Calculus Examples

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

Step 1

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

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Step 1.1

Differentiate $M$ with respect to $y$ .

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

Step 1.2

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

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

Step 1.3

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

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Step 1.3.1

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

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

Step 1.3.2

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

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

Step 1.3.3

Multiply $2$ by $2$ .

$\frac{\partial M}{\partial y} = 4 x y + \frac{d}{d y} [- y]$

Step 1.4

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

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Step 1.4.1

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

$\frac{\partial M}{\partial y} = 4 x y - \frac{d}{d y} [y]$

Step 1.4.2

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

$\frac{\partial M}{\partial y} = 4 x y - 1 \cdot 1$

Step 1.4.3

Multiply $- 1$ by $1$ .

$\frac{\partial M}{\partial y} = 4 x y - 1$

Step 2

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

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Step 2.1

Differentiate $N$ with respect to $x$ .

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

Step 2.2

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

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

Step 3

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

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Step 3.1

Substitute $4 x y - 1$ for $\frac{\partial M}{\partial y}$ and $1$ for $\frac{\partial N}{\partial x}$ .

$4 x y - 1 = 1$

Step 3.2

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

$4 x y - 1 = 1$ is not an identity.

Step 4

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 4.1

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

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

Step 4.2

Substitute $4 x y - 1$ for $\frac{\partial M}{\partial y}$ .

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

Step 4.3

Substitute $2 x y^{2} - y$ for $M$ .

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Step 4.3.1

Substitute $2 x y^{2} - y$ for $M$ .

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

Step 4.3.2

Simplify the numerator.

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Step 4.3.2.1

Apply the distributive property.

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

Step 4.3.2.2

Multiply $4$ by $- 1$ .

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

Step 4.3.2.3

Multiply $- 1$ by $- 1$ .

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

Step 4.3.2.4

Add $1$ and $1$ .

$\frac{- 4 x y + 2}{2 x y^{2} - y}$

Step 4.3.2.5

Factor $2$ out of $- 4 x y + 2$ .

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Step 4.3.2.5.1

Factor $2$ out of $- 4 x y$ .

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

Step 4.3.2.5.2

Factor $2$ out of $2$ .

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

Step 4.3.2.5.3

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

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

Step 4.3.3

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

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Step 4.3.3.1

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

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

Step 4.3.3.2

Factor $y$ out of $- y$ .

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

Step 4.3.3.3

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

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

Step 4.3.4

Cancel the common factor of $- 2 x y + 1$ and $2 x y - 1$ .

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Step 4.3.4.1

Factor $- 1$ out of $- 2 x y$ .

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

Step 4.3.4.2

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

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

Step 4.3.4.3

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

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

Step 4.3.4.4

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

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

Step 4.3.4.5

Cancel the common factor.

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

Step 4.3.4.6

Rewrite the expression.

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

Step 4.3.5

Multiply $2$ by $- 1$ .

$\frac{- 2}{y}$

Step 4.3.6

Substitute $2 x y^{2} - y$ for $M$ .

$- \frac{2}{y}$

Step 4.4

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 5

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

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Step 5.1

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 5.2

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 5.3

Multiply $2$ by $- 1$ .

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

Step 5.4

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

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

Step 5.5

Simplify.

$μ (x, y) = e^{- 2 \ln (| y |)} + C$

Step 5.6

Simplify each term.

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Step 5.6.1

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

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

Step 5.6.2

Exponentiation and log are inverse functions.

$μ (x, y) = {| y |}^{- 2} + C$

Step 5.6.3

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

$μ (x, y) = y^{- 2} + C$

Step 5.6.4

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

$μ (x, y) = \frac{1}{y^{2}} + C$

Step 6

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

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Step 6.1

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

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

Step 6.2

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

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

Step 6.3

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

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Step 6.3.1

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

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

Step 6.3.2

Factor $y$ out of $- y$ .

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

Step 6.3.3

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

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

Step 6.4

Cancel the common factors.

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Step 6.4.1

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

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

Step 6.4.2

Cancel the common factor.

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

Step 6.4.3

Rewrite the expression.

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

Step 6.5

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

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

Step 6.6

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

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

Step 7

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

$f (x, y) = \int \frac{x}{y^{2}} d y$

Step 8

Integrate $N (x, y) = \frac{x}{y^{2}}$ to find $f (x, y)$ .

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Step 8.1

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

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

Step 8.2

Move $y^{2}$ out of the denominator by raising it to the $- 1$ power.

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

Step 8.3

Multiply the exponents in ${(y^{2})}^{- 1}$ .

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Step 8.3.1

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

$f (x, y) = x \int y^{2 \cdot - 1} d y$

Step 8.3.2

Multiply $2$ by $- 1$ .

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

Step 8.4

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

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

Step 8.5

Simplify the answer.

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Step 8.5.1

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

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

Step 8.5.2

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

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

Step 9

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

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

Step 10

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

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

Step 11

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

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Step 11.1

Differentiate $f$ with respect to $x$ .

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

Step 11.2

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

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

Step 11.3

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

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Step 11.3.1

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

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

Step 11.3.2

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

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

Step 11.3.3

Multiply $- 1$ by $1$ .

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

Step 11.4

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

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

Step 11.5

Reorder terms.

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

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 $\frac{2 x y - 1}{y}$ from both sides of the equation.

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

Step 12.1.1.2

Combine the numerators over the common denominator.

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

Step 12.1.1.3

Simplify each term.

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Step 12.1.1.3.1

Apply the distributive property.

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

Step 12.1.1.3.2

Multiply $2$ by $- 1$ .

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

Step 12.1.1.3.3

Multiply $- 1$ by $- 1$ .

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

Step 12.1.1.4

Combine the opposite terms in $- 1 - 2 x y + 1$ .

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Step 12.1.1.4.1

Add $- 1$ and $1$ .

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

Step 12.1.1.4.2

Add $- 2 x y$ and $0$ .

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

Step 12.1.1.5

Cancel the common factor of $y$ .

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Step 12.1.1.5.1

Cancel the common factor.

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

Step 12.1.1.5.2

Divide $- 2 x$ by $1$ .

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

Step 12.1.2

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

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

Step 13

Find the antiderivative of $2 x$ to find $g (x)$ .

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Step 13.1

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

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

Step 13.2

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

$g (x) = \int 2 x d x$

Step 13.3

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

$g (x) = 2 \int x d x$

Step 13.4

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

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

Step 13.5

Simplify the answer.

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Step 13.5.1

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

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

Step 13.5.2

Simplify.

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Step 13.5.2.1

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

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

Step 13.5.2.2

Cancel the common factor of $2$ .

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Step 13.5.2.2.1

Cancel the common factor.

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

Step 13.5.2.2.2

Rewrite the expression.

$g (x) = 1 x^{2} + C$

Step 13.5.2.3

Multiply $x^{2}$ by $1$ .

$g (x) = x^{2} + C$

Step 14

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

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