Calculus Examples

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

Step 1

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

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Step 1.1

Differentiate $M$ with respect to $y$ .

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

Step 1.2

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

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

Step 1.3

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

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Step 1.3.1

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

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

Step 1.3.2

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

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

Step 1.3.3

Multiply $x$ by $1$ .

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

Step 1.4

Differentiate using the Constant Rule.

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Step 1.4.1

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

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

Step 1.4.2

Add $x$ and $0$ .

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

Step 2

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

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Step 2.1

Differentiate $N$ with respect to $x$ .

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

Step 2.2

Differentiate.

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Step 2.2.1

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

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

Step 2.2.2

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

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

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

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

Step 2.3.2

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

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

Step 2.3.3

Multiply $- 1$ by $1$ .

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

Step 3

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

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Step 3.1

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

$x = 2 x - y$

Step 3.2

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

$x = 2 x - y$ is not an identity.

Step 4

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

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Step 4.1

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

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

Step 4.2

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

$\frac{x - (2 x - y)}{N}$

Step 4.3

Substitute $x^{2} - x y$ for $N$ .

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Step 4.3.1

Substitute $x^{2} - x y$ for $N$ .

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

Step 4.3.2

Simplify the numerator.

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Step 4.3.2.1

Apply the distributive property.

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

Step 4.3.2.2

Multiply $2$ by $- 1$ .

$\frac{x - 2 x - - y}{x^{2} - x y}$

Step 4.3.2.3

Multiply $- - y$ .

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Step 4.3.2.3.1

Multiply $- 1$ by $- 1$ .

$\frac{x - 2 x + 1 y}{x^{2} - x y}$

Step 4.3.2.3.2

Multiply $y$ by $1$ .

$\frac{x - 2 x + y}{x^{2} - x y}$

Step 4.3.2.4

Subtract $2 x$ from $x$ .

$\frac{- x + y}{x^{2} - x y}$

Step 4.3.3

Simplify the denominator.

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Step 4.3.3.1

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

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Step 4.3.3.1.1

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

$\frac{- x + y}{x \cdot x - x y}$

Step 4.3.3.1.2

Factor $x$ out of $- x y$ .

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

Step 4.3.3.1.3

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

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

Step 4.3.3.2

Rewrite $- 1 y$ as $- y$ .

$\frac{- x + y}{x (x - y)}$

Step 4.3.4

Cancel the common factor of $- x + y$ and $x - y$ .

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Step 4.3.4.1

Factor $- 1$ out of $- x$ .

$\frac{- (x) + y}{x (x - y)}$

Step 4.3.4.2

Factor $- 1$ out of $y$ .

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

Step 4.3.4.3

Factor $- 1$ out of $- (x) - 1 (- y)$ .

$\frac{- (x - y)}{x (x - y)}$

Step 4.3.4.4

Rewrite $- (x - y)$ as $- 1 (x - y)$ .

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

Step 4.3.4.5

Cancel the common factor.

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

Step 4.3.4.6

Rewrite the expression.

$\frac{- 1}{x}$

Step 4.3.5

Move the negative in front of the fraction.

$- \frac{1}{x}$

Step 4.4

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

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

Step 5

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

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Step 5.1

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

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

Step 5.2

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

$μ (x, y) = e^{- (\ln (| x |) + C)}$

Step 5.3

Simplify.

$μ (x, y) = e^{- \ln (| x |)} + C$

Step 5.4

Simplify each term.

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Step 5.4.1

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

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

Step 5.4.2

Exponentiation and log are inverse functions.

$μ (x, y) = {| x |}^{- 1} + C$

Step 5.4.3

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

$μ (x, y) = \frac{1}{| x |} + C$

Step 6

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

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Step 6.1

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

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

Step 6.2

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

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

Step 6.3

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

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

Step 6.4

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

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

Step 6.5

Simplify the numerator.

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Step 6.5.1

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

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Step 6.5.1.1

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

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

Step 6.5.1.2

Factor $x$ out of $- x y$ .

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

Step 6.5.1.3

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

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

Step 6.5.2

Rewrite $- 1 y$ as $- y$ .

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

Step 6.6

Cancel the common factor of $x$ .

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Step 6.6.1

Cancel the common factor.

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

Step 6.6.2

Divide $x - y$ by $1$ .

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

Step 7

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

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

Step 8

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

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Step 8.1

Split the single integral into multiple integrals.

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

Step 8.2

Apply the constant rule.

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

Step 8.3

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

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

Step 8.4

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

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

Step 8.5

Simplify.

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

Step 9

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

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

Step 10

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

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

Step 11

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

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Step 11.1

Differentiate $f$ with respect to $x$ .

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

Step 11.2

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

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

Step 11.3

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

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Step 11.3.1

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

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

Step 11.3.2

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

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

Step 11.3.3

Multiply $y$ by $1$ .

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

Step 11.4

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 $0$ .

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

Step 11.5

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

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

Step 11.6

Simplify.

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Step 11.6.1

Add $y$ and $0$ .

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

Step 11.6.2

Reorder terms.

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

Step 12

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

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Step 12.1

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

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Step 12.1.1

Subtract $y$ from both sides of the equation.

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

Step 12.1.2

Simplify each term.

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Step 12.1.2.1

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

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

Step 12.1.2.2

Simplify each term.

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Step 12.1.2.2.1

Cancel the common factor of $x$ .

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Step 12.1.2.2.1.1

Cancel the common factor.

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

Step 12.1.2.2.1.2

Divide $y$ by $1$ .

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

Step 12.1.2.2.2

Move the negative in front of the fraction.

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

Step 12.1.3

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

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Step 12.1.3.1

Subtract $y$ from $y$ .

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

Step 12.1.3.2

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

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

Step 13

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

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Step 13.1

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

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

Step 13.2

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

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

Step 13.3

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

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

Step 13.4

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

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

Step 13.5

Simplify.

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

Step 14

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

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

Step 15

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

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