Calculus Examples

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

Step 1

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

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Step 1.1

Differentiate $M$ with respect to $y$ .

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

Step 1.2

Differentiate.

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Step 1.2.1

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

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

Step 1.2.2

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

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

Step 1.3

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

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Step 1.3.1

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 $g (y) = e^{\frac{y}{x}}$ .

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

Step 1.3.2

Differentiate using the chain rule, which states that $\frac{d}{d y} [f (g (y))]$ is $f' (g (y)) g' (y)$ where $f (y) = e^{y}$ and $g (y) = \frac{y}{x}$ .

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Step 1.3.2.1

To apply the Chain Rule, set $u$ as $\frac{y}{x}$ .

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

Step 1.3.2.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u} [a^{u}]$ is $a^{u} \ln (a)$ where $a$ = $e$ .

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

Step 1.3.2.3

Replace all occurrences of $u$ with $\frac{y}{x}$ .

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

Step 1.3.3

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

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

Step 1.3.4

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

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

Step 1.3.5

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

$\frac{\partial M}{\partial y} = 0 + y (e^{\frac{y}{x}} (\frac{1}{x} \cdot 1)) + e^{\frac{y}{x}} \cdot 1$

Step 1.3.6

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

$\frac{\partial M}{\partial y} = 0 + y (e^{\frac{y}{x}} \frac{1}{x}) + e^{\frac{y}{x}} \cdot 1$

Step 1.3.7

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

$\frac{\partial M}{\partial y} = 0 + y \frac{e^{\frac{y}{x}}}{x} + e^{\frac{y}{x}} \cdot 1$

Step 1.3.8

Combine $y$ and $\frac{e^{\frac{y}{x}}}{x}$ .

$\frac{\partial M}{\partial y} = 0 + \frac{y e^{\frac{y}{x}}}{x} + e^{\frac{y}{x}} \cdot 1$

Step 1.3.9

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

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

Step 1.4

Add $0$ and $\frac{y e^{\frac{y}{x}}}{x} + e^{\frac{y}{x}}$ .

$\frac{\partial M}{\partial y} = \frac{y e^{\frac{y}{x}}}{x} + e^{\frac{y}{x}}$

Step 2

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

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Step 2.1

Differentiate $N$ with respect to $x$ .

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

Step 2.2

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

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

Step 2.3

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

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

Step 2.4

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = e^{x}$ and $g (x) = \frac{y}{x}$ .

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Step 2.4.1

To apply the Chain Rule, set $u$ as $\frac{y}{x}$ .

$\frac{\partial N}{\partial x} = - (x (\frac{d}{d u} [e^{u}] \frac{d}{d x} [\frac{y}{x}]) + e^{\frac{y}{x}} \frac{d}{d x} [x])$

Step 2.4.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u} [a^{u}]$ is $a^{u} \ln (a)$ where $a$ = $e$ .

$\frac{\partial N}{\partial x} = - (x (e^{u} \frac{d}{d x} [\frac{y}{x}]) + e^{\frac{y}{x}} \frac{d}{d x} [x])$

Step 2.4.3

Replace all occurrences of $u$ with $\frac{y}{x}$ .

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

Step 2.5

Differentiate.

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Step 2.5.1

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

$\frac{\partial N}{\partial x} = - (x (e^{\frac{y}{x}} (y \frac{d}{d x} [\frac{1}{x}])) + e^{\frac{y}{x}} \frac{d}{d x} [x])$

Step 2.5.2

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

$\frac{\partial N}{\partial x} = - (x (e^{\frac{y}{x}} (y \frac{d}{d x} [x^{- 1}])) + e^{\frac{y}{x}} \frac{d}{d x} [x])$

Step 2.5.3

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

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

Step 2.6

Raise $x$ to the power of $1$ .

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

Step 2.7

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

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

Step 2.8

Simplify the expression.

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Step 2.8.1

Add $- 2$ and $1$ .

$\frac{\partial N}{\partial x} = - (x^{- 1} (e^{\frac{y}{x}} (y \cdot (- 1))) + e^{\frac{y}{x}} \frac{d}{d x} [x])$

Step 2.8.2

Move $- 1$ to the left of $y$ .

$\frac{\partial N}{\partial x} = - (x^{- 1} (e^{\frac{y}{x}} (- 1 \cdot y)) + e^{\frac{y}{x}} \frac{d}{d x} [x])$

Step 2.8.3

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

$\frac{\partial N}{\partial x} = - (x^{- 1} (e^{\frac{y}{x}} (- y)) + e^{\frac{y}{x}} \frac{d}{d x} [x])$

Step 2.9

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

$\frac{\partial N}{\partial x} = - (x^{- 1} (e^{\frac{y}{x}} (- y)) + e^{\frac{y}{x}} \cdot 1)$

Step 2.10

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

$\frac{\partial N}{\partial x} = - (x^{- 1} (e^{\frac{y}{x}} (- y)) + e^{\frac{y}{x}})$

Step 2.11

Simplify.

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Step 2.11.1

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

$\frac{\partial N}{\partial x} = - (\frac{1}{x} (e^{\frac{y}{x}} (- y)) + e^{\frac{y}{x}})$

Step 2.11.2

Apply the distributive property.

$\frac{\partial N}{\partial x} = - (\frac{1}{x} (e^{\frac{y}{x}} (- y))) - e^{\frac{y}{x}}$

Step 2.11.3

Combine terms.

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Step 2.11.3.1

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

$\frac{\partial N}{\partial x} = - (\frac{e^{\frac{y}{x}}}{x} (- y)) - e^{\frac{y}{x}}$

Step 2.11.3.2

Combine $\frac{e^{\frac{y}{x}}}{x}$ and $y$ .

$\frac{\partial N}{\partial x} = - - \frac{e^{\frac{y}{x}} y}{x} - e^{\frac{y}{x}}$

Step 2.11.3.3

Multiply $- 1$ by $- 1$ .

$\frac{\partial N}{\partial x} = 1 \frac{e^{\frac{y}{x}} y}{x} - e^{\frac{y}{x}}$

Step 2.11.3.4

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

$\frac{\partial N}{\partial x} = \frac{e^{\frac{y}{x}} y}{x} - e^{\frac{y}{x}}$

Step 2.11.4

Reorder factors in $\frac{e^{\frac{y}{x}} y}{x} - e^{\frac{y}{x}}$ .

$\frac{\partial N}{\partial x} = \frac{y e^{\frac{y}{x}}}{x} - e^{\frac{y}{x}}$

Step 3

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

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Step 3.1

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

$\frac{y e^{\frac{y}{x}}}{x} + e^{\frac{y}{x}} = \frac{y e^{\frac{y}{x}}}{x} - e^{\frac{y}{x}}$

Step 3.2

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

$\frac{y e^{\frac{y}{x}}}{x} + e^{\frac{y}{x}} = \frac{y e^{\frac{y}{x}}}{x} - e^{\frac{y}{x}}$ 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 $\frac{y e^{\frac{y}{x}}}{x} + e^{\frac{y}{x}}$ for $\frac{\partial M}{\partial y}$ .

$\frac{\frac{y e^{\frac{y}{x}}}{x} + e^{\frac{y}{x}} - \frac{\partial N}{\partial x}}{N}$

Step 4.2

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

$\frac{\frac{y e^{\frac{y}{x}}}{x} + e^{\frac{y}{x}} - (\frac{y e^{\frac{y}{x}}}{x} - e^{\frac{y}{x}})}{N}$

Step 4.3

Substitute $- x e^{\frac{y}{x}}$ for $N$ .

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Step 4.3.1

Substitute $- x e^{\frac{y}{x}}$ for $N$ .

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

Step 4.3.2

Simplify the numerator.

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Step 4.3.2.1

Apply the distributive property.

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

Step 4.3.2.2

Multiply $- - e^{\frac{y}{x}}$ .

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Step 4.3.2.2.1

Multiply $- 1$ by $- 1$ .

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

Step 4.3.2.2.2

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

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

Step 4.3.2.3

Subtract $\frac{y e^{\frac{y}{x}}}{x}$ from $\frac{y e^{\frac{y}{x}}}{x}$ .

$\frac{0 + e^{\frac{y}{x}} + e^{\frac{y}{x}}}{- x e^{\frac{y}{x}}}$

Step 4.3.2.4

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

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

Step 4.3.2.5

Add $e^{\frac{y}{x}}$ and $e^{\frac{y}{x}}$ .

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

Step 4.3.3

Cancel the common factor of $e^{\frac{y}{x}}$ .

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Step 4.3.3.1

Cancel the common factor.

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

Step 4.3.3.2

Rewrite the expression.

$\frac{2}{- x}$

Step 4.3.4

Move the negative in front of the fraction.

$- \frac{2}{x}$

Step 4.4

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

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

Step 5

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

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Step 5.1

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

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

Step 5.2

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 5.3

Multiply $2$ by $- 1$ .

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

Step 5.4

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

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

Step 5.5

Simplify.

$μ (x, y) = e^{- 2 \ln (| x |)} + C$

Step 5.6

Simplify each term.

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Step 5.6.1

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

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

Step 5.6.2

Exponentiation and log are inverse functions.

$μ (x, y) = {| x |}^{- 2} + C$

Step 5.6.3

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

$μ (x, y) = x^{- 2} + C$

Step 5.6.4

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

$μ (x, y) = \frac{1}{x^{2}} + C$

Step 6

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

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Step 6.1

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

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

Step 6.2

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

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

Step 6.3

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

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

Step 6.4

Cancel the common factor of $x$ .

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Step 6.4.1

Factor $x$ out of $- x e^{\frac{y}{x}}$ .

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

Step 6.4.2

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

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

Step 6.4.3

Cancel the common factor.

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

Step 6.4.4

Rewrite the expression.

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

Step 6.5

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

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

Step 6.6

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

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

Step 7

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

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

Step 8

Integrate $N (x, y) = - \frac{e^{\frac{y}{x}}}{x}$ 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.

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

Step 8.2

Since $\frac{1}{x}$ is constant with respect to $y$ , move $\frac{1}{x}$ out of the integral.

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

Step 8.3

Remove parentheses.

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

Step 8.4

Let $u = \frac{y}{x}$ . Then $d u = \frac{1}{x} d y$ , so $x d u = d y$ . Rewrite using $u$ and $d$ $u$ .

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Step 8.4.1

Let $u = \frac{y}{x}$ . Find $\frac{d u}{d y}$ .

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Step 8.4.1.1

Differentiate $\frac{y}{x}$ .

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

Step 8.4.1.2

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

$\frac{1}{x} \frac{d}{d y} [y]$

Step 8.4.1.3

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

$\frac{1}{x} \cdot 1$

Step 8.4.1.4

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

$\frac{1}{x}$

Step 8.4.2

Rewrite the problem using $u$ and $d u$ .

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

Step 8.5

Simplify.

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Step 8.5.1

Multiply by the reciprocal of the fraction to divide by $\frac{1}{x}$ .

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

Step 8.5.2

Multiply $x$ by $1$ .

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

Step 8.6

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

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

Step 8.7

Simplify.

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Step 8.7.1

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

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

Step 8.7.2

Cancel the common factor of $x$ .

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Step 8.7.2.1

Cancel the common factor.

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

Step 8.7.2.2

Rewrite the expression.

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

Step 8.7.3

Multiply $- 1$ by $1$ .

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

Step 8.8

The integral of $e^{u}$ with respect to $u$ is $e^{u}$ .

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

Step 8.9

Simplify.

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

Step 8.10

Replace all occurrences of $u$ with $\frac{y}{x}$ .

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

Step 9

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

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

Step 10

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

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

Step 11

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

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Step 11.1

Differentiate $f$ with respect to $x$ .

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

Step 11.2

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

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

Step 11.3

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

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Step 11.3.1

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

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

Step 11.3.2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = e^{x}$ and $g (x) = \frac{y}{x}$ .

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Step 11.3.2.1

To apply the Chain Rule, set $u$ as $\frac{y}{x}$ .

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

Step 11.3.2.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u} [a^{u}]$ is $a^{u} \ln (a)$ where $a$ = $e$ .

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

Step 11.3.2.3

Replace all occurrences of $u$ with $\frac{y}{x}$ .

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

Step 11.3.3

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

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

Step 11.3.4

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

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

Step 11.3.5

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

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

Step 11.3.6

Multiply $- 1$ by $- 1$ .

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

Step 11.3.7

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

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

Step 11.4

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

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

Step 11.5

Simplify.

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Step 11.5.1

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

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

Step 11.5.2

Combine terms.

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Step 11.5.2.1

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

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

Step 11.5.2.2

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

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

Step 11.5.3

Reorder terms.

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

Step 11.5.4

Reorder factors in $\frac{d g}{d x} + \frac{e^{\frac{y}{x}} y}{x^{2}}$ .

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

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

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

Step 12.1.1.2

Combine the numerators over the common denominator.

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

Step 12.1.1.3

Apply the distributive property.

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

Step 12.1.1.4

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

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Step 12.1.1.4.1

Subtract $y e^{\frac{y}{x}}$ from $y e^{\frac{y}{x}}$ .

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

Step 12.1.1.4.2

Add $- x$ and $0$ .

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

Step 12.1.1.5

Simplify each term.

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Step 12.1.1.5.1

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

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Step 12.1.1.5.1.1

Factor $x$ out of $- x$ .

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

Step 12.1.1.5.1.2

Cancel the common factors.

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Step 12.1.1.5.1.2.1

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

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

Step 12.1.1.5.1.2.2

Cancel the common factor.

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

Step 12.1.1.5.1.2.3

Rewrite the expression.

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

Step 12.1.1.5.2

Move the negative in front of the fraction.

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

Step 12.1.2

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

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

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

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

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

Step 14

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

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