Calculus Examples

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

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 $N (x, y) = 2 y - x$ .

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Step 2.1

Differentiate $N$ with respect to $x$ .

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

Step 2.2

Differentiate.

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Step 2.2.1

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

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

Step 2.2.2

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

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

Step 2.3

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

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Step 2.3.1

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

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

Step 2.3.2

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

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

Step 2.3.3

Multiply $- 1$ by $1$ .

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

Step 2.4

Subtract $1$ from $0$ .

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

Step 3

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

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Step 3.1

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

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Step 6.1

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

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

Step 6.2

Cancel the common factor of $y$ .

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Step 6.2.1

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

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

Step 6.2.2

Cancel the common factor.

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

Step 6.2.3

Rewrite the expression.

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

Step 6.3

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

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

Step 6.4

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

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

Step 6.5

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

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

Step 7

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

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

Step 8

Integrate $M (x, y) = \frac{1 + x y}{y}$ to find $f (x, y)$ .

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Step 8.1

Split the fraction into multiple fractions.

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

Step 8.2

Split the single integral into multiple integrals.

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

Step 8.3

Cancel the common factor of $y$ .

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Step 8.3.1

Cancel the common factor.

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

Step 8.3.2

Divide $x$ by $1$ .

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

Step 8.4

Apply the constant rule.

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

Step 8.5

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

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

Step 8.6

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

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

Step 8.7

Simplify.

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

Step 9

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

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

Step 10

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

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

Step 11

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

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Step 11.1

Differentiate $f$ with respect to $y$ .

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

Step 11.2

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

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

Step 11.3

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

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Step 11.3.1

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

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

Step 11.3.2

Rewrite $\frac{1}{y}$ as $y^{- 1}$ .

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

Step 11.3.3

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

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

Step 11.4

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 $0$ .

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

Step 11.5

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

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

Step 11.6

Simplify.

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Step 11.6.1

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

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

Step 11.6.2

Combine terms.

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Step 11.6.2.1

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

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

Step 11.6.2.2

Add $- \frac{x}{y^{2}}$ and $0$ .

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

Step 11.6.3

Reorder terms.

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

Step 12

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

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Step 12.1

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

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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 $\frac{2 y - x}{y^{2}}$ from both sides of the equation.

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

Step 12.1.1.2

Combine the numerators over the common denominator.

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

Step 12.1.1.3

Simplify each term.

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Step 12.1.1.3.1

Apply the distributive property.

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

Step 12.1.1.3.2

Multiply $2$ by $- 1$ .

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

Step 12.1.1.3.3

Multiply $- - x$ .

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Step 12.1.1.3.3.1

Multiply $- 1$ by $- 1$ .

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

Step 12.1.1.3.3.2

Multiply $x$ by $1$ .

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

Step 12.1.1.4

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

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Step 12.1.1.4.1

Add $- x$ and $x$ .

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

Step 12.1.1.4.2

Add $- 2 y$ and $0$ .

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

Step 12.1.1.5

Simplify each term.

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Step 12.1.1.5.1

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

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Step 12.1.1.5.1.1

Factor $y$ out of $- 2 y$ .

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

Step 12.1.1.5.1.2

Cancel the common factors.

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Step 12.1.1.5.1.2.1

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

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

Step 12.1.1.5.1.2.2

Cancel the common factor.

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

Step 12.1.1.5.1.2.3

Rewrite the expression.

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

Step 12.1.1.5.2

Move the negative in front of the fraction.

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

Step 12.1.2

Add $\frac{2}{y}$ to both sides of the equation.

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

Step 13

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

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Step 13.1

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 13.2

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

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

Step 13.3

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

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

Step 13.4

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

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

Step 13.5

Simplify.

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

Step 14

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

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

Step 15

Simplify each term.

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Step 15.1

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

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

Step 15.2

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

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

Step 15.3

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

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