Calculus Examples

$(x y + 1) d x + x (x + 4 y - 2) d y = 0$

Step 1

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

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Step 1.1

Differentiate $M$ with respect to $y$ .

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

Step 1.2

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

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

Step 1.3

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

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Step 1.3.1

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

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

Step 1.3.2

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

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

Step 1.3.3

Multiply $x$ by $1$ .

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

Step 1.4

Differentiate using the Constant Rule.

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Step 1.4.1

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

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

Step 1.4.2

Add $x$ and $0$ .

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

Step 2

Find $\frac{\partial N}{\partial x}$ where $N (x, y) = x (x + 4 y - 2)$ .

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Step 2.1

Differentiate $N$ with respect to $x$ .

$\frac{\partial N}{\partial x} = \frac{d}{d x} [x (x + 4 y - 2)]$

Step 2.2

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 $g (x) = x + 4 y - 2$ .

$\frac{\partial N}{\partial x} = x \frac{d}{d x} [x + 4 y - 2] + (x + 4 y - 2) \frac{d}{d x} [x]$

Step 2.3

Differentiate.

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Step 2.3.1

By the Sum Rule, the derivative of $x + 4 y - 2$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [4 y] + \frac{d}{d x} [- 2]$ .

$\frac{\partial N}{\partial x} = x (\frac{d}{d x} [x] + \frac{d}{d x} [4 y] + \frac{d}{d x} [- 2]) + (x + 4 y - 2) \frac{d}{d x} [x]$

Step 2.3.2

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

$\frac{\partial N}{\partial x} = x (1 + \frac{d}{d x} [4 y] + \frac{d}{d x} [- 2]) + (x + 4 y - 2) \frac{d}{d x} [x]$

Step 2.3.3

Since $4 y$ is constant with respect to $x$ , the derivative of $4 y$ with respect to $x$ is $0$ .

$\frac{\partial N}{\partial x} = x (1 + 0 + \frac{d}{d x} [- 2]) + (x + 4 y - 2) \frac{d}{d x} [x]$

Step 2.3.4

Add $1$ and $0$ .

$\frac{\partial N}{\partial x} = x (1 + \frac{d}{d x} [- 2]) + (x + 4 y - 2) \frac{d}{d x} [x]$

Step 2.3.5

Since $- 2$ is constant with respect to $x$ , the derivative of $- 2$ with respect to $x$ is $0$ .

$\frac{\partial N}{\partial x} = x (1 + 0) + (x + 4 y - 2) \frac{d}{d x} [x]$

Step 2.3.6

Simplify the expression.

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Step 2.3.6.1

Add $1$ and $0$ .

$\frac{\partial N}{\partial x} = x \cdot 1 + (x + 4 y - 2) \frac{d}{d x} [x]$

Step 2.3.6.2

Multiply $x$ by $1$ .

$\frac{\partial N}{\partial x} = x + (x + 4 y - 2) \frac{d}{d x} [x]$

Step 2.3.7

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

$\frac{\partial N}{\partial x} = x + (x + 4 y - 2) \cdot 1$

Step 2.3.8

Simplify by adding terms.

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Step 2.3.8.1

Multiply $x + 4 y - 2$ by $1$ .

$\frac{\partial N}{\partial x} = x + x + 4 y - 2$

Step 2.3.8.2

Add $x$ and $x$ .

$\frac{\partial N}{\partial x} = 2 x + 4 y - 2$

Step 3

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

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Step 3.1

Substitute $x$ for $\frac{\partial M}{\partial y}$ and $2 x + 4 y - 2$ for $\frac{\partial N}{\partial x}$ .

$x = 2 x + 4 y - 2$

Step 3.2

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

$x = 2 x + 4 y - 2$ 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 $x$ for $\frac{\partial M}{\partial y}$ .

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

Step 4.2

Substitute $2 x + 4 y - 2$ for $\frac{\partial N}{\partial x}$ .

$\frac{x - (2 x + 4 y - 2)}{N}$

Step 4.3

Substitute $x (x + 4 y - 2)$ for $N$ .

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Step 4.3.1

Substitute $x (x + 4 y - 2)$ for $N$ .

$\frac{x - (2 x + 4 y - 2)}{x (x + 4 y - 2)}$

Step 4.3.2

Simplify the numerator.

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Step 4.3.2.1

Apply the distributive property.

$\frac{x - (2 x) - (4 y) - - 2}{x (x + 4 y - 2)}$

Step 4.3.2.2

Simplify.

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Step 4.3.2.2.1

Multiply $2$ by $- 1$ .

$\frac{x - 2 x - (4 y) - - 2}{x (x + 4 y - 2)}$

Step 4.3.2.2.2

Multiply $4$ by $- 1$ .

$\frac{x - 2 x - 4 y - - 2}{x (x + 4 y - 2)}$

Step 4.3.2.2.3

Multiply $- 1$ by $- 2$ .

$\frac{x - 2 x - 4 y + 2}{x (x + 4 y - 2)}$

Step 4.3.2.3

Subtract $2 x$ from $x$ .

$\frac{- x - 4 y + 2}{x (x + 4 y - 2)}$

Step 4.3.3

Cancel the common factor of $- x - 4 y + 2$ and $x + 4 y - 2$ .

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Step 4.3.3.1

Factor $- 1$ out of $- x$ .

$\frac{- (x) - 4 y + 2}{x (x + 4 y - 2)}$

Step 4.3.3.2

Factor $- 1$ out of $- 4 y$ .

$\frac{- (x) - (4 y) + 2}{x (x + 4 y - 2)}$

Step 4.3.3.3

Factor $- 1$ out of $- (x) - (4 y)$ .

$\frac{- (x + 4 y) + 2}{x (x + 4 y - 2)}$

Step 4.3.3.4

Rewrite $2$ as $- 1 (- 2)$ .

$\frac{- (x + 4 y) - 1 (- 2)}{x (x + 4 y - 2)}$

Step 4.3.3.5

Factor $- 1$ out of $- (x + 4 y) - 1 (- 2)$ .

$\frac{- (x + 4 y - 2)}{x (x + 4 y - 2)}$

Step 4.3.3.6

Rewrite $- (x + 4 y - 2)$ as $- 1 (x + 4 y - 2)$ .

$\frac{- 1 (x + 4 y - 2)}{x (x + 4 y - 2)}$

Step 4.3.3.7

Cancel the common factor.

$\frac{- 1 (x + 4 y - 2)}{x (x + 4 y - 2)}$

Step 4.3.3.8

Rewrite the expression.

$\frac{- 1}{x}$

Step 4.3.4

Move the negative in front of the fraction.

$- \frac{1}{x}$

Step 4.4

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

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

Step 5

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

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Step 5.1

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

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

Step 5.2

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

$μ (x, y) = e^{- (\ln (| x |) + C)}$

Step 5.3

Simplify.

$μ (x, y) = e^{- \ln (| x |)} + C$

Step 5.4

Simplify each term.

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Step 5.4.1

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

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

Step 5.4.2

Exponentiation and log are inverse functions.

$μ (x, y) = {| x |}^{- 1} + C$

Step 5.4.3

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

$μ (x, y) = \frac{1}{| x |} + C$

Step 6

Multiply both sides of $(x y + 1) d x + x (x + 4 y - 2) d y = 0$ by the integration factor $\frac{1}{x}$ .

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Step 6.1

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

$(x y + 1) \frac{1}{x} d x + x (x + 4 y - 2) d y = 0$

Step 6.2

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

$\frac{x y + 1}{x} d x + x (x + 4 y - 2) d y = 0$

Step 6.3

Multiply $x (x + 4 y - 2)$ by $\frac{1}{x}$ .

$\frac{x y + 1}{x} d x + x (x + 4 y - 2) \frac{1}{x} d y = 0$

Step 6.4

Cancel the common factor of $x$ .

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Step 6.4.1

Cancel the common factor.

$\frac{x y + 1}{x} d x + x (x + 4 y - 2) \frac{1}{x} d y = 0$

Step 6.4.2

Rewrite the expression.

$\frac{x y + 1}{x} d x + (x + 4 y - 2) d y = 0$

Step 7

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

$f (x, y) = \int x + 4 y - 2 d y$

Step 8

Integrate $N (x, y) = x + 4 y - 2$ to find $f (x, y)$ .

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Step 8.1

Split the single integral into multiple integrals.

$f (x, y) = \int x d y + \int 4 y d y + \int - 2 d y$

Step 8.2

Apply the constant rule.

$f (x, y) = x y + C + \int 4 y d y + \int - 2 d y$

Step 8.3

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

$f (x, y) = x y + C + 4 \int y d y + \int - 2 d y$

Step 8.4

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

$f (x, y) = x y + C + 4 (\frac{1}{2} y^{2} + C) + \int - 2 d y$

Step 8.5

Apply the constant rule.

$f (x, y) = x y + C + 4 (\frac{1}{2} y^{2} + C) - 2 y + C$

Step 8.6

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

$f (x, y) = x y + C + 4 (\frac{y^{2}}{2} + C) - 2 y + C$

Step 8.7

Simplify.

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

Step 9

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

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

Step 10

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

$\frac{\partial f}{\partial x} = \frac{x y + 1}{x}$

Step 11

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

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Step 11.1

Differentiate $f$ with respect to $x$ .

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

Step 11.2

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

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

Step 11.3

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

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Step 11.3.1

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

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

Step 11.3.2

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

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

Step 11.3.3

Multiply $y$ by $1$ .

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

Step 11.4

Differentiate using the Constant Rule.

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Step 11.4.1

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

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

Step 11.4.2

Since $- 2 y$ is constant with respect to $x$ , the derivative of $- 2 y$ with respect to $x$ is $0$ .

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

Step 11.5

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

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

Step 11.6

Simplify.

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Step 11.6.1

Combine terms.

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Step 11.6.1.1

Add $y$ and $0$ .

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

Step 11.6.1.2

Add $y$ and $0$ .

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

Step 11.6.2

Reorder terms.

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

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

Subtract $y$ from both sides of the equation.

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

Step 12.1.2

Simplify each term.

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Step 12.1.2.1

Split the fraction $\frac{x y + 1}{x}$ into two fractions.

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

Step 12.1.2.2

Cancel the common factor of $x$ .

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Step 12.1.2.2.1

Cancel the common factor.

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

Step 12.1.2.2.2

Divide $y$ by $1$ .

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

Step 12.1.3

Combine the opposite terms in $y + \frac{1}{x} - y$ .

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Step 12.1.3.1

Subtract $y$ from $y$ .

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

Step 12.1.3.2

Add $\frac{1}{x}$ and $0$ .

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

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

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