Calculus Examples

$(x \sin (x) + 1) d y + (y \sin (x) + x y \cos (x)) d x = 0$

Step 1

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

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Step 1.1

Rewrite.

$(y \sin (x) + x y \cos (x)) d x + (x \sin (x) + 1) d y = 0$

Step 2

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

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Step 2.1

Differentiate $M$ with respect to $y$ .

$\frac{\partial M}{\partial y} = \frac{d}{d y} [y \sin (x) + x y \cos (x)]$

Step 2.2

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

$\frac{\partial M}{\partial y} = \frac{d}{d y} [y \sin (x)] + \frac{d}{d y} [x y \cos (x)]$

Step 2.3

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

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Step 2.3.1

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

$\frac{\partial M}{\partial y} = \sin (x) \frac{d}{d y} [y] + \frac{d}{d y} [x y \cos (x)]$

Step 2.3.2

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

$\frac{\partial M}{\partial y} = \sin (x) \cdot 1 + \frac{d}{d y} [x y \cos (x)]$

Step 2.3.3

Multiply $\sin (x)$ by $1$ .

$\frac{\partial M}{\partial y} = \sin (x) + \frac{d}{d y} [x y \cos (x)]$

Step 2.4

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

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Step 2.4.1

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

$\frac{\partial M}{\partial y} = \sin (x) + x \cos (x) \frac{d}{d y} [y]$

Step 2.4.2

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

$\frac{\partial M}{\partial y} = \sin (x) + x \cos (x) \cdot 1$

Step 2.4.3

Multiply $x$ by $1$ .

$\frac{\partial M}{\partial y} = \sin (x) + x \cos (x)$

Step 2.5

Reorder terms.

$\frac{\partial M}{\partial y} = x \cos (x) + \sin (x)$

Step 3

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

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Step 3.1

Differentiate $N$ with respect to $x$ .

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

Step 3.2

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

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

Step 3.3

Evaluate $\frac{d}{d x} [x \sin (x)]$ .

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Step 3.3.1

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 $g (x) = \sin (x)$ .

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

Step 3.3.2

The derivative of $\sin (x)$ with respect to $x$ is $\cos (x)$ .

$\frac{\partial N}{\partial x} = x \cos (x) + \sin (x) \frac{d}{d x} [x] + \frac{d}{d x} [1]$

Step 3.3.3

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

$\frac{\partial N}{\partial x} = x \cos (x) + \sin (x) \cdot 1 + \frac{d}{d x} [1]$

Step 3.3.4

Multiply $\sin (x)$ by $1$ .

$\frac{\partial N}{\partial x} = x \cos (x) + \sin (x) + \frac{d}{d x} [1]$

Step 3.4

Differentiate using the Constant Rule.

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Step 3.4.1

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

$\frac{\partial N}{\partial x} = x \cos (x) + \sin (x) + 0$

Step 3.4.2

Add $x \cos (x) + \sin (x)$ and $0$ .

$\frac{\partial N}{\partial x} = x \cos (x) + \sin (x)$

Step 4

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

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Step 4.1

Substitute $x \cos (x) + \sin (x)$ for $\frac{\partial M}{\partial y}$ and $x \cos (x) + \sin (x)$ for $\frac{\partial N}{\partial x}$ .

$x \cos (x) + \sin (x) = x \cos (x) + \sin (x)$

Step 4.2

Since the two sides have been shown to be equivalent, the equation is an identity.

$x \cos (x) + \sin (x) = x \cos (x) + \sin (x)$ is an identity.

Step 5

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

$f (x, y) = \int x \sin (x) + 1 d y$

Step 6

Integrate $N (x, y) = x \sin (x) + 1$ to find $f (x, y)$ .

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Step 6.1

Apply the constant rule.

$f (x, y) = (x \sin (x) + 1) y + C$

Step 7

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

$f (x, y) = (x \sin (x) + 1) y + g (x)$

Step 8

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

$\frac{\partial f}{\partial x} = y \sin (x) + x y \cos (x)$

Step 9

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

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Step 9.1

Differentiate $f$ with respect to $x$ .

$\frac{d}{d x} [(x \sin (x) + 1) y + g (x)] = y \sin (x) + x y \cos (x)$

Step 9.2

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

$\frac{d}{d x} [(x \sin (x) + 1) y] + \frac{d}{d x} [g (x)] = y \sin (x) + x y \cos (x)$

Step 9.3

Evaluate $\frac{d}{d x} [(x \sin (x) + 1) y]$ .

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Step 9.3.1

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

$y \frac{d}{d x} [x \sin (x) + 1] + \frac{d}{d x} [g (x)] = y \sin (x) + x y \cos (x)$

Step 9.3.2

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

$y (\frac{d}{d x} [x \sin (x)] + \frac{d}{d x} [1]) + \frac{d}{d x} [g (x)] = y \sin (x) + x y \cos (x)$

Step 9.3.3

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 $g (x) = \sin (x)$ .

$y (x \frac{d}{d x} [\sin (x)] + \sin (x) \frac{d}{d x} [x] + \frac{d}{d x} [1]) + \frac{d}{d x} [g (x)] = y \sin (x) + x y \cos (x)$

Step 9.3.4

The derivative of $\sin (x)$ with respect to $x$ is $\cos (x)$ .

$y (x \cos (x) + \sin (x) \frac{d}{d x} [x] + \frac{d}{d x} [1]) + \frac{d}{d x} [g (x)] = y \sin (x) + x y \cos (x)$

Step 9.3.5

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

$y (x \cos (x) + \sin (x) \cdot 1 + \frac{d}{d x} [1]) + \frac{d}{d x} [g (x)] = y \sin (x) + x y \cos (x)$

Step 9.3.6

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

$y (x \cos (x) + \sin (x) \cdot 1 + 0) + \frac{d}{d x} [g (x)] = y \sin (x) + x y \cos (x)$

Step 9.3.7

Multiply $\sin (x)$ by $1$ .

$y (x \cos (x) + \sin (x) + 0) + \frac{d}{d x} [g (x)] = y \sin (x) + x y \cos (x)$

Step 9.3.8

Add $x \cos (x) + \sin (x)$ and $0$ .

$y (x \cos (x) + \sin (x)) + \frac{d}{d x} [g (x)] = y \sin (x) + x y \cos (x)$

Step 9.4

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

$y (x \cos (x) + \sin (x)) + \frac{d g}{d x} = y \sin (x) + x y \cos (x)$

Step 9.5

Simplify.

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Step 9.5.1

Apply the distributive property.

$y (x \cos (x)) + y \sin (x) + \frac{d g}{d x} = y \sin (x) + x y \cos (x)$

Step 9.5.2

Reorder terms.

$\frac{d g}{d x} + x y \cos (x) + y \sin (x) = y \sin (x) + x y \cos (x)$

Step 10

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

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Step 10.1

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

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Step 10.1.1

Subtract $x y \cos (x)$ from both sides of the equation.

$\frac{d g}{d x} + y \sin (x) = y \sin (x) + x y \cos (x) - x y \cos (x)$

Step 10.1.2

Subtract $y \sin (x)$ from both sides of the equation.

$\frac{d g}{d x} = y \sin (x) + x y \cos (x) - x y \cos (x) - y \sin (x)$

Step 10.1.3

Combine the opposite terms in $y \sin (x) + x y \cos (x) - x y \cos (x) - y \sin (x)$ .

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Step 10.1.3.1

Subtract $x y \cos (x)$ from $x y \cos (x)$ .

$\frac{d g}{d x} = y \sin (x) + 0 - y \sin (x)$

Step 10.1.3.2

Add $y \sin (x)$ and $0$ .

$\frac{d g}{d x} = y \sin (x) - y \sin (x)$

Step 10.1.3.3

Subtract $y \sin (x)$ from $y \sin (x)$ .

$\frac{d g}{d x} = 0$

Step 11

Find the antiderivative of $0$ to find $g (x)$ .

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Step 11.1

Integrate both sides of $\frac{d g}{d x} = 0$ .

$\int \frac{d g}{d x} d x = \int 0 d x$

Step 11.2

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

$g (x) = \int 0 d x$

Step 11.3

The integral of $0$ with respect to $x$ is $0$ .

$g (x) = 0 + C$

Step 11.4

Add $0$ and $C$ .

$g (x) = C$

Step 12

Substitute for $g (x)$ in $f (x, y) = (x \sin (x) + 1) y + g (x)$ .

$f (x, y) = (x \sin (x) + 1) y + C$

Step 13

Simplify $f (x, y) = (x \sin (x) + 1) y + C$ .

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Step 13.1

Simplify each term.

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Step 13.1.1

Apply the distributive property.

$f (x, y) = x \sin (x) y + 1 y + C$

Step 13.1.2

Multiply $y$ by $1$ .

$f (x, y) = x \sin (x) y + y + C$

Step 13.2

Reorder factors in $f (x, y) = x \sin (x) y + y + C$ .

$f (x, y) = x y \sin (x) + y + C$