Calculus Examples

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

Step 1

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

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Step 1.1

Differentiate $M$ with respect to $y$ .

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

Step 1.2

Differentiate using the Product Rule which states that $\frac{d}{d y} [f (y) g (y)]$ is $f (y) \frac{d}{d y} [g (y)] + g (y) \frac{d}{d y} [f (y)]$ where $f (y) = y$ and $g (y) = 1 + x y$ .

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

Step 1.3

Differentiate.

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Step 1.3.1

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

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

Step 1.3.2

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

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

Step 1.3.3

Add $0$ and $\frac{d}{d y} [x y]$ .

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

Step 1.3.4

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

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

Step 1.3.5

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

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

Step 1.3.6

Multiply $x$ by $1$ .

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

Step 1.3.7

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

$\frac{\partial M}{\partial y} = y x + (1 + x y) \cdot 1$

Step 1.3.8

Multiply $1 + x y$ by $1$ .

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

Step 1.4

Add $y x$ and $x y$ .

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Step 1.4.1

Reorder $y$ and $x$ .

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

Step 1.4.2

Add $x y$ and $x y$ .

$\frac{\partial M}{\partial y} = 2 x y + 1$

Step 2

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

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Step 2.1

Differentiate $N$ with respect to $x$ .

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

Step 2.2

Since $- 1$ is constant with respect to $x$ , the derivative of $- x$ with respect to $x$ is $- \frac{d}{d x} [x]$ .

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

Step 2.3

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

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

Step 2.4

Multiply $- 1$ by $1$ .

$\frac{\partial N}{\partial x} = - 1$

Step 3

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

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Step 3.1

Substitute $2 x y + 1$ for $\frac{\partial M}{\partial y}$ and $- 1$ for $\frac{\partial N}{\partial x}$ .

$2 x y + 1 = - 1$

Step 3.2

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

$2 x y + 1 = - 1$ 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 $- 1$ for $\frac{\partial N}{\partial x}$ .

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

Step 4.2

Substitute $2 x y + 1$ for $\frac{\partial M}{\partial y}$ .

$\frac{- 1 - (2 x y + 1)}{M}$

Step 4.3

Substitute $y (1 + x y)$ for $M$ .

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Step 4.3.1

Substitute $y (1 + x y)$ for $M$ .

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

Step 4.3.2

Simplify the numerator.

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Step 4.3.2.1

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

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Step 4.3.2.1.1

Reorder $- 1$ and $- (2 x y + 1)$ .

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

Step 4.3.2.1.2

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

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

Step 4.3.2.1.3

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

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

Step 4.3.2.2

Add $1$ and $1$ .

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

Step 4.3.2.3

Factor $2$ out of $2 x y + 2$ .

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Step 4.3.2.3.1

Factor $2$ out of $2 x y$ .

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

Step 4.3.2.3.2

Factor $2$ out of $2$ .

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

Step 4.3.2.3.3

Factor $2$ out of $2 (x y) + 2 (1)$ .

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

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

Step 4.3.2.4

Multiply $- 1$ by $2$ .

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

Step 4.3.3

Cancel the common factor of $x y + 1$ and $1 + x y$ .

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Step 4.3.3.1

Reorder terms.

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

Step 4.3.3.2

Cancel the common factor.

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

Step 4.3.3.3

Rewrite the expression.

$\frac{- 2}{y}$

Step 4.3.4

Substitute $y (1 + x y)$ for $M$ .

$- \frac{2}{y}$

Step 4.4

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

$μ (x, y) = e^{\int - \frac{2}{y} d y}$

Step 5

Evaluate the integral $e^{\int - \frac{2}{y} d y}$ .

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Step 5.1

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

$μ (x, y) = e^{- \int \frac{2}{y} d y}$

Step 5.2

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

$μ (x, y) = e^{- (2 \int \frac{1}{y} d y)}$

Step 5.3

Multiply $2$ by $- 1$ .

$μ (x, y) = e^{- 2 \int \frac{1}{y} d y}$

Step 5.4

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

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

Step 5.5

Simplify.

$μ (x, y) = e^{- 2 \ln (| y |)} + C$

Step 5.6

Simplify each term.

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Step 5.6.1

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

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

Step 5.6.2

Exponentiation and log are inverse functions.

$μ (x, y) = {| y |}^{- 2} + C$

Step 5.6.3

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

$μ (x, y) = y^{- 2} + C$

Step 5.6.4

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

$μ (x, y) = \frac{1}{y^{2}} + C$

Step 6

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

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Step 6.1

Multiply $y (1 + x y)$ by $\frac{1}{y^{2}}$ .

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

Step 6.2

Cancel the common factor of $y$ .

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Step 6.2.1

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

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

Step 6.2.2

Cancel the common factor.

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

Step 6.2.3

Rewrite the expression.

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

Step 6.3

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

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

Step 6.4

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

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

Step 6.5

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

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

Step 7

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

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

Step 8

Integrate $N (x, y) = - \frac{x}{y^{2}}$ to find $f (x, y)$ .

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Step 8.1

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

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

Step 8.2

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

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

Step 8.3

Remove parentheses.

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

Step 8.4

Move $y^{2}$ out of the denominator by raising it to the $- 1$ power.

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

Step 8.5

Multiply the exponents in ${(y^{2})}^{- 1}$ .

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Step 8.5.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

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

Step 8.5.2

Multiply $2$ by $- 1$ .

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

Step 8.6

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

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

Step 8.7

Simplify the answer.

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Step 8.7.1

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

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

Step 8.7.2

Simplify.

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Step 8.7.2.1

Multiply $- 1$ by $- 1$ .

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

Step 8.7.2.2

Multiply $x$ by $1$ .

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

Step 8.7.2.3

Combine $x$ and $\frac{1}{y}$ .

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

Step 9

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

$f (x, y) = \frac{x}{y} + g (x)$

Step 10

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

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

Step 11

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

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Step 11.1

Differentiate $f$ with respect to $x$ .

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

Step 11.2

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

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

Step 11.3

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

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Step 11.3.1

Since $\frac{1}{y}$ is constant with respect to $x$ , the derivative of $\frac{x}{y}$ with respect to $x$ is $\frac{1}{y} \frac{d}{d x} [x]$ .

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

Step 11.3.2

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

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

Step 11.3.3

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

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

Step 11.4

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

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

Step 11.5

Reorder terms.

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

Step 12

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

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Step 12.1

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

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Step 12.1.1

Move all terms containing variables to the left side of the equation.

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Step 12.1.1.1

Subtract $\frac{1 + x y}{y}$ from both sides of the equation.

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

Step 12.1.1.2

Combine the numerators over the common denominator.

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

Step 12.1.1.3

Simplify each term.

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Step 12.1.1.3.1

Apply the distributive property.

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

Step 12.1.1.3.2

Multiply $- 1$ by $1$ .

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

Step 12.1.1.4

Subtract $1$ from $1$ .

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

Step 12.1.1.5

Subtract $x y$ from $0$ .

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

Step 12.1.1.6

Cancel the common factor of $y$ .

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Step 12.1.1.6.1

Cancel the common factor.

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

Step 12.1.1.6.2

Divide $- x$ by $1$ .

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

Step 12.1.2

Add $x$ to both sides of the equation.

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

Step 13

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

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Step 13.1

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

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

Step 13.2

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

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

Step 13.3

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

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

Step 14

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

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

Step 15

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

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