Calculus Examples

$y d x = x (1 + x y^{4}) d y$

Step 1

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

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Step 1.1

Subtract $x (1 + x y^{4}) d y$ from both sides of the equation.

$y d x - x (1 + x y^{4}) d y = 0$

Step 2

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

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Step 2.1

Differentiate $M$ with respect to $y$ .

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

Step 2.2

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

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

Step 3

Find $\frac{\partial N}{\partial x}$ where $N (x, y) = - x (1 + x y^{4})$ .

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Step 3.1

Differentiate $N$ with respect to $x$ .

$\frac{\partial N}{\partial x} = \frac{d}{d x} [- x (1 + x y^{4})]$

Step 3.2

Since $- 1$ is constant with respect to $x$ , the derivative of $- x (1 + x y^{4})$ with respect to $x$ is $- \frac{d}{d x} [x (1 + x y^{4})]$ .

$\frac{\partial N}{\partial x} = - \frac{d}{d x} [x (1 + x y^{4})]$

Step 3.3

Differentiate using the Product Rule which states that $\frac{d}{d x} [f (x) g (x)]$ is $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ where $f (x) = x$ and $g (x) = 1 + x y^{4}$ .

$\frac{\partial N}{\partial x} = - (x \frac{d}{d x} [1 + x y^{4}] + (1 + x y^{4}) \frac{d}{d x} [x])$

Step 3.4

Differentiate.

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Step 3.4.1

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

$\frac{\partial N}{\partial x} = - (x (\frac{d}{d x} [1] + \frac{d}{d x} [x y^{4}]) + (1 + x y^{4}) \frac{d}{d x} [x])$

Step 3.4.2

Since $1$ is constant with respect to $x$ , the derivative of $1$ with respect to $x$ is $0$ .

$\frac{\partial N}{\partial x} = - (x (0 + \frac{d}{d x} [x y^{4}]) + (1 + x y^{4}) \frac{d}{d x} [x])$

Step 3.4.3

Add $0$ and $\frac{d}{d x} [x y^{4}]$ .

$\frac{\partial N}{\partial x} = - (x \frac{d}{d x} [x y^{4}] + (1 + x y^{4}) \frac{d}{d x} [x])$

Step 3.4.4

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

$\frac{\partial N}{\partial x} = - (x (y^{4} \frac{d}{d x} [x]) + (1 + x y^{4}) \frac{d}{d x} [x])$

Step 3.4.5

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

$\frac{\partial N}{\partial x} = - (x (y^{4} \cdot 1) + (1 + x y^{4}) \frac{d}{d x} [x])$

Step 3.4.6

Multiply $y^{4}$ by $1$ .

$\frac{\partial N}{\partial x} = - (x y^{4} + (1 + x y^{4}) \frac{d}{d x} [x])$

Step 3.4.7

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

$\frac{\partial N}{\partial x} = - (x y^{4} + (1 + x y^{4}) \cdot 1)$

Step 3.4.8

Simplify by adding terms.

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Step 3.4.8.1

Multiply $1 + x y^{4}$ by $1$ .

$\frac{\partial N}{\partial x} = - (x y^{4} + 1 + x y^{4})$

Step 3.4.8.2

Add $x y^{4}$ and $x y^{4}$ .

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

Step 3.5

Simplify.

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Step 3.5.1

Apply the distributive property.

$\frac{\partial N}{\partial x} = - (2 x y^{4}) - 1 \cdot 1$

Step 3.5.2

Combine terms.

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Step 3.5.2.1

Multiply $2$ by $- 1$ .

$\frac{\partial N}{\partial x} = - 2 (x y^{4}) - 1 \cdot 1$

Step 3.5.2.2

Multiply $- 1$ by $1$ .

$\frac{\partial N}{\partial x} = - 2 x y^{4} - 1$

Step 3.5.3

Reorder terms.

$\frac{\partial N}{\partial x} = - 2 y^{4} x - 1$

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 $- 2 y^{4} x - 1$ for $\frac{\partial N}{\partial x}$ .

$1 = - 2 y^{4} x - 1$

Step 4.2

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

$1 = - 2 y^{4} x - 1$ 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 $- 2 y^{4} x - 1$ for $\frac{\partial N}{\partial x}$ .

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

Step 5.3

Substitute $- x (1 + x y^{4})$ for $N$ .

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Step 5.3.1

Substitute $- x (1 + x y^{4})$ for $N$ .

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

Step 5.3.2

Simplify the numerator.

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Step 5.3.2.1

Apply the distributive property.

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

Step 5.3.2.2

Multiply $- 2$ by $- 1$ .

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

Step 5.3.2.3

Multiply $- 1$ by $- 1$ .

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

Step 5.3.2.4

Add $1$ and $1$ .

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

Step 5.3.2.5

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

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Step 5.3.2.5.1

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

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

Step 5.3.2.5.2

Factor $2$ out of $2$ .

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

Step 5.3.2.5.3

Factor $2$ out of $2 (y^{4} x) + 2 (1)$ .

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

Step 5.3.3

Cancel the common factor of $y^{4} x + 1$ and $1 + x y^{4}$ .

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Step 5.3.3.1

Reorder terms.

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

Step 5.3.3.2

Cancel the common factor.

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

Step 5.3.3.3

Rewrite the expression.

$\frac{2}{- x}$

Step 5.3.4

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 $y d x - x (1 + x y^{4}) d y = 0$ by the integration factor $\frac{1}{x^{2}}$ .

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Step 7.1

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

$y \frac{1}{x^{2}} d x - x (1 + x y^{4}) d y = 0$

Step 7.2

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

$\frac{y}{x^{2}} d x - x (1 + x y^{4}) d y = 0$

Step 7.3

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

$\frac{y}{x^{2}} d x - x (1 + x y^{4}) \frac{1}{x^{2}} d y = 0$

Step 7.4

Cancel the common factor of $x$ .

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Step 7.4.1

Factor $x$ out of $- x (1 + x y^{4})$ .

$\frac{y}{x^{2}} d x + x (- 1 (1 + x y^{4})) \frac{1}{x^{2}} d y = 0$

Step 7.4.2

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

$\frac{y}{x^{2}} d x + x (- 1 (1 + x y^{4})) \frac{1}{x \cdot x} d y = 0$

Step 7.4.3

Cancel the common factor.

$\frac{y}{x^{2}} d x + x (- 1 (1 + x y^{4})) \frac{1}{x \cdot x} d y = 0$

Step 7.4.4

Rewrite the expression.

$\frac{y}{x^{2}} d x - 1 (1 + x y^{4}) \frac{1}{x} d y = 0$

Step 7.5

Apply the distributive property.

$\frac{y}{x^{2}} d x + (- 1 \cdot 1 - 1 (x y^{4})) \frac{1}{x} d y = 0$

Step 7.6

Multiply $- 1$ by $1$ .

$\frac{y}{x^{2}} d x + (- 1 - 1 (x y^{4})) \frac{1}{x} d y = 0$

Step 7.7

Rewrite $- 1 (x y^{4})$ as $- (x y^{4})$ .

$\frac{y}{x^{2}} d x + (- 1 - (x y^{4})) \frac{1}{x} d y = 0$

Step 7.8

Multiply $- 1 - x y^{4}$ by $\frac{1}{x}$ .

$\frac{y}{x^{2}} d x + \frac{- 1 - x y^{4}}{x} d y = 0$

Step 7.9

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

$\frac{y}{x^{2}} d x + \frac{- 1 (1) - x y^{4}}{x} d y = 0$

Step 7.10

Factor $- 1$ out of $- x y^{4}$ .

$\frac{y}{x^{2}} d x + \frac{- 1 (1) - (x y^{4})}{x} d y = 0$

Step 7.11

Factor $- 1$ out of $- 1 (1) - (x y^{4})$ .

$\frac{y}{x^{2}} d x + \frac{- 1 (1 + x y^{4})}{x} d y = 0$

Step 7.12

Move the negative in front of the fraction.

$\frac{y}{x^{2}} d x - \frac{1 + x y^{4}}{x} d y = 0$

Step 8

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

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

Step 9

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

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Step 9.1

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

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

Step 9.2

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

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

Step 9.3

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

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Step 9.3.1

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

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

Step 9.3.2

Multiply $2$ by $- 1$ .

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

Step 9.4

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

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

Step 9.5

Simplify the answer.

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Step 9.5.1

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

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

Step 9.5.2

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

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

Step 10

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

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

Step 11

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

$\frac{\partial f}{\partial y} = - \frac{1 + x y^{4}}{x}$

Step 12

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

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Step 12.1

Differentiate $f$ with respect to $y$ .

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

Step 12.2

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

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

Step 12.3

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

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Step 12.3.1

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

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

Step 12.3.2

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

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

Step 12.3.3

Multiply $- 1$ by $1$ .

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

Step 12.4

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

$- \frac{1}{x} + \frac{d g}{d y} = - \frac{1 + x y^{4}}{x}$

Step 12.5

Reorder terms.

$\frac{d g}{d y} - \frac{1}{x} = - \frac{1 + x y^{4}}{x}$

Step 13

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

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Step 13.1

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

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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 $\frac{1 + x y^{4}}{x}$ to both sides of the equation.

$\frac{d g}{d y} - \frac{1}{x} + \frac{1 + x y^{4}}{x} = 0$

Step 13.1.1.2

Combine the numerators over the common denominator.

$\frac{d g}{d y} + \frac{- 1 + 1 + x y^{4}}{x} = 0$

Step 13.1.1.3

Add $- 1$ and $1$ .

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

Step 13.1.1.4

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

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

Step 13.1.1.5

Cancel the common factor of $x$ .

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Step 13.1.1.5.1

Cancel the common factor.

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

Step 13.1.1.5.2

Divide $y^{4}$ by $1$ .

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

Step 13.1.2

Subtract $y^{4}$ from both sides of the equation.

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

Step 14

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

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Step 14.1

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

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

Step 14.2

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

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

Step 14.3

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

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

Step 14.4

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

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

Step 14.5

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

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

Step 15

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

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

Step 16

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

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