Calculus Examples

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

Step 1

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

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Step 1.1

Rewrite.

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

Step 2

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

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Step 2.1

Differentiate $M$ with respect to $y$ .

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

Step 2.2

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

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

Step 2.3

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

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

Step 2.4

Multiply $- 1$ by $1$ .

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

Step 3

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

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Step 3.1

Differentiate $N$ with respect to $x$ .

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

Step 3.2

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

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

Step 3.3

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

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

Step 3.4

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

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

Step 3.5

Add $1$ and $0$ .

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

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

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Step 5.1

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

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

Step 5.2

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

$\frac{1 + 1}{M}$

Step 5.3

Substitute $- y$ for $M$ .

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Step 5.3.1

Substitute $- y$ for $M$ .

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

Step 5.3.2

Add $1$ and $1$ .

$\frac{2}{- y}$

Step 5.3.3

Substitute $- y$ for $M$ .

$- \frac{2}{y}$

Step 5.4

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 6

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

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Step 6.1

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 6.2

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 6.3

Multiply $2$ by $- 1$ .

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

Step 6.4

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

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

Step 6.5

Simplify.

$μ (x, y) = e^{- 2 \ln (| y |)} + C$

Step 6.6

Simplify each term.

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Step 6.6.1

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

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

Step 6.6.2

Exponentiation and log are inverse functions.

$μ (x, y) = {| y |}^{- 2} + C$

Step 6.6.3

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

$μ (x, y) = y^{- 2} + C$

Step 6.6.4

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

$μ (x, y) = \frac{1}{y^{2}} + C$

Step 7

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

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Step 7.1

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

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

Step 7.2

Cancel the common factor of $y$ .

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Step 7.2.1

Factor $y$ out of $- y$ .

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

Step 7.2.2

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

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

Step 7.2.3

Cancel the common factor.

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

Step 7.2.4

Rewrite the expression.

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

Step 7.3

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

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

Step 7.4

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

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

Step 8

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

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

Step 9

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

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Step 9.1

Apply the constant rule.

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

Step 9.2

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

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

Step 10

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

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

Step 11

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

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

Step 12

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

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Step 12.1

Differentiate $f$ with respect to $y$ .

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

Step 12.2

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

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

Step 12.3

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

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Step 12.3.1

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

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

Step 12.3.2

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

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

Step 12.3.3

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

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

Step 12.3.4

Multiply $- 1$ by $- 1$ .

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

Step 12.3.5

Multiply $x$ by $1$ .

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

Step 12.4

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

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

Step 12.5

Simplify.

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Step 12.5.1

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

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

Step 12.5.2

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

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

Step 12.5.3

Reorder terms.

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

Step 13

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

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Step 13.1

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

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

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

Step 13.1.1.2

Combine the numerators over the common denominator.

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

Step 13.1.1.3

Apply the distributive property.

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

Step 13.1.1.4

Subtract $x$ from $x$ .

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

Step 13.1.1.5

Subtract $y$ from $0$ .

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

Step 13.1.1.6

Simplify each term.

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Step 13.1.1.6.1

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

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Step 13.1.1.6.1.1

Factor $y$ out of $- y$ .

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

Step 13.1.1.6.1.2

Cancel the common factors.

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Step 13.1.1.6.1.2.1

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

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

Step 13.1.1.6.1.2.2

Cancel the common factor.

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

Step 13.1.1.6.1.2.3

Rewrite the expression.

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

Step 13.1.1.6.2

Move the negative in front of the fraction.

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

Step 13.1.2

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

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

Step 14

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

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Step 14.1

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 14.2

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

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

Step 14.3

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

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

Step 15

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

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