$

Step 5

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

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Step 5.1

Apply the constant rule.

$μ (x, y) = e^{- y + C}$

Step 5.2

Simplify.

$μ (x, y) = e^{- y} + C$

Step 6

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

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Step 6.1

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

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

Step 6.2

Multiply $x - x y + 2$ by $e^{- y}$ .

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

Step 6.3

Apply the distributive property.

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

Step 7

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

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

Step 8

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

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Step 8.1

Apply the constant rule.

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

Step 9

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

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

Step 10

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

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

Step 11

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

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Step 11.1

Differentiate $f$ with respect to $y$ .

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

Step 11.2

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

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

Step 11.3

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

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Step 11.3.1

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

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

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

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

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 $g (y) = - y$ .

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Step 11.3.3.1

To apply the Chain Rule, set $u$ as $- y$ .

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

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 $a$ = $e$ .

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

Step 11.3.3.3

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

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

Step 11.3.4

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

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

Step 11.3.5

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

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

Step 11.3.6

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

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

Step 11.3.7

Multiply $- 1$ by $1$ .

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

Step 11.3.8

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

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

Step 11.3.9

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

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

Step 11.3.10

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

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

Step 11.4

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

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

Step 11.5

Simplify.

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Step 11.5.1

Apply the distributive property.

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

Step 11.5.2

Reorder terms.

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

Step 11.5.3

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

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

Step 12

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

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Step 12.1

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

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Step 12.1.1

Add $x y e^{- y}$ to both sides of the equation.

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

Step 12.1.2

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

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

Step 12.1.3

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

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Step 12.1.3.1

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

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

Step 12.1.3.2

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

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

Step 12.1.3.3

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

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

Step 12.1.3.4

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

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

Step 13

Find the antiderivative of $2 e^{- y}$ to find $g (y)$ .

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Step 13.1

Integrate both sides of $\frac{d g}{d y} = 2 e^{- y}$ .

$\int \frac{d g}{d y} d y = \int 2 e^{- y} d y$

Step 13.2

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

$g (y) = \int 2 e^{- y} d y$

Step 13.3

Since $2$ is constant with respect to $y$ , move $2$ out of the integral.

$g (y) = 2 \int e^{- y} d y$

Step 13.4

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

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Step 13.4.1

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

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Step 13.4.1.1

Differentiate $- y$ .

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

Step 13.4.1.2

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 13.4.1.3

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 13.4.1.4

Multiply $- 1$ by $1$ .

$- 1$

Step 13.4.2

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

$g (y) = 2 \int - e^{u} d u$

Step 13.5

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

$g (y) = 2 (- \int e^{u} d u)$

Step 13.6

Multiply $- 1$ by $2$ .

$g (y) = - 2 \int e^{u} d u$

Step 13.7

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

$g (y) = - 2 (e^{u} + C)$

Step 13.8

Simplify.

$g (y) = - 2 e^{u} + C$

Step 13.9

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

$g (y) = - 2 e^{- y} + C$

Step 14

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

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

Step 15

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

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