Calculus Examples

$(x + y) d x = x d y$

Step 1

Rewrite the differential equation to fit the Exact differential equation technique.

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Step 1.1

Subtract $x d y$ from both sides of the equation.

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

Step 2

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

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Step 2.1

Differentiate $M$ with respect to $y$ .

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

Step 2.2

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

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

Step 2.3

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

$\frac{\partial M}{\partial y} = 0 + \frac{d}{d y} [y]$

Step 2.4

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

$\frac{\partial M}{\partial y} = 0 + 1$

Step 2.5

Add $0$ and $1$ .

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

Step 3

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

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Step 3.1

Differentiate $N$ with respect to $x$ .

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

Step 3.2

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 3.3

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 3.4

Multiply $- 1$ by $1$ .

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

Step 4

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

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Step 4.1

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

$1 = - 1$

Step 4.2

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

$1 = - 1$ is not an identity.

Step 5

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 5.1

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

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

Step 5.2

Substitute $- 1$ for $\frac{\partial N}{\partial x}$ .

$\frac{1 + 1}{N}$

Step 5.3

Substitute $- x$ for $N$ .

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Step 5.3.1

Substitute $- x$ for $N$ .

$\frac{1 + 1}{- x}$

Step 5.3.2

Add $1$ and $1$ .

$\frac{2}{- x}$

Step 5.3.3

Move the negative in front of the fraction.

$- \frac{2}{x}$

Step 5.4

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

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

Step 6

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

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Step 6.1

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

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

Step 6.2

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

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

Step 6.3

Multiply $2$ by $- 1$ .

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

Step 6.4

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

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

Step 6.5

Simplify.

$μ (x, y) = e^{- 2 \ln (| x |)} + C$

Step 6.6

Simplify each term.

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Step 6.6.1

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

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

Step 6.6.2

Exponentiation and log are inverse functions.

$μ (x, y) = {| x |}^{- 2} + C$

Step 6.6.3

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

$μ (x, y) = x^{- 2} + C$

Step 6.6.4

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

$μ (x, y) = \frac{1}{x^{2}} + C$

Step 7

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

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Step 7.1

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

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

Step 7.2

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

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

Step 7.3

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

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

Step 7.4

Cancel the common factor of $x$ .

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Step 7.4.1

Factor $x$ out of $- x$ .

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

Step 7.4.2

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

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

Step 7.4.3

Cancel the common factor.

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

Step 7.4.4

Rewrite the expression.

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

Step 8

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

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

Step 9

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

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Step 9.1

Apply the constant rule.

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

Step 9.2

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

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

Step 10

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

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

Step 11

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

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

Step 12

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

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Step 12.1

Differentiate $f$ with respect to $x$ .

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

Step 12.2

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

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

Step 12.3

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

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Step 12.3.1

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

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

Step 12.3.2

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

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

Step 12.3.3

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

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

Step 12.3.4

Multiply $- 1$ by $- 1$ .

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

Step 12.3.5

Multiply $y$ by $1$ .

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

Step 12.4

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

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

Step 12.5

Simplify.

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Step 12.5.1

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

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

Step 12.5.2

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

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

Step 12.5.3

Reorder terms.

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

Step 13

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

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Step 13.1

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

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Step 13.1.1

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

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Step 13.1.1.1

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

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

Step 13.1.1.2

Combine the numerators over the common denominator.

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

Step 13.1.1.3

Apply the distributive property.

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

Step 13.1.1.4

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

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Step 13.1.1.4.1

Subtract $y$ from $y$ .

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

Step 13.1.1.4.2

Add $- x$ and $0$ .

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

Step 13.1.1.5

Simplify each term.

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Step 13.1.1.5.1

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

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Step 13.1.1.5.1.1

Factor $x$ out of $- x$ .

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

Step 13.1.1.5.1.2

Cancel the common factors.

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Step 13.1.1.5.1.2.1

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

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

Step 13.1.1.5.1.2.2

Cancel the common factor.

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

Step 13.1.1.5.1.2.3

Rewrite the expression.

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

Step 13.1.1.5.2

Move the negative in front of the fraction.

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

Step 13.1.2

Add $\frac{1}{x}$ to both sides of the equation.

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

Step 14

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

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Step 14.1

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 14.2

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

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

Step 14.3

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

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

Step 15

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

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