Calculus Examples

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

Step 1

Find $\frac{\partial M}{\partial y}$ where $M (x, y) = 2 y^{3} - x^{3}$ .

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Step 1.1

Differentiate $M$ with respect to $y$ .

$\frac{\partial M}{\partial y} = \frac{d}{d y} [2 y^{3} - x^{3}]$

Step 1.2

By the Sum Rule, the derivative of $2 y^{3} - x^{3}$ with respect to $y$ is $\frac{d}{d y} [2 y^{3}] + \frac{d}{d y} [- x^{3}]$ .

$\frac{\partial M}{\partial y} = \frac{d}{d y} [2 y^{3}] + \frac{d}{d y} [- x^{3}]$

Step 1.3

Evaluate $\frac{d}{d y} [2 y^{3}]$ .

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Step 1.3.1

Since $2$ is constant with respect to $y$ , the derivative of $2 y^{3}$ with respect to $y$ is $2 \frac{d}{d y} [y^{3}]$ .

$\frac{\partial M}{\partial y} = 2 \frac{d}{d y} [y^{3}] + \frac{d}{d y} [- x^{3}]$

Step 1.3.2

Differentiate using the Power Rule which states that $\frac{d}{d y} [y^{n}]$ is $n y^{n - 1}$ where $n = 3$ .

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

Step 1.3.3

Multiply $3$ by $2$ .

$\frac{\partial M}{\partial y} = 6 y^{2} + \frac{d}{d y} [- x^{3}]$

Step 1.4

Differentiate using the Constant Rule.

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Step 1.4.1

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

$\frac{\partial M}{\partial y} = 6 y^{2} + 0$

Step 1.4.2

Add $6 y^{2}$ and $0$ .

$\frac{\partial M}{\partial y} = 6 y^{2}$

Step 2

Find $\frac{\partial N}{\partial x}$ where $N (x, y) = 3 x y^{2}$ .

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Step 2.1

Differentiate $N$ with respect to $x$ .

$\frac{\partial N}{\partial x} = \frac{d}{d x} [3 x y^{2}]$

Step 2.2

Since $3 y^{2}$ is constant with respect to $x$ , the derivative of $3 x y^{2}$ with respect to $x$ is $3 y^{2} \frac{d}{d x} [x]$ .

$\frac{\partial N}{\partial x} = 3 y^{2} \frac{d}{d x} [x]$

Step 2.3

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

$\frac{\partial N}{\partial x} = 3 y^{2} \cdot 1$

Step 2.4

Multiply $3$ by $1$ .

$\frac{\partial N}{\partial x} = 3 y^{2}$

Step 3

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

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Step 3.1

Substitute $6 y^{2}$ for $\frac{\partial M}{\partial y}$ and $3 y^{2}$ for $\frac{\partial N}{\partial x}$ .

$6 y^{2} = 3 y^{2}$

Step 3.2

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

$6 y^{2} = 3 y^{2}$ 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 $6 y^{2}$ for $\frac{\partial M}{\partial y}$ .

$\frac{6 y^{2} - \frac{\partial N}{\partial x}}{N}$

Step 4.2

Substitute $3 y^{2}$ for $\frac{\partial N}{\partial x}$ .

$\frac{6 y^{2} - (3 y^{2})}{N}$

Step 4.3

Substitute $3 x y^{2}$ for $N$ .

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Step 4.3.1

Substitute $3 x y^{2}$ for $N$ .

$\frac{6 y^{2} - (3 y^{2})}{3 x y^{2}}$

Step 4.3.2

Simplify the numerator.

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Step 4.3.2.1

Factor $y^{2}$ out of $6 y^{2} - 1 \cdot 3 y^{2}$ .

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Step 4.3.2.1.1

Factor $y^{2}$ out of $6 y^{2}$ .

$\frac{y^{2} \cdot 6 - 1 \cdot 3 y^{2}}{3 x y^{2}}$

Step 4.3.2.1.2

Factor $y^{2}$ out of $- 1 \cdot 3 y^{2}$ .

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

Step 4.3.2.1.3

Factor $y^{2}$ out of $y^{2} \cdot 6 + y^{2} (- 1 \cdot 3)$ .

$\frac{y^{2} (6 - 1 \cdot 3)}{3 x y^{2}}$

Step 4.3.2.2

Multiply $- 1$ by $3$ .

$\frac{y^{2} (6 - 3)}{3 x y^{2}}$

Step 4.3.2.3

Subtract $3$ from $6$ .

$\frac{y^{2} \cdot 3}{3 x y^{2}}$

Step 4.3.3

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

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Step 4.3.3.1

Cancel the common factor.

$\frac{y^{2} \cdot 3}{3 x y^{2}}$

Step 4.3.3.2

Rewrite the expression.

$\frac{3}{3 x}$

Step 4.3.4

Cancel the common factor of $3$ .

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Step 4.3.4.1

Cancel the common factor.

$\frac{3}{3 x}$

Step 4.3.4.2

Rewrite the expression.

$\frac{1}{x}$

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

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

$μ (x, y) = e^{\ln (| x |) + C}$

Step 5.2

Simplify the answer.

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Step 5.2.1

Simplify.

$μ (x, y) = e^{\ln (| x |)} + C$

Step 5.2.2

Exponentiation and log are inverse functions.

$μ (x, y) = | x | + C$

Step 6

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

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Step 6.1

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

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

Step 6.2

Apply the distributive property.

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

Step 6.3

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

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Step 6.3.1

Move $x$ .

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

Step 6.3.2

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

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Step 6.3.2.1

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

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

Step 6.3.2.2

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

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

Step 6.3.3

Add $1$ and $3$ .

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

Step 6.4

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

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

Step 6.5

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

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Step 6.5.1

Move $x$ .

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

Step 6.5.2

Multiply $x$ by $x$ .

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

Step 7

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

$f (x, y) = \int 3 x^{2} y^{2} d y$

Step 8

Integrate $N (x, y) = 3 x^{2} y^{2}$ to find $f (x, y)$ .

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Step 8.1

Since $3 x^{2}$ is constant with respect to $y$ , move $3 x^{2}$ out of the integral.

$f (x, y) = 3 x^{2} \int y^{2} d y$

Step 8.2

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

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

Step 8.3

Simplify the answer.

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Step 8.3.1

Rewrite $3 x^{2} (\frac{1}{3} y^{3} + C)$ as $3 x^{2} \frac{1}{3} y^{3} + C$ .

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

Step 8.3.2

Simplify.

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Step 8.3.2.1

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

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

Step 8.3.2.2

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

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

Step 8.3.2.3

Move $3$ to the left of $x^{2}$ .

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

Step 8.3.2.4

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

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

Step 8.3.2.5

Cancel the common factor of $3$ .

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Step 8.3.2.5.1

Cancel the common factor.

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

Step 8.3.2.5.2

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

$f (x, y) = x^{2} y^{3} + C$

Step 9

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

$f (x, y) = x^{2} y^{3} + g (x)$

Step 10

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

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

Step 11

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

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Step 11.1

Differentiate $f$ with respect to $x$ .

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

Step 11.2

By the Sum Rule, the derivative of $x^{2} y^{3} + g (x)$ with respect to $x$ is $\frac{d}{d x} [x^{2} y^{3}] + \frac{d}{d x} [g (x)]$ .

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

Step 11.3

Evaluate $\frac{d}{d x} [x^{2} y^{3}]$ .

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Step 11.3.1

Since $y^{3}$ is constant with respect to $x$ , the derivative of $x^{2} y^{3}$ with respect to $x$ is $y^{3} \frac{d}{d x} [x^{2}]$ .

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

Step 11.3.2

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

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

Step 11.3.3

Move $2$ to the left of $y^{3}$ .

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

Step 11.4

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

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

Step 11.5

Reorder terms.

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

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

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

Step 12.1.2

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

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Step 12.1.2.1

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

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

Step 12.1.2.2

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

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

Step 13

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

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Step 13.1

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

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

Step 13.2

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

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

Step 13.3

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

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

Step 13.4

By the Power Rule, the integral of $x^{4}$ with respect to $x$ is $\frac{1}{5} x^{5}$ .

$g (x) = - (\frac{1}{5} x^{5} + C)$

Step 13.5

Rewrite $- (\frac{1}{5} x^{5} + C)$ as $- \frac{1}{5} x^{5} + C$ .

$g (x) = - \frac{1}{5} x^{5} + C$

Step 14

Substitute for $g (x)$ in $f (x, y) = x^{2} y^{3} + g (x)$ .

$f (x, y) = x^{2} y^{3} - \frac{1}{5} x^{5} + C$

Step 15

Combine $x^{5}$ and $\frac{1}{5}$ .

$f (x, y) = x^{2} y^{3} - \frac{x^{5}}{5} + C$