Calculus Examples

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

Step 1

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

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Step 1.1

Rewrite.

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

Step 2

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

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Step 2.1

Differentiate $M$ with respect to $y$ .

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

Step 2.2

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

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

Step 2.3

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

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Step 2.3.1

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

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

Step 2.3.2

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

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

Step 2.3.3

Multiply $x$ by $1$ .

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

Step 2.4

Differentiate.

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Step 2.4.1

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

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

Step 2.4.2

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

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

Step 2.4.3

Add $x + 1$ and $0$ .

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

Step 3

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

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Step 3.1

Differentiate $N$ with respect to $x$ .

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

Step 3.2

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

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

Step 4

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

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Step 4.1

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

$x + 1 = 1$

Step 4.2

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

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

Step 5

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 5.1

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

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

Step 5.2

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

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

Step 5.3

Substitute $x$ for $N$ .

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Step 5.3.1

Substitute $x$ for $N$ .

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

Step 5.3.2

Simplify the numerator.

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Step 5.3.2.1

Subtract $1$ from $1$ .

$\frac{x + 0}{x}$

Step 5.3.2.2

Add $x$ and $0$ .

$\frac{x}{x}$

Step 5.3.3

Cancel the common factor of $x$ .

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Step 5.3.3.1

Cancel the common factor.

$\frac{x}{x}$

Step 5.3.3.2

Rewrite the expression.

$1$

Step 5.4

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 6

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

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Step 6.1

Apply the constant rule.

$μ (x, y) = e^{x + C}$

Step 6.2

Simplify.

$μ (x, y) = e^{x} + C$

Step 7

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

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Step 7.1

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

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

Step 7.2

Apply the distributive property.

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

Step 7.3

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

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

Step 7.4

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

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

Step 8

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

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

Step 9

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

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Step 9.1

Apply the constant rule.

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

Step 10

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

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

Step 11

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

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

Step 12

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

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Step 12.1

Differentiate $f$ with respect to $x$ .

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

Step 12.2

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

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

Step 12.3

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

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Step 12.3.1

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

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

Step 12.3.2

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

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

Step 12.3.3

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

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

Step 12.3.4

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

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

Step 12.3.5

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

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

Step 12.4

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

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

Step 12.5

Simplify.

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Step 12.5.1

Apply the distributive property.

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

Step 12.5.2

Reorder terms.

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

Step 12.5.3

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

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

Step 13

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

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Step 13.1

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

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Step 13.1.1

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

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

Step 13.1.2

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

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

Step 13.1.3

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

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Step 13.1.3.1

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

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

Step 13.1.3.2

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

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

Step 13.1.3.3

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

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

Step 13.1.3.4

Subtract $e^{x}$ from $0$ .

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

Step 14

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

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Step 14.1

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

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

Step 14.2

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

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

Step 14.3

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

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

Step 14.4

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

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

Step 14.5

Simplify.

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

Step 15

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

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

Step 16

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

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