Calculus Examples

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

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 $N (x, y) = 2 x^{2} + y^{2}$ .

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Step 2.1

Differentiate $N$ with respect to $x$ .

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

Step 2.2

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

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

Step 2.3

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

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Step 2.3.1

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

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

Step 2.3.2

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

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

Step 2.3.3

Multiply $2$ by $2$ .

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

Step 2.4

Differentiate using the Constant Rule.

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Step 2.4.1

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

$\frac{\partial N}{\partial x} = 4 x + 0$

Step 2.4.2

Add $4 x$ and $0$ .

$\frac{\partial N}{\partial x} = 4 x$

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 $4 x$ for $\frac{\partial N}{\partial x}$ .

$x = 4 x$

Step 3.2

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

$x = 4 x$ 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 $4 x$ for $\frac{\partial N}{\partial x}$ .

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

Step 4.2

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

$\frac{4 x - x}{M}$

Step 4.3

Substitute $x y$ for $M$ .

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Step 4.3.1

Substitute $x y$ for $M$ .

$\frac{4 x - x}{x y}$

Step 4.3.2

Subtract $x$ from $4 x$ .

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

Step 4.3.3

Substitute $x y$ for $M$ .

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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 4.4

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

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

Step 5.2

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

$μ (x, y) = e^{3 (\ln (| y |) + C)}$

Step 5.3

Simplify.

$μ (x, y) = e^{3 \ln (| y |)} + C$

Step 5.4

Simplify each term.

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Step 5.4.1

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

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

Step 5.4.2

Exponentiation and log are inverse functions.

$μ (x, y) = {| y |}^{3} + C$

Step 6

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

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Step 6.1

Multiply $x y$ by $y^{3}$ .

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

Step 6.2

Multiply $y$ by $y^{3}$ by adding the exponents.

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Step 6.2.1

Move $y^{3}$ .

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

Step 6.2.2

Multiply $y^{3}$ by $y$ .

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Step 6.2.2.1

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

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

Step 6.2.2.2

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

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

Step 6.2.3

Add $3$ and $1$ .

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

Step 6.3

Multiply $2 x^{2} + y^{2}$ by $y^{3}$ .

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

Step 6.4

Apply the distributive property.

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

Step 6.5

Multiply $y^{2}$ by $y^{3}$ by adding the exponents.

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Step 6.5.1

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

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

Step 6.5.2

Add $2$ and $3$ .

$x y^{4} d x + (2 x^{2} y^{3} + y^{5}) d y = 0$

Step 7

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

$f (x, y) = \int x y^{4} d x$

Step 8

Integrate $M (x, y) = x y^{4}$ to find $f (x, y)$ .

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Step 8.1

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

$f (x, y) = y^{4} \int x d x$

Step 8.2

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

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

Step 8.3

Simplify the answer.

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Step 8.3.1

Rewrite $y^{4} (\frac{1}{2} x^{2} + C)$ as $y^{4} \frac{1}{2} x^{2} + C$ .

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

Step 8.3.2

Simplify.

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Step 8.3.2.1

Combine $y^{4}$ and $\frac{1}{2}$ .

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

Step 8.3.2.2

Combine $\frac{y^{4}}{2}$ and $x^{2}$ .

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

Step 8.3.3

Reorder terms.

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

Step 9

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

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

Step 10

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

$\frac{\partial f}{\partial y} = 2 x^{2} y^{3} + y^{5}$

Step 11

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

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Step 11.1

Differentiate $f$ with respect to $y$ .

$\frac{d}{d y} [\frac{1}{2} y^{4} x^{2} + g (y)] = 2 x^{2} y^{3} + y^{5}$

Step 11.2

By the Sum Rule, the derivative of $\frac{1}{2} y^{4} x^{2} + g (y)$ with respect to $y$ is $\frac{d}{d y} [\frac{1}{2} y^{4} x^{2}] + \frac{d}{d y} [g (y)]$ .

$\frac{d}{d y} [\frac{1}{2} y^{4} x^{2}] + \frac{d}{d y} [g (y)] = 2 x^{2} y^{3} + y^{5}$

Step 11.3

Evaluate $\frac{d}{d y} [\frac{1}{2} y^{4} x^{2}]$ .

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Step 11.3.1

Combine $\frac{1}{2}$ and $y^{4}$ .

$\frac{d}{d y} [\frac{y^{4}}{2} x^{2}] + \frac{d}{d y} [g (y)] = 2 x^{2} y^{3} + y^{5}$

Step 11.3.2

Combine $\frac{y^{4}}{2}$ and $x^{2}$ .

$\frac{d}{d y} [\frac{y^{4} x^{2}}{2}] + \frac{d}{d y} [g (y)] = 2 x^{2} y^{3} + y^{5}$

Step 11.3.3

Since $\frac{x^{2}}{2}$ is constant with respect to $y$ , the derivative of $\frac{y^{4} x^{2}}{2}$ with respect to $y$ is $\frac{x^{2}}{2} \frac{d}{d y} [y^{4}]$ .

$\frac{x^{2}}{2} \frac{d}{d y} [y^{4}] + \frac{d}{d y} [g (y)] = 2 x^{2} y^{3} + y^{5}$

Step 11.3.4

Differentiate using the Power Rule which states that $\frac{d}{d y} [y^{n}]$ is $n y^{n - 1}$ where $n = 4$ .

$\frac{x^{2}}{2} (4 y^{3}) + \frac{d}{d y} [g (y)] = 2 x^{2} y^{3} + y^{5}$

Step 11.3.5

Combine $4$ and $\frac{x^{2}}{2}$ .

$\frac{4 x^{2}}{2} y^{3} + \frac{d}{d y} [g (y)] = 2 x^{2} y^{3} + y^{5}$

Step 11.3.6

Combine $\frac{4 x^{2}}{2}$ and $y^{3}$ .

$\frac{4 x^{2} y^{3}}{2} + \frac{d}{d y} [g (y)] = 2 x^{2} y^{3} + y^{5}$

Step 11.3.7

Cancel the common factor of $4$ and $2$ .

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Step 11.3.7.1

Factor $2$ out of $4 x^{2} y^{3}$ .

$\frac{2 (2 x^{2} y^{3})}{2} + \frac{d}{d y} [g (y)] = 2 x^{2} y^{3} + y^{5}$

Step 11.3.7.2

Cancel the common factors.

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Step 11.3.7.2.1

Factor $2$ out of $2$ .

$\frac{2 (2 x^{2} y^{3})}{2 (1)} + \frac{d}{d y} [g (y)] = 2 x^{2} y^{3} + y^{5}$

Step 11.3.7.2.2

Cancel the common factor.

$\frac{2 (2 x^{2} y^{3})}{2 \cdot 1} + \frac{d}{d y} [g (y)] = 2 x^{2} y^{3} + y^{5}$

Step 11.3.7.2.3

Rewrite the expression.

$\frac{2 x^{2} y^{3}}{1} + \frac{d}{d y} [g (y)] = 2 x^{2} y^{3} + y^{5}$

Step 11.3.7.2.4

Divide $2 x^{2} y^{3}$ by $1$ .

$2 x^{2} y^{3} + \frac{d}{d y} [g (y)] = 2 x^{2} y^{3} + y^{5}$

Step 11.4

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

$2 x^{2} y^{3} + \frac{d g}{d y} = 2 x^{2} y^{3} + y^{5}$

Step 11.5

Reorder terms.

$\frac{d g}{d y} + 2 y^{3} x^{2} = 2 x^{2} y^{3} + y^{5}$

Step 12

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

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Step 12.1

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

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Step 12.1.1

Subtract $2 y^{3} x^{2}$ from both sides of the equation.

$\frac{d g}{d y} = 2 x^{2} y^{3} + y^{5} - 2 y^{3} x^{2}$

Step 12.1.2

Combine the opposite terms in $2 x^{2} y^{3} + y^{5} - 2 y^{3} x^{2}$ .

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Step 12.1.2.1

Reorder the factors in the terms $2 x^{2} y^{3}$ and $- 2 y^{3} x^{2}$ .

$\frac{d g}{d y} = 2 y^{3} x^{2} + y^{5} - 2 y^{3} x^{2}$

Step 12.1.2.2

Subtract $2 y^{3} x^{2}$ from $2 y^{3} x^{2}$ .

$\frac{d g}{d y} = y^{5} + 0$

Step 12.1.2.3

Add $y^{5}$ and $0$ .

$\frac{d g}{d y} = y^{5}$

Step 13

Find the antiderivative of $y^{5}$ to find $g (y)$ .

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Step 13.1

Integrate both sides of $\frac{d g}{d y} = y^{5}$ .

$\int \frac{d g}{d y} d y = \int y^{5} d y$

Step 13.2

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

$g (y) = \int y^{5} d y$

Step 13.3

By the Power Rule, the integral of $y^{5}$ with respect to $y$ is $\frac{1}{6} y^{6}$ .

$g (y) = \frac{1}{6} y^{6} + C$

Step 14

Substitute for $g (y)$ in $f (x, y) = \frac{1}{2} y^{4} x^{2} + g (y)$ .

$f (x, y) = \frac{1}{2} y^{4} x^{2} + \frac{1}{6} y^{6} + C$

Step 15

Simplify each term.

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Step 15.1

Combine $\frac{1}{2}$ and $y^{4}$ .

$f (x, y) = \frac{y^{4}}{2} x^{2} + \frac{1}{6} y^{6} + C$

Step 15.2

Combine $\frac{y^{4}}{2}$ and $x^{2}$ .

$f (x, y) = \frac{y^{4} x^{2}}{2} + \frac{1}{6} y^{6} + C$

Step 15.3

Combine $\frac{1}{6}$ and $y^{6}$ .

$f (x, y) = \frac{y^{4} x^{2}}{2} + \frac{y^{6}}{6} + C$