Calculus Examples

$(x - 2 y - 1) d x - (x - 3) d y = 0$

Step 1

Find $\frac{\partial M}{\partial y}$ where $M (x, y) = x - 2 y - 1$ .

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Step 1.1

Differentiate $M$ with respect to $y$ .

$\frac{\partial M}{\partial y} = \frac{d}{d y} [x - 2 y - 1]$

Step 1.2

Differentiate.

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Step 1.2.1

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

$\frac{\partial M}{\partial y} = \frac{d}{d y} [x] + \frac{d}{d y} [- 2 y] + \frac{d}{d y} [- 1]$

Step 1.2.2

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

$\frac{\partial M}{\partial y} = 0 + \frac{d}{d y} [- 2 y] + \frac{d}{d y} [- 1]$

Step 1.3

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

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Step 1.3.1

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

$\frac{\partial M}{\partial y} = 0 - 2 \frac{d}{d y} [y] + \frac{d}{d y} [- 1]$

Step 1.3.2

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

$\frac{\partial M}{\partial y} = 0 - 2 \cdot 1 + \frac{d}{d y} [- 1]$

Step 1.3.3

Multiply $- 2$ by $1$ .

$\frac{\partial M}{\partial y} = 0 - 2 + \frac{d}{d y} [- 1]$

Step 1.4

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

$\frac{\partial M}{\partial y} = 0 - 2 + 0$

Step 1.5

Combine terms.

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Step 1.5.1

Subtract $2$ from $0$ .

$\frac{\partial M}{\partial y} = - 2 + 0$

Step 1.5.2

Add $- 2$ and $0$ .

$\frac{\partial M}{\partial y} = - 2$

Step 2

Find $\frac{\partial N}{\partial x}$ where $N (x, y) = - (x - 3)$ .

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Step 2.1

Differentiate $N$ with respect to $x$ .

$\frac{\partial N}{\partial x} = \frac{d}{d x} [- (x - 3)]$

Step 2.2

Since $- 1$ is constant with respect to $x$ , the derivative of $- (x - 3)$ with respect to $x$ is $- \frac{d}{d x} [x - 3]$ .

$\frac{\partial N}{\partial x} = - \frac{d}{d x} [x - 3]$

Step 2.3

By the Sum Rule, the derivative of $x - 3$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [- 3]$ .

$\frac{\partial N}{\partial x} = - (\frac{d}{d x} [x] + \frac{d}{d x} [- 3])$

Step 2.4

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

$\frac{\partial N}{\partial x} = - (1 + \frac{d}{d x} [- 3])$

Step 2.5

Since $- 3$ is constant with respect to $x$ , the derivative of $- 3$ with respect to $x$ is $0$ .

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

Step 2.6

Simplify the expression.

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Step 2.6.1

Add $1$ and $0$ .

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

Step 2.6.2

Multiply $- 1$ by $1$ .

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

Step 3

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

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Step 3.1

Substitute $- 2$ for $\frac{\partial M}{\partial y}$ and $- 1$ for $\frac{\partial N}{\partial x}$ .

$- 2 = - 1$

Step 3.2

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

$- 2 = - 1$ 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 $- 2$ for $\frac{\partial M}{\partial y}$ .

$\frac{- 2 - \frac{\partial N}{\partial x}}{N}$

Step 4.2

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

$\frac{- 2 + 1}{N}$

Step 4.3

Substitute $- (x - 3)$ for $N$ .

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Step 4.3.1

Substitute $- (x - 3)$ for $N$ .

$\frac{- 2 + 1}{- (x - 3)}$

Step 4.3.2

Add $- 2$ and $1$ .

$\frac{- 1}{- (x - 3)}$

Step 4.3.3

Dividing two negative values results in a positive value.

$\frac{1}{x - 3}$

Step 4.4

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

$μ (x, y) = e^{\int \frac{1}{x - 3} d x}$

Step 5

Evaluate the integral $e^{\int \frac{1}{x - 3} d x}$ .

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Step 5.1

Let $u = x - 3$ . Then $d u = d x$ . Rewrite using $u$ and $d$ $u$ .

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Step 5.1.1

Let $u = x - 3$ . Find $\frac{d u}{d x}$ .

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Step 5.1.1.1

Differentiate $x - 3$ .

$\frac{d}{d x} [x - 3]$

Step 5.1.1.2

By the Sum Rule, the derivative of $x - 3$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [- 3]$ .

$\frac{d}{d x} [x] + \frac{d}{d x} [- 3]$

Step 5.1.1.3

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

$1 + \frac{d}{d x} [- 3]$

Step 5.1.1.4

Since $- 3$ is constant with respect to $x$ , the derivative of $- 3$ with respect to $x$ is $0$ .

$1 + 0$

Step 5.1.1.5

Add $1$ and $0$ .

$1$

Step 5.1.2

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

$μ (x, y) = e^{\int \frac{1}{u} d u}$

Step 5.2

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

$μ (x, y) = e^{\ln (| u |) + C}$

Step 5.3

Simplify.

$μ (x, y) = e^{\ln (| u |)} + C$

Step 5.4

Exponentiation and log are inverse functions.

$μ (x, y) = | u | + C$

Step 5.5

Replace all occurrences of $u$ with $x - 3$ .

$μ (x, y) = | x - 3 | + C$

Step 6

Multiply both sides of $(x - 2 y - 1) d x - (x - 3) d y = 0$ by the integration factor $x - 3$ .

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Step 6.1

Multiply $x - 2 y - 1$ by $x - 3$ .

$(x - 2 y - 1) (x - 3) d x - (x - 3) d y = 0$

Step 6.2

Expand $(x - 2 y - 1) (x - 3)$ by multiplying each term in the first expression by each term in the second expression.

$(x \cdot x + x \cdot - 3 - 2 y x - 2 y \cdot - 3 - 1 x - 1 \cdot - 3) d x - (x - 3) d y = 0$

Step 6.3

Simplify each term.

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Step 6.3.1

Multiply $x$ by $x$ .

$(x^{2} + x \cdot - 3 - 2 y x - 2 y \cdot - 3 - 1 x - 1 \cdot - 3) d x - (x - 3) d y = 0$

Step 6.3.2

Move $- 3$ to the left of $x$ .

$(x^{2} - 3 \cdot x - 2 y x - 2 y \cdot - 3 - 1 x - 1 \cdot - 3) d x - (x - 3) d y = 0$

Step 6.3.3

Multiply $- 3$ by $- 2$ .

$(x^{2} - 3 x - 2 y x + 6 y - 1 x - 1 \cdot - 3) d x - (x - 3) d y = 0$

Step 6.3.4

Rewrite $- 1 x$ as $- x$ .

$(x^{2} - 3 x - 2 y x + 6 y - x - 1 \cdot - 3) d x - (x - 3) d y = 0$

Step 6.3.5

Multiply $- 1$ by $- 3$ .

$(x^{2} - 3 x - 2 y x + 6 y - x + 3) d x - (x - 3) d y = 0$

Step 6.4

Subtract $x$ from $- 3 x$ .

$(x^{2} - 4 x - 2 y x + 6 y + 3) d x - (x - 3) d y = 0$

Step 6.5

Multiply $- (x - 3)$ by $x - 3$ .

$(x^{2} - 4 x - 2 y x + 6 y + 3) d x - (x - 3) (x - 3) d y = 0$

Step 6.6

Apply the distributive property.

$(x^{2} - 4 x - 2 y x + 6 y + 3) d x + (- x - - 3) (x - 3) d y = 0$

Step 6.7

Multiply $- 1$ by $- 3$ .

$(x^{2} - 4 x - 2 y x + 6 y + 3) d x + (- x + 3) (x - 3) d y = 0$

Step 6.8

Expand $(- x + 3) (x - 3)$ using the FOIL Method.

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Step 6.8.1

Apply the distributive property.

$(x^{2} - 4 x - 2 y x + 6 y + 3) d x + (- x (x - 3) + 3 (x - 3)) d y = 0$

Step 6.8.2

Apply the distributive property.

$(x^{2} - 4 x - 2 y x + 6 y + 3) d x + (- x \cdot x - x \cdot - 3 + 3 (x - 3)) d y = 0$

Step 6.8.3

Apply the distributive property.

$(x^{2} - 4 x - 2 y x + 6 y + 3) d x + (- x \cdot x - x \cdot - 3 + 3 x + 3 \cdot - 3) d y = 0$

Step 6.9

Simplify and combine like terms.

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Step 6.9.1

Simplify each term.

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Step 6.9.1.1

Multiply $x$ by $x$ by adding the exponents.

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Step 6.9.1.1.1

Move $x$ .

$(x^{2} - 4 x - 2 y x + 6 y + 3) d x + (- (x \cdot x) - x \cdot - 3 + 3 x + 3 \cdot - 3) d y = 0$

Step 6.9.1.1.2

Multiply $x$ by $x$ .

$(x^{2} - 4 x - 2 y x + 6 y + 3) d x + (- x^{2} - x \cdot - 3 + 3 x + 3 \cdot - 3) d y = 0$

Step 6.9.1.2

Multiply $- 3$ by $- 1$ .

$(x^{2} - 4 x - 2 y x + 6 y + 3) d x + (- x^{2} + 3 x + 3 x + 3 \cdot - 3) d y = 0$

Step 6.9.1.3

Multiply $3$ by $- 3$ .

$(x^{2} - 4 x - 2 y x + 6 y + 3) d x + (- x^{2} + 3 x + 3 x - 9) d y = 0$

Step 6.9.2

Add $3 x$ and $3 x$ .

$(x^{2} - 4 x - 2 y x + 6 y + 3) d x + (- x^{2} + 6 x - 9) d y = 0$

Step 7

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

$f (x, y) = \int - x^{2} + 6 x - 9 d y$

Step 8

Integrate $N (x, y) = - x^{2} + 6 x - 9$ to find $f (x, y)$ .

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Step 8.1

Apply the constant rule.

$f (x, y) = (- x^{2} + 6 x - 9) y + C$

Step 9

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

$f (x, y) = (- x^{2} + 6 x - 9) y + g (x)$

Step 10

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

$\frac{\partial f}{\partial x} = x^{2} - 4 x - 2 y x + 6 y + 3$

Step 11

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

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Step 11.1

Differentiate $f$ with respect to $x$ .

$\frac{d}{d x} [(- x^{2} + 6 x - 9) y + g (x)] = x^{2} - 4 x - 2 y x + 6 y + 3$

Step 11.2

By the Sum Rule, the derivative of $(- x^{2} + 6 x - 9) y + g (x)$ with respect to $x$ is $\frac{d}{d x} [(- x^{2} + 6 x - 9) y] + \frac{d}{d x} [g (x)]$ .

$\frac{d}{d x} [(- x^{2} + 6 x - 9) y] + \frac{d}{d x} [g (x)] = x^{2} - 4 x - 2 y x + 6 y + 3$

Step 11.3

Evaluate $\frac{d}{d x} [(- x^{2} + 6 x - 9) y]$ .

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Step 11.3.1

Since $y$ is constant with respect to $x$ , the derivative of $(- x^{2} + 6 x - 9) y$ with respect to $x$ is $y \frac{d}{d x} [- x^{2} + 6 x - 9]$ .

$y \frac{d}{d x} [- x^{2} + 6 x - 9] + \frac{d}{d x} [g (x)] = x^{2} - 4 x - 2 y x + 6 y + 3$

Step 11.3.2

By the Sum Rule, the derivative of $- x^{2} + 6 x - 9$ with respect to $x$ is $\frac{d}{d x} [- x^{2}] + \frac{d}{d x} [6 x] + \frac{d}{d x} [- 9]$ .

$y (\frac{d}{d x} [- x^{2}] + \frac{d}{d x} [6 x] + \frac{d}{d x} [- 9]) + \frac{d}{d x} [g (x)] = x^{2} - 4 x - 2 y x + 6 y + 3$

Step 11.3.3

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

$y (- \frac{d}{d x} [x^{2}] + \frac{d}{d x} [6 x] + \frac{d}{d x} [- 9]) + \frac{d}{d x} [g (x)] = x^{2} - 4 x - 2 y x + 6 y + 3$

Step 11.3.4

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 2$ .

$y (- (2 x) + \frac{d}{d x} [6 x] + \frac{d}{d x} [- 9]) + \frac{d}{d x} [g (x)] = x^{2} - 4 x - 2 y x + 6 y + 3$

Step 11.3.5

Since $6$ is constant with respect to $x$ , the derivative of $6 x$ with respect to $x$ is $6 \frac{d}{d x} [x]$ .

$y (- (2 x) + 6 \frac{d}{d x} [x] + \frac{d}{d x} [- 9]) + \frac{d}{d x} [g (x)] = x^{2} - 4 x - 2 y x + 6 y + 3$

Step 11.3.6

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

$y (- (2 x) + 6 \cdot 1 + \frac{d}{d x} [- 9]) + \frac{d}{d x} [g (x)] = x^{2} - 4 x - 2 y x + 6 y + 3$

Step 11.3.7

Since $- 9$ is constant with respect to $x$ , the derivative of $- 9$ with respect to $x$ is $0$ .

$y (- (2 x) + 6 \cdot 1 + 0) + \frac{d}{d x} [g (x)] = x^{2} - 4 x - 2 y x + 6 y + 3$

Step 11.3.8

Multiply $2$ by $- 1$ .

$y (- 2 x + 6 \cdot 1 + 0) + \frac{d}{d x} [g (x)] = x^{2} - 4 x - 2 y x + 6 y + 3$

Step 11.3.9

Multiply $6$ by $1$ .

$y (- 2 x + 6 + 0) + \frac{d}{d x} [g (x)] = x^{2} - 4 x - 2 y x + 6 y + 3$

Step 11.3.10

Add $- 2 x + 6$ and $0$ .

$y (- 2 x + 6) + \frac{d}{d x} [g (x)] = x^{2} - 4 x - 2 y x + 6 y + 3$

Step 11.4

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

$y (- 2 x + 6) + \frac{d g}{d x} = x^{2} - 4 x - 2 y x + 6 y + 3$

Step 11.5

Simplify.

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Step 11.5.1

Apply the distributive property.

$y (- 2 x) + y \cdot 6 + \frac{d g}{d x} = x^{2} - 4 x - 2 y x + 6 y + 3$

Step 11.5.2

Move $6$ to the left of $y$ .

$y (- 2 x) + 6 y + \frac{d g}{d x} = x^{2} - 4 x - 2 y x + 6 y + 3$

Step 11.5.3

Reorder terms.

$\frac{d g}{d x} - 2 x y + 6 y = x^{2} - 4 x - 2 y x + 6 y + 3$

Step 12

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

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Step 12.1

Move all terms not containing $\frac{d g}{d x}$ to the right side of the equation.

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Step 12.1.1

Add $2 x y$ to both sides of the equation.

$\frac{d g}{d x} + 6 y = x^{2} - 4 x - 2 y x + 6 y + 3 + 2 x y$

Step 12.1.2

Subtract $6 y$ from both sides of the equation.

$\frac{d g}{d x} = x^{2} - 4 x - 2 y x + 6 y + 3 + 2 x y - 6 y$

Step 12.1.3

Combine the opposite terms in $x^{2} - 4 x - 2 y x + 6 y + 3 + 2 x y - 6 y$ .

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Step 12.1.3.1

Reorder the factors in the terms $- 2 y x$ and $2 x y$ .

$\frac{d g}{d x} = x^{2} - 4 x - 2 x y + 6 y + 3 + 2 x y - 6 y$

Step 12.1.3.2

Add $- 2 x y$ and $2 x y$ .

$\frac{d g}{d x} = x^{2} - 4 x + 6 y + 3 + 0 - 6 y$

Step 12.1.3.3

Add $x^{2} - 4 x + 6 y + 3$ and $0$ .

$\frac{d g}{d x} = x^{2} - 4 x + 6 y + 3 - 6 y$

Step 12.1.3.4

Subtract $6 y$ from $6 y$ .

$\frac{d g}{d x} = x^{2} - 4 x + 0 + 3$

Step 12.1.3.5

Add $x^{2} - 4 x$ and $0$ .

$\frac{d g}{d x} = x^{2} - 4 x + 3$

Step 13

Find the antiderivative of $x^{2} - 4 x + 3$ to find $g (x)$ .

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Step 13.1

Integrate both sides of $\frac{d g}{d x} = x^{2} - 4 x + 3$ .

$\int \frac{d g}{d x} d x = \int x^{2} - 4 x + 3 d x$

Step 13.2

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

$g (x) = \int x^{2} - 4 x + 3 d x$

Step 13.3

Split the single integral into multiple integrals.

$g (x) = \int x^{2} d x + \int - 4 x d x + \int 3 d x$

Step 13.4

By the Power Rule, the integral of $x^{2}$ with respect to $x$ is $\frac{1}{3} x^{3}$ .

$g (x) = \frac{1}{3} x^{3} + C + \int - 4 x d x + \int 3 d x$

Step 13.5

Since $- 4$ is constant with respect to $x$ , move $- 4$ out of the integral.

$g (x) = \frac{1}{3} x^{3} + C - 4 \int x d x + \int 3 d x$

Step 13.6

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

$g (x) = \frac{1}{3} x^{3} + C - 4 (\frac{1}{2} x^{2} + C) + \int 3 d x$

Step 13.7

Apply the constant rule.

$g (x) = \frac{1}{3} x^{3} + C - 4 (\frac{1}{2} x^{2} + C) + 3 x + C$

Step 13.8

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

$g (x) = \frac{1}{3} x^{3} + C - 4 (\frac{x^{2}}{2} + C) + 3 x + C$

Step 13.9

Simplify.

$g (x) = \frac{x^{3}}{3} - 2 x^{2} + 3 x + C$

Step 13.10

Reorder terms.

$g (x) = \frac{1}{3} x^{3} - 2 x^{2} + 3 x + C$

Step 14

Substitute for $g (x)$ in $f (x, y) = (- x^{2} + 6 x - 9) y + g (x)$ .

$f (x, y) = (- x^{2} + 6 x - 9) y + \frac{1}{3} x^{3} - 2 x^{2} + 3 x + C$

Step 15

Simplify each term.

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Step 15.1

Apply the distributive property.

$f (x, y) = - x^{2} y + 6 x y - 9 y + \frac{1}{3} x^{3} - 2 x^{2} + 3 x + C$

Step 15.2

Combine $\frac{1}{3}$ and $x^{3}$ .

$f (x, y) = - x^{2} y + 6 x y - 9 y + \frac{x^{3}}{3} - 2 x^{2} + 3 x + C$