Calculus Examples

$(y + \sin (x)) d y + (y \cos (x) - 2 x) d x = 0$

Step 1

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

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Step 1.1

Rewrite.

$(y \cos (x) - 2 x) d x + (y + \sin (x)) d y = 0$

Step 2

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

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Step 2.1

Differentiate $M$ with respect to $y$ .

$\frac{\partial M}{\partial y} = \frac{d}{d y} [y \cos (x) - 2 x]$

Step 2.2

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

$\frac{\partial M}{\partial y} = \frac{d}{d y} [y \cos (x)] + \frac{d}{d y} [- 2 x]$

Step 2.3

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

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Step 2.3.1

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

$\frac{\partial M}{\partial y} = \cos (x) \frac{d}{d y} [y] + \frac{d}{d y} [- 2 x]$

Step 2.3.2

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

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

Step 2.3.3

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

$\frac{\partial M}{\partial y} = \cos (x) + \frac{d}{d y} [- 2 x]$

Step 2.4

Differentiate using the Constant Rule.

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Step 2.4.1

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

$\frac{\partial M}{\partial y} = \cos (x) + 0$

Step 2.4.2

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

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

Step 3

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

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Step 3.1

Differentiate $N$ with respect to $x$ .

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

Step 3.2

Differentiate.

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Step 3.2.1

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

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

Step 3.2.2

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

$\frac{\partial N}{\partial x} = 0 + \frac{d}{d x} [\sin (x)]$

Step 3.3

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

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

Step 3.4

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

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

Step 4

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

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Step 4.1

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

$\cos (x) = \cos (x)$

Step 4.2

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

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

Step 5

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

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

Step 6

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

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Step 6.1

Split the single integral into multiple integrals.

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

Step 6.2

By the Power Rule, the integral of $y$ with respect to $y$ is $\frac{1}{2} y^{2}$ .

$f (x, y) = \frac{1}{2} y^{2} + C + \int \sin (x) d y$

Step 6.3

Apply the constant rule.

$f (x, y) = \frac{1}{2} y^{2} + C + \sin (x) y + C$

Step 6.4

Simplify.

$f (x, y) = \frac{1}{2} y^{2} + \sin (x) y + C$

Step 7

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

$f (x, y) = \frac{1}{2} y^{2} + \sin (x) y + g (x)$

Step 8

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

$\frac{\partial f}{\partial x} = y \cos (x) - 2 x$

Step 9

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

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Step 9.1

Differentiate $f$ with respect to $x$ .

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

Step 9.2

Differentiate.

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Step 9.2.1

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

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

Step 9.2.2

Since $\frac{1}{2} y^{2}$ is constant with respect to $x$ , the derivative of $\frac{1}{2} y^{2}$ with respect to $x$ is $0$ .

$0 + \frac{d}{d x} [\sin (x) y] + \frac{d}{d x} [g (x)] = y \cos (x) - 2 x$

Step 9.3

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

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Step 9.3.1

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

$0 + y \frac{d}{d x} [\sin (x)] + \frac{d}{d x} [g (x)] = y \cos (x) - 2 x$

Step 9.3.2

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

$0 + y \cos (x) + \frac{d}{d x} [g (x)] = y \cos (x) - 2 x$

Step 9.4

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

$0 + y \cos (x) + \frac{d g}{d x} = y \cos (x) - 2 x$

Step 9.5

Simplify.

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Step 9.5.1

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

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

Step 9.5.2

Reorder terms.

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

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

$\frac{d g}{d x} = y \cos (x) - 2 x - y \cos (x)$

Step 10.1.2

Combine the opposite terms in $y \cos (x) - 2 x - y \cos (x)$ .

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Step 10.1.2.1

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

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

Step 10.1.2.2

Add $- 2 x$ and $0$ .

$\frac{d g}{d x} = - 2 x$

Step 11

Find the antiderivative of $- 2 x$ to find $g (x)$ .

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Step 11.1

Integrate both sides of $\frac{d g}{d x} = - 2 x$ .

$\int \frac{d g}{d x} d x = \int - 2 x d x$

Step 11.2

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

$g (x) = \int - 2 x d x$

Step 11.3

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

$g (x) = - 2 \int x d x$

Step 11.4

By the Power Rule, the integral of $x$ with respect to $x$ is $\frac{1}{2} x^{2}$ .

$g (x) = - 2 (\frac{1}{2} x^{2} + C)$

Step 11.5

Simplify the answer.

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Step 11.5.1

Rewrite $- 2 (\frac{1}{2} x^{2} + C)$ as $- 2 (\frac{1}{2}) x^{2} + C$ .

$g (x) = - 2 (\frac{1}{2}) x^{2} + C$

Step 11.5.2

Simplify.

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Step 11.5.2.1

Combine $- 2$ and $\frac{1}{2}$ .

$g (x) = \frac{- 2}{2} x^{2} + C$

Step 11.5.2.2

Cancel the common factor of $- 2$ and $2$ .

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Step 11.5.2.2.1

Factor $2$ out of $- 2$ .

$g (x) = \frac{2 \cdot - 1}{2} x^{2} + C$

Step 11.5.2.2.2

Cancel the common factors.

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Step 11.5.2.2.2.1

Factor $2$ out of $2$ .

$g (x) = \frac{2 \cdot - 1}{2 (1)} x^{2} + C$

Step 11.5.2.2.2.2

Cancel the common factor.

$g (x) = \frac{2 \cdot - 1}{2 \cdot 1} x^{2} + C$

Step 11.5.2.2.2.3

Rewrite the expression.

$g (x) = \frac{- 1}{1} x^{2} + C$

Step 11.5.2.2.2.4

Divide $- 1$ by $1$ .

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

Step 12

Substitute for $g (x)$ in $f (x, y) = \frac{1}{2} y^{2} + \sin (x) y + g (x)$ .

$f (x, y) = \frac{1}{2} y^{2} + \sin (x) y - x^{2} + C$

Step 13

Simplify $f (x, y) = \frac{1}{2} y^{2} + \sin (x) y - x^{2} + C$ .

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Step 13.1

Combine $\frac{1}{2}$ and $y^{2}$ .

$f (x, y) = \frac{y^{2}}{2} + \sin (x) y - x^{2} + C$

Step 13.2

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

$f (x, y) = \frac{y^{2}}{2} + y \sin (x) - x^{2} + C$