Calculus Examples

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

Step 1

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

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Step 1.1

Differentiate $M$ with respect to $y$ .

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

Step 1.2

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

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

Step 1.3

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

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Step 1.3.1

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

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

Step 1.3.2

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

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

Step 1.3.3

Multiply $x$ by $1$ .

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

Step 1.4

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

$\frac{\partial M}{\partial y} = x + 0 + \frac{d}{d y} [2 y] + \frac{d}{d y} [1]$

Step 1.5

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

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Step 1.5.1

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

$\frac{\partial M}{\partial y} = x + 0 + 2 \frac{d}{d y} [y] + \frac{d}{d y} [1]$

Step 1.5.2

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

$\frac{\partial M}{\partial y} = x + 0 + 2 \cdot 1 + \frac{d}{d y} [1]$

Step 1.5.3

Multiply $2$ by $1$ .

$\frac{\partial M}{\partial y} = x + 0 + 2 + \frac{d}{d y} [1]$

Step 1.6

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

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

Step 1.7

Combine terms.

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Step 1.7.1

Add $x$ and $0$ .

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

Step 1.7.2

Add $x + 2$ and $0$ .

$\frac{\partial M}{\partial y} = x + 2$

Step 2

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

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Step 2.1

Differentiate $N$ with respect to $x$ .

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

Step 2.2

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

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

Step 2.3

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

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

Step 2.4

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

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

Step 2.5

Add $1$ and $0$ .

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

Step 3

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

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Step 3.1

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

$x + 2 = 1$

Step 3.2

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

$x + 2 = 1$ is not an identity.

Step 4

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

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Step 4.1

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

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

Step 4.2

Substitute $1$ for $\frac{\partial N}{\partial x}$ .

$\frac{x + 2 - 1}{N}$

Step 4.3

Substitute $x + 1$ for $N$ .

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Step 4.3.1

Substitute $x + 1$ for $N$ .

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

Step 4.3.2

Subtract $1$ from $2$ .

$\frac{x + 1}{x + 1}$

Step 4.3.3

Cancel the common factor of $x + 1$ .

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Step 4.3.3.1

Cancel the common factor.

$\frac{x + 1}{x + 1}$

Step 4.3.3.2

Rewrite the expression.

$1$

Step 4.4

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

$μ (x, y) = e^{\int d x}$

Step 5

Evaluate the integral $e^{\int d x}$ .

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Step 5.1

Apply the constant rule.

$μ (x, y) = e^{x + C}$

Step 5.2

Simplify.

$μ (x, y) = e^{x} + C$

Step 6

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

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Step 6.1

Multiply $x y + x + 2 y + 1$ by $e^{x}$ .

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

Step 6.2

Apply the distributive property.

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

Step 6.3

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

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

Step 6.4

Multiply $x + 1$ by $e^{x}$ .

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

Step 6.5

Apply the distributive property.

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

Step 6.6

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

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

Step 7

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

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

Step 8

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

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Step 8.1

Apply the constant rule.

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

Step 9

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

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

Step 10

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

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

Step 11

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

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Step 11.1

Differentiate $f$ with respect to $x$ .

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

Step 11.2

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

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

Step 11.3

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

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Step 11.3.1

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

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

Step 11.3.2

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

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

Step 11.3.3

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 $g (x) = e^{x}$ .

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

Step 11.3.4

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

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

Step 11.3.5

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

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

Step 11.3.6

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

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

Step 11.3.7

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

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

Step 11.3.8

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

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

Step 11.4

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

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

Step 11.5

Simplify.

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Step 11.5.1

Apply the distributive property.

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

Step 11.5.2

Move $2$ to the left of $y$ .

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

Step 11.5.3

Reorder terms.

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

Step 11.5.4

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

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

Step 12

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

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Step 12.1

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

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Step 12.1.1

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

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

Step 12.1.2

Subtract $2 y e^{x}$ from both sides of the equation.

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

Step 12.1.3

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

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Step 12.1.3.1

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

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

Step 12.1.3.2

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

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

Step 12.1.3.3

Subtract $2 y e^{x}$ from $2 y e^{x}$ .

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

Step 12.1.3.4

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

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

Step 13

Find the antiderivative of $x e^{x} + e^{x}$ to find $g (x)$ .

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Step 13.1

Integrate both sides of $\frac{d g}{d x} = x e^{x} + e^{x}$ .

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

Step 13.2

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

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

Step 13.3

Split the single integral into multiple integrals.

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

Step 13.4

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

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

Step 13.5

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

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

Step 13.6

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

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

Step 13.7

Simplify.

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

Step 13.8

Simplify.

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Step 13.8.1

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

$g (x) = x e^{x} + 0 + C$

Step 13.8.2

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

$g (x) = x e^{x} + C$

Step 14

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

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

Step 15

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

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Step 15.1

Apply the distributive property.

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

Step 15.2

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

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