Calculus Examples

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

Step 1

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

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Step 1.1

Differentiate $M$ with respect to $y$ .

$\frac{\partial M}{\partial y} = \frac{d}{d y} [1]$

Step 1.2

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

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

Step 2

Find $\frac{\partial N}{\partial x}$ where $N (x, y) = - (y - 2 x y)$ .

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Step 2.1

Differentiate $N$ with respect to $x$ .

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

Step 2.2

Since $- 1$ is constant with respect to $x$ , the derivative of $- (y - 2 x y)$ with respect to $x$ is $- \frac{d}{d x} [y - 2 x y]$ .

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

Step 2.3

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

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

Step 2.4

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

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

Step 2.5

Add $0$ and $\frac{d}{d x} [- 2 x y]$ .

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

Step 2.6

Since $- 2 y$ is constant with respect to $x$ , the derivative of $- 2 x y$ with respect to $x$ is $- 2 y \frac{d}{d x} [x]$ .

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

Step 2.7

Multiply $- 2$ by $- 1$ .

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

Step 2.8

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

$\frac{\partial N}{\partial x} = 2 y \cdot 1$

Step 2.9

Multiply $2$ by $1$ .

$\frac{\partial N}{\partial x} = 2 y$

Step 3

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

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Step 3.1

Substitute $0$ for $\frac{\partial M}{\partial y}$ and $2 y$ for $\frac{\partial N}{\partial x}$ .

$0 = 2 y$

Step 3.2

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

$0 = 2 y$ 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 $0$ for $\frac{\partial M}{\partial y}$ .

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

Step 4.2

Substitute $2 y$ for $\frac{\partial N}{\partial x}$ .

$\frac{0 - (2 y)}{N}$

Step 4.3

Substitute $- (y - 2 x y)$ for $N$ .

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Step 4.3.1

Substitute $- (y - 2 x y)$ for $N$ .

$\frac{0 - (2 y)}{- (y - 2 x y)}$

Step 4.3.2

Simplify the numerator.

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Step 4.3.2.1

Multiply $- 1$ by $2$ .

$\frac{0 - 2 y}{- (y - 2 x y)}$

Step 4.3.2.2

Subtract $2 y$ from $0$ .

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

Step 4.3.3

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

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Step 4.3.3.1

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

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

Step 4.3.3.2

Factor $y$ out of $y^{1}$ .

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

Step 4.3.3.3

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

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

Step 4.3.3.4

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

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

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

Step 4.3.4

Cancel the common factor of $y$ .

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Step 4.3.4.1

Cancel the common factor.

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

Step 4.3.4.2

Rewrite the expression.

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

Step 4.3.5

Dividing two negative values results in a positive value.

$\frac{2}{1 - 2 x}$

Step 4.4

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

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

Step 5

Evaluate the integral $e^{\int \frac{2}{1 - 2 x} d x}$ .

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Step 5.1

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

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

Step 5.2

Let $u = 1 - 2 x$ . Then $d u = - 2 d x$ , so $- \frac{1}{2} d u = d x$ . Rewrite using $u$ and $d$ $u$ .

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Step 5.2.1

Let $u = 1 - 2 x$ . Find $\frac{d u}{d x}$ .

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Step 5.2.1.1

Differentiate $1 - 2 x$ .

$\frac{d}{d x} [1 - 2 x]$

Step 5.2.1.2

Differentiate.

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Step 5.2.1.2.1

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

$\frac{d}{d x} [1] + \frac{d}{d x} [- 2 x]$

Step 5.2.1.2.2

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

$0 + \frac{d}{d x} [- 2 x]$

Step 5.2.1.3

Evaluate $\frac{d}{d x} [- 2 x]$ .

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Step 5.2.1.3.1

Since $- 2$ is constant with respect to $x$ , the derivative of $- 2 x$ with respect to $x$ is $- 2 \frac{d}{d x} [x]$ .

$0 - 2 \frac{d}{d x} [x]$

Step 5.2.1.3.2

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

$0 - 2 \cdot 1$

Step 5.2.1.3.3

Multiply $- 2$ by $1$ .

$0 - 2$

Step 5.2.1.4

Subtract $2$ from $0$ .

$- 2$

Step 5.2.2

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

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

Step 5.3

Simplify.

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Step 5.3.1

Move the negative in front of the fraction.

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

Step 5.3.2

Multiply $\frac{1}{u}$ by $\frac{1}{2}$ .

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

Step 5.3.3

Move $2$ to the left of $u$ .

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

Step 5.4

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

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

Step 5.5

Multiply $- 1$ by $2$ .

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

Step 5.6

Since $\frac{1}{2}$ is constant with respect to $u$ , move $\frac{1}{2}$ out of the integral.

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

Step 5.7

Simplify.

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Step 5.7.1

Combine $\frac{1}{2}$ and $- 2$ .

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

Step 5.7.2

Cancel the common factor of $- 2$ and $2$ .

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Step 5.7.2.1

Factor $2$ out of $- 2$ .

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

Step 5.7.2.2

Cancel the common factors.

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Step 5.7.2.2.1

Factor $2$ out of $2$ .

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

Step 5.7.2.2.2

Cancel the common factor.

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

Step 5.7.2.2.3

Rewrite the expression.

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

Step 5.7.2.2.4

Divide $- 1$ by $1$ .

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

Step 5.8

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

$μ (x, y) = e^{- (\ln (| u |) + C)}$

Step 5.9

Simplify.

$μ (x, y) = e^{- \ln (| u |)} + C$

Step 5.10

Replace all occurrences of $u$ with $1 - 2 x$ .

$μ (x, y) = e^{- \ln (| 1 - 2 x |)} + C$

Step 5.11

Simplify each term.

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Step 5.11.1

Simplify $- \ln (| 1 - 2 x |)$ by moving $- 1$ inside the logarithm.

$μ (x, y) = e^{\ln ({| 1 - 2 x |}^{- 1})} + C$

Step 5.11.2

Exponentiation and log are inverse functions.

$μ (x, y) = {| 1 - 2 x |}^{- 1} + C$

Step 5.11.3

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

$μ (x, y) = \frac{1}{| 1 - 2 x |} + C$

Step 6

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

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

Step 7

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

$f (x, y) = \int - y d y$

Step 8

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

$f (x, y) = - \int y d y$

Step 8.2

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

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

Step 8.3

Rewrite $- (\frac{1}{2} y^{2} + C)$ as $- \frac{1}{2} y^{2} + C$ .

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

Step 9

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

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

Step 10

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

$\frac{\partial f}{\partial x} = \frac{1}{1 - 2 x}$

Step 11

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

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Step 11.1

Differentiate $f$ with respect to $x$ .

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

Step 11.2

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

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

Step 11.3

Since $- \frac{1}{2} y^{2}$ is constant with respect to $x$ , the derivative of $- \frac{1}{2} y^{2}$ with respect to $x$ is $0$ .

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

Step 11.4

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

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

Step 11.5

Add $0$ and $\frac{d g}{d x}$ .

$\frac{d g}{d x} = \frac{1}{1 - 2 x}$

Step 12

Find the antiderivative of $\frac{1}{1 - 2 x}$ to find $g (x)$ .

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Step 12.1

Integrate both sides of $\frac{d g}{d x} = \frac{1}{1 - 2 x}$ .

$\int \frac{d g}{d x} d x = \int \frac{1}{1 - 2 x} d x$

Step 12.2

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

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

Step 12.3

Let $u = 1 - 2 x$ . Then $d u = - 2 d x$ , so $- \frac{1}{2} d u = d x$ . Rewrite using $u$ and $d$ $u$ .

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Step 12.3.1

Let $u = 1 - 2 x$ . Find $\frac{d u}{d x}$ .

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Step 12.3.1.1

Differentiate $1 - 2 x$ .

$\frac{d}{d x} [1 - 2 x]$

Step 12.3.1.2

Differentiate.

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Step 12.3.1.2.1

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

$\frac{d}{d x} [1] + \frac{d}{d x} [- 2 x]$

Step 12.3.1.2.2

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

$0 + \frac{d}{d x} [- 2 x]$

Step 12.3.1.3

Evaluate $\frac{d}{d x} [- 2 x]$ .

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Step 12.3.1.3.1

Since $- 2$ is constant with respect to $x$ , the derivative of $- 2 x$ with respect to $x$ is $- 2 \frac{d}{d x} [x]$ .

$0 - 2 \frac{d}{d x} [x]$

Step 12.3.1.3.2

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

$0 - 2 \cdot 1$

Step 12.3.1.3.3

Multiply $- 2$ by $1$ .

$0 - 2$

Step 12.3.1.4

Subtract $2$ from $0$ .

$- 2$

Step 12.3.2

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

$g (x) = \int \frac{1}{u} \cdot \frac{1}{- 2} d u$

Step 12.4

Simplify.

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Step 12.4.1

Move the negative in front of the fraction.

$g (x) = \int \frac{1}{u} (- \frac{1}{2}) d u$

Step 12.4.2

Multiply $\frac{1}{u}$ by $\frac{1}{2}$ .

$g (x) = \int - \frac{1}{u \cdot 2} d u$

Step 12.4.3

Move $2$ to the left of $u$ .

$g (x) = \int - \frac{1}{2 \cdot u} d u$

Step 12.4.4

Multiply $2$ by $u$ .

$g (x) = \int - \frac{1}{2 u} d u$

Step 12.5

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

$g (x) = - \int \frac{1}{2 u} d u$

Step 12.6

Since $\frac{1}{2}$ is constant with respect to $u$ , move $\frac{1}{2}$ out of the integral.

$g (x) = - (\frac{1}{2} \int \frac{1}{u} d u)$

Step 12.7

Remove parentheses.

$g (x) = - \frac{1}{2} \int \frac{1}{u} d u$

Step 12.8

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

$g (x) = - \frac{1}{2} (\ln (| u |) + C)$

Step 12.9

Simplify.

$g (x) = - \frac{1}{2} \ln (| u |) + C$

Step 12.10

Replace all occurrences of $u$ with $1 - 2 x$ .

$g (x) = - \frac{1}{2} \ln (| 1 - 2 x |) + C$

Step 13

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

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

Step 14

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

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Step 14.1

Simplify each term.

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Step 14.1.1

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

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

Step 14.1.2

Multiply $- \frac{1}{2} \ln (| 1 - 2 x |)$ .

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Step 14.1.2.1

Reorder $\ln (| 1 - 2 x |)$ and $\frac{1}{2}$ .

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

Step 14.1.2.2

Simplify $\frac{1}{2} \ln (| 1 - 2 x |)$ by moving $\frac{1}{2}$ inside the logarithm.

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

Step 14.2

To write $- \ln ({| 1 - 2 x |}^{\frac{1}{2}})$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

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

Step 14.3

Combine $- \ln ({| 1 - 2 x |}^{\frac{1}{2}})$ and $\frac{2}{2}$ .

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

Step 14.4

Combine the numerators over the common denominator.

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

Step 14.5

Simplify the numerator.

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Step 14.5.1

Multiply $- \ln ({| 1 - 2 x |}^{\frac{1}{2}}) \cdot 2$ .

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Step 14.5.1.1

Multiply $2$ by $- 1$ .

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

Step 14.5.1.2

Simplify $- 2 \ln ({| 1 - 2 x |}^{\frac{1}{2}})$ by moving $2$ inside the logarithm.

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

Step 14.5.2

Multiply the exponents in ${({| 1 - 2 x |}^{\frac{1}{2}})}^{2}$ .

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Step 14.5.2.1

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

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

Step 14.5.2.2

Cancel the common factor of $2$ .

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Step 14.5.2.2.1

Cancel the common factor.

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

Step 14.5.2.2.2

Rewrite the expression.

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

Step 14.5.3

Simplify.

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