Calculus Examples

$y^{2} d x + (x y - 1) d y = 0$

Step 1

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

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Step 1.1

Differentiate $M$ with respect to $y$ .

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

Step 1.2

Differentiate using the Power Rule which states that $\frac{d}{d y} [y^{n}]$ is $n y^{n - 1}$ where $n = 2$ .

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

Step 2

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

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Step 2.1

Differentiate $N$ with respect to $x$ .

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

Step 2.2

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

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

Step 2.3

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

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Step 2.3.1

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

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

Step 2.3.2

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

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

Step 2.3.3

Multiply $y$ by $1$ .

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

Step 2.4

Differentiate using the Constant Rule.

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Step 2.4.1

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

$\frac{\partial N}{\partial x} = y + 0$

Step 2.4.2

Add $y$ and $0$ .

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

Step 3

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

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Step 3.1

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

$2 y = y$

Step 3.2

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

$2 y = y$ is not an identity.

Step 4

Find the integration factor $μ (x, y) = e^{\int \frac{\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y}}{M} d y}$ .

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Step 4.1

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

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

Step 4.2

Substitute $2 y$ for $\frac{\partial M}{\partial y}$ .

$\frac{y - (2 y)}{M}$

Step 4.3

Substitute $y^{2}$ for $M$ .

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Step 4.3.1

Substitute $y^{2}$ for $M$ .

$\frac{y - (2 y)}{y^{2}}$

Step 4.3.2

Simplify the numerator.

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Step 4.3.2.1

Factor $y$ out of $y - 1 \cdot 2 y$ .

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Step 4.3.2.1.1

Raise $y$ to the power of $1$ .

$\frac{y^{1} - 1 \cdot 2 y}{y^{2}}$

Step 4.3.2.1.2

Factor $y$ out of $y^{1}$ .

$\frac{y \cdot 1 - 1 \cdot 2 y}{y^{2}}$

Step 4.3.2.1.3

Factor $y$ out of $- 1 \cdot 2 y$ .

$\frac{y \cdot 1 + y (- 1 \cdot 2)}{y^{2}}$

Step 4.3.2.1.4

Factor $y$ out of $y \cdot 1 + y (- 1 \cdot 2)$ .

$\frac{y (1 - 1 \cdot 2)}{y^{2}}$

Step 4.3.2.2

Multiply $- 1$ by $2$ .

$\frac{y (1 - 2)}{y^{2}}$

Step 4.3.2.3

Subtract $2$ from $1$ .

$\frac{y \cdot - 1}{y^{2}}$

Step 4.3.3

Cancel the common factor of $y$ and $y^{2}$ .

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Step 4.3.3.1

Factor $y$ out of $y \cdot - 1$ .

$\frac{y (- 1)}{y^{2}}$

Step 4.3.3.2

Cancel the common factors.

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Step 4.3.3.2.1

Factor $y$ out of $y^{2}$ .

$\frac{y (- 1)}{y \cdot y}$

Step 4.3.3.2.2

Cancel the common factor.

$\frac{y \cdot - 1}{y \cdot y}$

Step 4.3.3.2.3

Rewrite the expression.

$\frac{- 1}{y}$

Step 4.3.4

Substitute $y^{2}$ for $M$ .

$- \frac{1}{y}$

Step 4.4

Find the integration factor $μ (x, y) = e^{\int \frac{\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y}}{M} d y}$ .

$μ (x, y) = e^{\int - \frac{1}{y} d y}$

Step 5

Evaluate the integral $e^{\int - \frac{1}{y} d y}$ .

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Step 5.1

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

$μ (x, y) = e^{- \int \frac{1}{y} d y}$

Step 5.2

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

$μ (x, y) = e^{- (\ln (| y |) + C)}$

Step 5.3

Simplify.

$μ (x, y) = e^{- \ln (| y |)} + C$

Step 5.4

Simplify each term.

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Step 5.4.1

Simplify $- \ln (| y |)$ by moving $- 1$ inside the logarithm.

$μ (x, y) = e^{\ln ({| y |}^{- 1})} + C$

Step 5.4.2

Exponentiation and log are inverse functions.

$μ (x, y) = {| y |}^{- 1} + C$

Step 5.4.3

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$μ (x, y) = \frac{1}{| y |} + C$

Step 6

Multiply both sides of $y^{2} d x + (x y - 1) d y = 0$ by the integration factor $\frac{1}{y}$ .

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Step 6.1

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

$y^{2} \frac{1}{y} d x + (x y - 1) d y = 0$

Step 6.2

Cancel the common factor of $y$ .

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Step 6.2.1

Factor $y$ out of $y^{2}$ .

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

Step 6.2.2

Cancel the common factor.

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

Step 6.2.3

Rewrite the expression.

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

Step 6.3

Multiply $x y - 1$ by $\frac{1}{y}$ .

$y d x + (x y - 1) \frac{1}{y} d y = 0$

Step 6.4

Multiply $x y - 1$ by $\frac{1}{y}$ .

$y d x + \frac{x y - 1}{y} d y = 0$

Step 7

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

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

Step 8

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

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Step 8.1

Apply the constant rule.

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

Step 9

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 10

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

$\frac{\partial f}{\partial y} = \frac{x y - 1}{y}$

Step 11

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

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Step 11.1

Differentiate $f$ with respect to $y$ .

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

Step 11.2

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

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

Step 11.3

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

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Step 11.3.1

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

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

Step 11.3.2

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

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

Step 11.3.3

Multiply $x$ by $1$ .

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

Step 11.4

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

$x + \frac{d g}{d y} = \frac{x y - 1}{y}$

Step 11.5

Reorder terms.

$\frac{d g}{d y} + x = \frac{x y - 1}{y}$

Step 12

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

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Step 12.1

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

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Step 12.1.1

Subtract $x$ from both sides of the equation.

$\frac{d g}{d y} = \frac{x y - 1}{y} - x$

Step 12.1.2

Simplify each term.

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Step 12.1.2.1

Split the fraction $\frac{x y - 1}{y}$ into two fractions.

$\frac{d g}{d y} = \frac{x y}{y} + \frac{- 1}{y} - x$

Step 12.1.2.2

Simplify each term.

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Step 12.1.2.2.1

Cancel the common factor of $y$ .

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Step 12.1.2.2.1.1

Cancel the common factor.

$\frac{d g}{d y} = \frac{x y}{y} + \frac{- 1}{y} - x$

Step 12.1.2.2.1.2

Divide $x$ by $1$ .

$\frac{d g}{d y} = x + \frac{- 1}{y} - x$

Step 12.1.2.2.2

Move the negative in front of the fraction.

$\frac{d g}{d y} = x - \frac{1}{y} - x$

Step 12.1.3

Combine the opposite terms in $x - \frac{1}{y} - x$ .

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Step 12.1.3.1

Subtract $x$ from $x$ .

$\frac{d g}{d y} = - \frac{1}{y} + 0$

Step 12.1.3.2

Add $- \frac{1}{y}$ and $0$ .

$\frac{d g}{d y} = - \frac{1}{y}$

Step 13

Find the antiderivative of $- \frac{1}{y}$ to find $g (y)$ .

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Step 13.1

Integrate both sides of $\frac{d g}{d y} = - \frac{1}{y}$ .

$\int \frac{d g}{d y} d y = \int - \frac{1}{y} d y$

Step 13.2

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

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

Step 13.3

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

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

Step 13.4

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

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

Step 13.5

Simplify.

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

Step 14

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

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