Calculus Examples

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

Step 1

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

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Step 1.1

Differentiate $M$ with respect to $y$ .

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

Step 1.2

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

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

Step 2

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

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Step 2.1

Differentiate $N$ with respect to $x$ .

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

Step 2.2

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

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

Step 2.3

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

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

Step 2.4

Since $\frac{2}{y}$ is constant with respect to $x$ , the derivative of $\frac{2}{y}$ 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 $1$ for $\frac{\partial M}{\partial y}$ and $1$ for $\frac{\partial N}{\partial x}$ .

$1 = 1$

Step 3.2

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

$1 = 1$ is an identity.

Step 4

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

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

Step 5

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

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Step 5.1

Apply the constant rule.

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

Step 6

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

$f (x, y) = y x + g (y)$

Step 7

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

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

Step 8

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

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Step 8.1

Differentiate $f$ with respect to $y$ .

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

Step 8.2

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

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

Step 8.3

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

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Step 8.3.1

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

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

Step 8.3.2

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

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

Step 8.3.3

Multiply $x$ by $1$ .

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

Step 8.4

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

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

Step 8.5

Reorder terms.

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

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

Subtract $x$ from both sides of the equation.

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

Step 9.1.2

Combine the opposite terms in $x + \frac{2}{y} - x$ .

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Step 9.1.2.1

Subtract $x$ from $x$ .

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

Step 9.1.2.2

Add $\frac{2}{y}$ and $0$ .

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

Step 10

Find the antiderivative of $\frac{2}{y}$ to find $g (y)$ .

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Step 10.1

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

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

Step 10.2

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

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

Step 10.3

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

$g (y) = 2 \int \frac{1}{y} d y$

Step 10.4

The integral of $\frac{1}{y}$ with respect to $y$ is $\ln (| y |)$ .

$g (y) = 2 (\ln (| y |) + C)$

Step 10.5

Simplify.

$g (y) = 2 \ln (| y |) + C$

Step 11

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

$f (x, y) = y x + 2 \ln (| y |) + C$

Step 12

Simplify each term.

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Step 12.1

Simplify $2 \ln (| y |)$ by moving $2$ inside the logarithm.

$f (x, y) = y x + \ln ({| y |}^{2}) + C$

Step 12.2

Remove the absolute value in ${| y |}^{2}$ because exponentiations with even powers are always positive.

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