Calculus Examples

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

Step 1

Rewrite the differential equation to fit the Exact differential equation technique.

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Step 1.1

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

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

Step 1.2

Rewrite.

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

Step 2

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

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Step 2.1

Differentiate $M$ with respect to $y$ .

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

Step 2.2

Since $- 1$ is constant with respect to $y$ , the derivative of $- (y + 2 x^{3})$ with respect to $y$ is $- \frac{d}{d y} [y + 2 x^{3}]$ .

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

Step 2.3

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

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

Step 2.4

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

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

Step 2.5

Since $2 x^{3}$ is constant with respect to $y$ , the derivative of $2 x^{3}$ with respect to $y$ is $0$ .

$\frac{\partial M}{\partial y} = - (1 + 0)$

Step 2.6

Simplify the expression.

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Step 2.6.1

Add $1$ and $0$ .

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

Step 2.6.2

Multiply $- 1$ by $1$ .

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

Step 3

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

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Step 3.1

Differentiate $N$ with respect to $x$ .

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

Step 3.2

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

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

Step 4

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

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Step 4.1

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

$- 1 = 1$

Step 4.2

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

$- 1 = 1$ is not an identity.

Step 5

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 5.1

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

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

Step 5.2

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

$\frac{- 1 - 1}{N}$

Step 5.3

Substitute $x$ for $N$ .

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Step 5.3.1

Substitute $x$ for $N$ .

$\frac{- 1 - 1}{x}$

Step 5.3.2

Subtract $1$ from $- 1$ .

$\frac{- 2}{x}$

Step 5.3.3

Move the negative in front of the fraction.

$- \frac{2}{x}$

Step 5.4

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

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

Step 6

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

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Step 6.1

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

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

Step 6.2

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

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

Step 6.3

Multiply $2$ by $- 1$ .

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

Step 6.4

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

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

Step 6.5

Simplify.

$μ (x, y) = e^{- 2 \ln (| x |)} + C$

Step 6.6

Simplify each term.

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Step 6.6.1

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

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

Step 6.6.2

Exponentiation and log are inverse functions.

$μ (x, y) = {| x |}^{- 2} + C$

Step 6.6.3

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

$μ (x, y) = x^{- 2} + C$

Step 6.6.4

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

$μ (x, y) = \frac{1}{x^{2}} + C$

Step 7

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

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Step 7.1

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

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

Step 7.2

Apply the distributive property.

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

Step 7.3

Multiply $2$ by $- 1$ .

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

Step 7.4

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

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

Step 7.5

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

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

Step 7.6

Factor $- 1$ out of $- 2 x^{3}$ .

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

Step 7.7

Factor $- 1$ out of $- (y) - (2 x^{3})$ .

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

Step 7.8

Rewrite $- (y + 2 x^{3})$ as $- 1 (y + 2 x^{3})$ .

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

Step 7.9

Move the negative in front of the fraction.

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

Step 7.10

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

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

Step 7.11

Cancel the common factor of $x$ .

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Step 7.11.1

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

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

Step 7.11.2

Cancel the common factor.

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

Step 7.11.3

Rewrite the expression.

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

Step 8

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

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

Step 9

Integrate $N (x, y) = \frac{1}{x}$ to find $f (x, y)$ .

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Step 9.1

Apply the constant rule.

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

Step 9.2

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

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

Step 10

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

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

Step 11

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

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

Step 12

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

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Step 12.1

Differentiate $f$ with respect to $x$ .

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

Step 12.2

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

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

Step 12.3

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

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Step 12.3.1

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

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

Step 12.3.2

Rewrite $\frac{1}{x}$ as $x^{- 1}$ .

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

Step 12.3.3

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

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

Step 12.4

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

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

Step 12.5

Simplify.

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Step 12.5.1

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

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

Step 12.5.2

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

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

Step 12.5.3

Reorder terms.

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

Step 13

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

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Step 13.1

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

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Step 13.1.1

Move all terms containing variables to the left side of the equation.

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Step 13.1.1.1

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

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

Step 13.1.1.2

Combine the numerators over the common denominator.

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

Step 13.1.1.3

Add $- y$ and $y$ .

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

Step 13.1.1.4

Add $0$ and $2 x^{3}$ .

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

Step 13.1.1.5

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

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Step 13.1.1.5.1

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

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

Step 13.1.1.5.2

Cancel the common factors.

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Step 13.1.1.5.2.1

Multiply by $1$ .

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

Step 13.1.1.5.2.2

Cancel the common factor.

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

Step 13.1.1.5.2.3

Rewrite the expression.

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

Step 13.1.1.5.2.4

Divide $2 x$ by $1$ .

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

Step 13.1.2

Subtract $2 x$ from both sides of the equation.

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

Step 14

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

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Step 14.1

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 14.2

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

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

Step 14.3

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

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

Step 14.4

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 14.5

Simplify the answer.

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Step 14.5.1

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 14.5.2

Simplify.

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Step 14.5.2.1

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

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

Step 14.5.2.2

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

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Step 14.5.2.2.1

Factor $2$ out of $- 2$ .

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

Step 14.5.2.2.2

Cancel the common factors.

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Step 14.5.2.2.2.1

Factor $2$ out of $2$ .

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

Step 14.5.2.2.2.2

Cancel the common factor.

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

Step 14.5.2.2.2.3

Rewrite the expression.

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

Step 14.5.2.2.2.4

Divide $- 1$ by $1$ .

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

Step 15

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

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