Calculus Examples

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

Step 1

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

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Step 1.1

Differentiate $M$ with respect to $y$ .

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

Step 1.2

Differentiate.

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Step 1.2.1

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

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

Step 1.2.2

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

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

Step 1.3

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

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Step 1.3.1

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

$\frac{\partial M}{\partial y} = 0 - 2 \frac{d}{d y} [y]$

Step 1.3.2

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

$\frac{\partial M}{\partial y} = 0 - 2 \cdot 1$

Step 1.3.3

Multiply $- 2$ by $1$ .

$\frac{\partial M}{\partial y} = 0 - 2$

Step 1.4

Subtract $2$ from $0$ .

$\frac{\partial M}{\partial y} = - 2$

Step 2

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

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Step 2.1

Differentiate $N$ with respect to $x$ .

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

Step 2.2

Differentiate.

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Step 2.2.1

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

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

Step 2.2.2

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

$\frac{\partial N}{\partial x} = 0 + \frac{d}{d x} [- 2 x]$

Step 2.3

Evaluate $\frac{d}{d x} [- 2 x]$ .

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Step 2.3.1

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

$\frac{\partial N}{\partial x} = 0 - 2 \frac{d}{d x} [x]$

Step 2.3.2

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

$\frac{\partial N}{\partial x} = 0 - 2 \cdot 1$

Step 2.3.3

Multiply $- 2$ by $1$ .

$\frac{\partial N}{\partial x} = 0 - 2$

Step 2.4

Subtract $2$ from $0$ .

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

Step 3

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

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Step 3.1

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

$- 2 = - 2$

Step 3.2

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

$- 2 = - 2$ is an identity.

Step 4

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

$f (x, y) = \int x - 2 y d x$

Step 5

Integrate $M (x, y) = x - 2 y$ to find $f (x, y)$ .

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Step 5.1

Split the single integral into multiple integrals.

$f (x, y) = \int x d x + \int - 2 y d x$

Step 5.2

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

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

Step 5.3

Apply the constant rule.

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

Step 5.4

Simplify.

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

Step 6

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

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

Step 7

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

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

Step 8

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

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Step 8.1

Differentiate $f$ with respect to $y$ .

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

Step 8.2

Differentiate.

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Step 8.2.1

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

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

Step 8.2.2

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

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

Step 8.3

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

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Step 8.3.1

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

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

Step 8.3.2

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

$0 - 2 x \cdot 1 + \frac{d}{d y} [g (y)] = 2 y - 2 x$

Step 8.3.3

Multiply $- 2$ by $1$ .

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

Step 8.4

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

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

Step 8.5

Simplify.

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Step 8.5.1

Subtract $2 x$ from $0$ .

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

Step 8.5.2

Reorder terms.

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

Step 9

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

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Step 9.1

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

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Step 9.1.1

Add $2 x$ to both sides of the equation.

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

Step 9.1.2

Combine the opposite terms in $2 y - 2 x + 2 x$ .

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Step 9.1.2.1

Add $- 2 x$ and $2 x$ .

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

Step 9.1.2.2

Add $2 y$ and $0$ .

$\frac{d g}{d y} = 2 y$

Step 10

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

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Step 10.1

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

$\int \frac{d g}{d y} d y = \int 2 y d y$

Step 10.2

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

$g (y) = \int 2 y d y$

Step 10.3

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

$g (y) = 2 \int y d y$

Step 10.4

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

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

Step 10.5

Simplify the answer.

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Step 10.5.1

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

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

Step 10.5.2

Simplify.

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Step 10.5.2.1

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

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

Step 10.5.2.2

Cancel the common factor of $2$ .

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Step 10.5.2.2.1

Cancel the common factor.

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

Step 10.5.2.2.2

Rewrite the expression.

$g (y) = 1 y^{2} + C$

Step 10.5.2.3

Multiply $y^{2}$ by $1$ .

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

Step 11

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

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

Step 12

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

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