Calculus Examples

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

Step 1

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

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Step 1.1

Differentiate $M$ with respect to $y$ .

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

Step 1.2

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

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

Step 1.3

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

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

Step 1.4

Multiply $x$ by $1$ .

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

Step 2

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

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Step 2.1

Differentiate $N$ with respect to $x$ .

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

Step 2.2

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

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

Step 2.3

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

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

Step 2.4

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

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

Step 2.5

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

$\frac{\partial N}{\partial x} = - (2 x + 0)$

Step 2.6

Simplify the expression.

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Step 2.6.1

Add $2 x$ and $0$ .

$\frac{\partial N}{\partial x} = - (2 x)$

Step 2.6.2

Multiply $2$ by $- 1$ .

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

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

$x = - 2 x$

Step 3.2

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

$x = - 2 x$ 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 $- 2 x$ for $\frac{\partial N}{\partial x}$ .

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

Step 4.2

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

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

Step 4.3

Substitute $x y$ for $M$ .

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Step 4.3.1

Substitute $x y$ for $M$ .

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

Step 4.3.2

Subtract $x$ from $- 2 x$ .

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

Step 4.3.3

Cancel the common factor of $x$ .

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Step 4.3.3.1

Cancel the common factor.

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

Step 4.3.3.2

Rewrite the expression.

$\frac{- 3}{y}$

Step 4.3.4

Substitute $x y$ for $M$ .

$- \frac{3}{y}$

Step 4.4

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

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

Step 5

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

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Step 5.1

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

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

Step 5.2

Since $3$ is constant with respect to $y$ , move $3$ out of the integral.

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

Step 5.3

Multiply $3$ by $- 1$ .

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

Step 5.4

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

$μ (x, y) = e^{- 3 (\ln (| y |) + C)}$

Step 5.5

Simplify.

$μ (x, y) = e^{- 3 \ln (| y |)} + C$

Step 5.6

Simplify each term.

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Step 5.6.1

Simplify $- 3 \ln (| y |)$ by moving $- 3$ inside the logarithm.

$μ (x, y) = e^{\ln ({| y |}^{- 3})} + C$

Step 5.6.2

Exponentiation and log are inverse functions.

$μ (x, y) = {| y |}^{- 3} + C$

Step 5.6.3

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

$μ (x, y) = \frac{1}{{| y |}^{3}} + C$

Step 6

Multiply both sides of $x y d x - (x^{2} + y^{2}) d y = 0$ by the integration factor $\frac{1}{y^{3}}$ .

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Step 6.1

Multiply $x y$ by $\frac{1}{y^{3}}$ .

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

Step 6.2

Cancel the common factor of $y$ .

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Step 6.2.1

Factor $y$ out of $x y$ .

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

Step 6.2.2

Factor $y$ out of $y^{3}$ .

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

Step 6.2.3

Cancel the common factor.

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

Step 6.2.4

Rewrite the expression.

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

Step 6.3

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

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

Step 6.4

Multiply $- (x^{2} + y^{2})$ by $\frac{1}{y^{3}}$ .

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

Step 6.5

Apply the distributive property.

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

Step 6.6

Multiply $- x^{2} - y^{2}$ by $\frac{1}{y^{3}}$ .

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

Step 6.7

Factor $- 1$ out of $- x^{2}$ .

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

Step 6.8

Factor $- 1$ out of $- y^{2}$ .

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

Step 6.9

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

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

Step 6.10

Rewrite $- (x^{2} + y^{2})$ as $- 1 (x^{2} + y^{2})$ .

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

Step 6.11

Move the negative in front of the fraction.

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

Step 7

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

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

Step 8

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

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Step 8.1

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

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

Step 8.2

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

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

Step 8.3

Simplify the answer.

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Step 8.3.1

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

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

Step 8.3.2

Simplify.

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Step 8.3.2.1

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

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

Step 8.3.2.2

Move $2$ to the left of $y^{2}$ .

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

Step 8.3.2.3

Multiply $2$ by $y^{2}$ .

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

Step 8.3.2.4

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

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

Step 9

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

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

Step 10

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

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

Step 11

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

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Step 11.1

Differentiate $f$ with respect to $y$ .

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

Step 11.2

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

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

Step 11.3

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

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Step 11.3.1

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

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

Step 11.3.2

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

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

Step 11.3.3

Differentiate using the chain rule, which states that $\frac{d}{d y} [f (g (y))]$ is $f' (g (y)) g' (y)$ where $f (y) = y^{- 1}$ and $g (y) = y^{2}$ .

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Step 11.3.3.1

To apply the Chain Rule, set $u$ as $y^{2}$ .

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

Step 11.3.3.2

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

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

Step 11.3.3.3

Replace all occurrences of $u$ with $y^{2}$ .

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

Step 11.3.4

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

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

Step 11.3.5

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

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Step 11.3.5.1

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

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

Step 11.3.5.2

Multiply $2$ by $- 2$ .

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

Step 11.3.6

Multiply $2$ by $- 1$ .

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

Step 11.3.7

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

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

Step 11.3.8

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

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

Step 11.3.9

Subtract $4$ from $1$ .

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

Step 11.3.10

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

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

Step 11.3.11

Combine $\frac{- 2 x^{2}}{2}$ and $y^{- 3}$ .

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

Step 11.3.12

Move $y^{- 3}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

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

Step 11.3.13

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

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Step 11.3.13.1

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

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

Step 11.3.13.2

Cancel the common factors.

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Step 11.3.13.2.1

Factor $2$ out of $2 y^{3}$ .

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

Step 11.3.13.2.2

Cancel the common factor.

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

Step 11.3.13.2.3

Rewrite the expression.

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

Step 11.3.14

Move the negative in front of the fraction.

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

Step 11.4

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

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

Step 11.5

Reorder terms.

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

Step 12

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

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Step 12.1

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

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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

Add $\frac{x^{2} + y^{2}}{y^{3}}$ to both sides of the equation.

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

Step 12.1.1.2

Combine the numerators over the common denominator.

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

Step 12.1.1.3

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

$\frac{d g}{d y} + \frac{0 + y^{2}}{y^{3}} = 0$

Step 12.1.1.4

Add $0$ and $y^{2}$ .

$\frac{d g}{d y} + \frac{y^{2}}{y^{3}} = 0$

Step 12.1.1.5

Cancel the common factor of $y^{2}$ and $y^{3}$ .

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Step 12.1.1.5.1

Multiply by $1$ .

$\frac{d g}{d y} + \frac{y^{2} \cdot 1}{y^{3}} = 0$

Step 12.1.1.5.2

Cancel the common factors.

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Step 12.1.1.5.2.1

Factor $y^{2}$ out of $y^{3}$ .

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

Step 12.1.1.5.2.2

Cancel the common factor.

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

Step 12.1.1.5.2.3

Rewrite the expression.

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

Step 12.1.2

Subtract $\frac{1}{y}$ from both sides of the equation.

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

Step 13

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

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Step 13.1

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

$\int \frac{d g}{d y} d y = \int - \frac{1}{y} d y$

Step 13.2

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

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

Step 13.3

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

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

Step 13.4

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

$g (y) = - (\ln (| y |) + C)$

Step 13.5

Simplify.

$g (y) = - \ln (| y |) + C$

Step 14

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

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