Calculus Examples

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

Step 1

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

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Step 1.1

Rewrite.

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

Step 2

Find $\frac{\partial M}{\partial y}$ where $M (x, y) = - 5 x$ .

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Step 2.1

Differentiate $M$ with respect to $y$ .

$\frac{\partial M}{\partial y} = \frac{d}{d y} [- 5 x]$

Step 2.2

Since $- 5 x$ is constant with respect to $y$ , the derivative of $- 5 x$ with respect to $y$ is $0$ .

$\frac{\partial M}{\partial y} = 0$

Step 3

Find $\frac{\partial N}{\partial x}$ where $N (x, y) = x y^{2} + 4 y^{2}$ .

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Step 3.1

Differentiate $N$ with respect to $x$ .

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

Step 3.2

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

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

Step 3.3

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

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Step 3.3.1

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

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

Step 3.3.2

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

$\frac{\partial N}{\partial x} = y^{2} \cdot 1 + \frac{d}{d x} [4 y^{2}]$

Step 3.3.3

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

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

Step 3.4

Differentiate using the Constant Rule.

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Step 3.4.1

Since $4 y^{2}$ is constant with respect to $x$ , the derivative of $4 y^{2}$ with respect to $x$ is $0$ .

$\frac{\partial N}{\partial x} = y^{2} + 0$

Step 3.4.2

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

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

Step 4

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

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Step 4.1

Substitute $0$ for $\frac{\partial M}{\partial y}$ and $y^{2}$ for $\frac{\partial N}{\partial x}$ .

$0 = y^{2}$

Step 4.2

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

$0 = y^{2}$ 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 $0$ for $\frac{\partial M}{\partial y}$ .

$\frac{0 - \frac{\partial N}{\partial x}}{N}$

Step 5.2

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

$\frac{0 - y^{2}}{N}$

Step 5.3

Substitute $x y^{2} + 4 y^{2}$ for $N$ .

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Step 5.3.1

Substitute $x y^{2} + 4 y^{2}$ for $N$ .

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

Step 5.3.2

Subtract $y^{2}$ from $0$ .

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

Step 5.3.3

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

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Step 5.3.3.1

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

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

Step 5.3.3.2

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

$\frac{- y^{2}}{y^{2} x + y^{2} \cdot 4}$

Step 5.3.3.3

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

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

Step 5.3.4

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

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Step 5.3.4.1

Cancel the common factor.

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

Step 5.3.4.2

Rewrite the expression.

$\frac{- 1}{x + 4}$

Step 5.3.5

Move the negative in front of the fraction.

$- \frac{1}{x + 4}$

Step 5.4

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

$μ (x, y) = e^{\int - \frac{1}{x + 4} d x}$

Step 6

Evaluate the integral $e^{\int - \frac{1}{x + 4} d x}$ .

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Step 6.1

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

$μ (x, y) = e^{- \int \frac{1}{x + 4} d x}$

Step 6.2

Let $u = x + 4$ . Then $d u = d x$ . Rewrite using $u$ and $d$ $u$ .

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Step 6.2.1

Let $u = x + 4$ . Find $\frac{d u}{d x}$ .

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Step 6.2.1.1

Differentiate $x + 4$ .

$\frac{d}{d x} [x + 4]$

Step 6.2.1.2

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

$\frac{d}{d x} [x] + \frac{d}{d x} [4]$

Step 6.2.1.3

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

$1 + \frac{d}{d x} [4]$

Step 6.2.1.4

Since $4$ is constant with respect to $x$ , the derivative of $4$ with respect to $x$ is $0$ .

$1 + 0$

Step 6.2.1.5

Add $1$ and $0$ .

$1$

Step 6.2.2

Rewrite the problem using $u$ and $d u$ .

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

Step 6.3

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

$μ (x, y) = e^{- (\ln (| u |) + C)}$

Step 6.4

Simplify.

$μ (x, y) = e^{- \ln (| u |)} + C$

Step 6.5

Replace all occurrences of $u$ with $x + 4$ .

$μ (x, y) = e^{- \ln (| x + 4 |)} + C$

Step 6.6

Simplify each term.

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Step 6.6.1

Simplify $- \ln (| x + 4 |)$ by moving $- 1$ inside the logarithm.

$μ (x, y) = e^{\ln ({| x + 4 |}^{- 1})} + C$

Step 6.6.2

Exponentiation and log are inverse functions.

$μ (x, y) = {| x + 4 |}^{- 1} + C$

Step 6.6.3

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

$μ (x, y) = \frac{1}{| x + 4 |} + C$

Step 7

Multiply both sides of $- 5 x d x + (x y^{2} + 4 y^{2}) d y = 0$ by the integration factor $\frac{1}{x + 4}$ .

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Step 7.1

Multiply $- 5 x$ by $\frac{1}{x + 4}$ .

$- 5 x \frac{1}{x + 4} d x + (x y^{2} + 4 y^{2}) d y = 0$

Step 7.2

Multiply $- 5 x \frac{1}{x + 4}$ .

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Step 7.2.1

Combine $- 5$ and $\frac{1}{x + 4}$ .

$x \frac{- 5}{x + 4} d x + (x y^{2} + 4 y^{2}) d y = 0$

Step 7.2.2

Combine $x$ and $\frac{- 5}{x + 4}$ .

$\frac{x \cdot - 5}{x + 4} d x + (x y^{2} + 4 y^{2}) d y = 0$

Step 7.3

Move $- 5$ to the left of $x$ .

$\frac{- 5 \cdot x}{x + 4} d x + (x y^{2} + 4 y^{2}) d y = 0$

Step 7.4

Move the negative in front of the fraction.

$- \frac{5 x}{x + 4} d x + (x y^{2} + 4 y^{2}) d y = 0$

Step 7.5

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

$- \frac{5 x}{x + 4} d x + (x y^{2} + 4 y^{2}) \frac{1}{x + 4} d y = 0$

Step 7.6

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

$- \frac{5 x}{x + 4} d x + \frac{x y^{2} + 4 y^{2}}{x + 4} d y = 0$

Step 7.7

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

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Step 7.7.1

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

$- \frac{5 x}{x + 4} d x + \frac{y^{2} x + 4 y^{2}}{x + 4} d y = 0$

Step 7.7.2

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

$- \frac{5 x}{x + 4} d x + \frac{y^{2} x + y^{2} \cdot 4}{x + 4} d y = 0$

Step 7.7.3

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

$- \frac{5 x}{x + 4} d x + \frac{y^{2} (x + 4)}{x + 4} d y = 0$

Step 7.8

Cancel the common factor of $x + 4$ .

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Step 7.8.1

Cancel the common factor.

$- \frac{5 x}{x + 4} d x + \frac{y^{2} (x + 4)}{x + 4} d y = 0$

Step 7.8.2

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

$- \frac{5 x}{x + 4} d x + y^{2} d y = 0$

Step 8

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

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

Step 9

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

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Step 9.1

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

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

Step 10

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

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

Step 11

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

$\frac{\partial f}{\partial x} = - \frac{5 x}{x + 4}$

Step 12

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

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Step 12.1

Differentiate $f$ with respect to $x$ .

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

Step 12.2

By the Sum Rule, the derivative of $\frac{1}{3} y^{3} + g (x)$ with respect to $x$ is $\frac{d}{d x} [\frac{1}{3} y^{3}] + \frac{d}{d x} [g (x)]$ .

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

Step 12.3

Since $\frac{1}{3} y^{3}$ is constant with respect to $x$ , the derivative of $\frac{1}{3} y^{3}$ with respect to $x$ is $0$ .

$0 + \frac{d}{d x} [g (x)] = - \frac{5 x}{x + 4}$

Step 12.4

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

$0 + \frac{d g}{d x} = - \frac{5 x}{x + 4}$

Step 12.5

Add $0$ and $\frac{d g}{d x}$ .

$\frac{d g}{d x} = - \frac{5 x}{x + 4}$

Step 13

Find the antiderivative of $- \frac{5 x}{x + 4}$ to find $g (x)$ .

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Step 13.1

Integrate both sides of $\frac{d g}{d x} = - \frac{5 x}{x + 4}$ .

$\int \frac{d g}{d x} d x = \int - \frac{5 x}{x + 4} d x$

Step 13.2

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

$g (x) = \int - \frac{5 x}{x + 4} d x$

Step 13.3

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

$g (x) = - \int \frac{5 x}{x + 4} d x$

Step 13.4

Since $5$ is constant with respect to $x$ , move $5$ out of the integral.

$g (x) = - (5 \int \frac{x}{x + 4} d x)$

Step 13.5

Multiply $5$ by $- 1$ .

$g (x) = - 5 \int \frac{x}{x + 4} d x$

Step 13.6

Divide $x$ by $x + 4$ .

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Step 13.6.1

Set up the polynomials to be divided. If there is not a term for every exponent, insert one with a value of $0$ .

$x$

$4$

$x$

$0$

Step 13.6.2

Divide the highest order term in the dividend $x$ by the highest order term in divisor $x$ .

					$1$
	$x$	+	$4$		$x$	+	$0$

Step 13.6.3

Multiply the new quotient term by the divisor.

				$1$
$x$	+	$4$		$x$	+	$0$
			+	$x$	+	$4$

Step 13.6.4

The expression needs to be subtracted from the dividend, so change all the signs in $x + 4$

				$1$
$x$	+	$4$		$x$	+	$0$
			-	$x$	-	$4$

Step 13.6.5

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

				$1$
$x$	+	$4$		$x$	+	$0$
			-	$x$	-	$4$
					-	$4$

Step 13.6.6

The final answer is the quotient plus the remainder over the divisor.

$g (x) = - 5 \int 1 - \frac{4}{x + 4} d x$

Step 13.7

Split the single integral into multiple integrals.

$g (x) = - 5 (\int d x + \int - \frac{4}{x + 4} d x)$

Step 13.8

Apply the constant rule.

$g (x) = - 5 (x + C + \int - \frac{4}{x + 4} d x)$

Step 13.9

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

$g (x) = - 5 (x + C - \int \frac{4}{x + 4} d x)$

Step 13.10

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

$g (x) = - 5 (x + C - (4 \int \frac{1}{x + 4} d x))$

Step 13.11

Multiply $4$ by $- 1$ .

$g (x) = - 5 (x + C - 4 \int \frac{1}{x + 4} d x)$

Step 13.12

Let $u = x + 4$ . Then $d u = d x$ . Rewrite using $u$ and $d$ $u$ .

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Step 13.12.1

Let $u = x + 4$ . Find $\frac{d u}{d x}$ .

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Step 13.12.1.1

Differentiate $x + 4$ .

$\frac{d}{d x} [x + 4]$

Step 13.12.1.2

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

$\frac{d}{d x} [x] + \frac{d}{d x} [4]$

Step 13.12.1.3

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

$1 + \frac{d}{d x} [4]$

Step 13.12.1.4

Since $4$ is constant with respect to $x$ , the derivative of $4$ with respect to $x$ is $0$ .

$1 + 0$

Step 13.12.1.5

Add $1$ and $0$ .

$1$

Step 13.12.2

Rewrite the problem using $u$ and $d u$ .

$g (x) = - 5 (x + C - 4 \int \frac{1}{u} d u)$

Step 13.13

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

$g (x) = - 5 (x + C - 4 (\ln (| u |) + C))$

Step 13.14

Simplify.

$g (x) = - 5 (x - 4 \ln (| u |)) + C$

Step 13.15

Replace all occurrences of $u$ with $x + 4$ .

$g (x) = - 5 (x - 4 \ln (| x + 4 |)) + C$

Step 14

Substitute for $g (x)$ in $f (x, y) = \frac{1}{3} y^{3} + g (x)$ .

$f (x, y) = \frac{1}{3} y^{3} - 5 (x - 4 \ln (| x + 4 |)) + C$

Step 15

Simplify $f (x, y) = \frac{1}{3} y^{3} - 5 (x - 4 \ln (| x + 4 |)) + C$ .

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Step 15.1

Simplify each term.

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Step 15.1.1

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

$f (x, y) = \frac{y^{3}}{3} - 5 (x - 4 \ln (| x + 4 |)) + C$

Step 15.1.2

Simplify each term.

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Step 15.1.2.1

Simplify $- 4 \ln (| x + 4 |)$ by moving $4$ inside the logarithm.

$f (x, y) = \frac{y^{3}}{3} - 5 (x - \ln ({| x + 4 |}^{4})) + C$

Step 15.1.2.2

Remove the absolute value in ${| x + 4 |}^{4}$ because exponentiations with even powers are always positive.

$f (x, y) = \frac{y^{3}}{3} - 5 (x - \ln ({(x + 4)}^{4})) + C$

Step 15.1.3

Apply the distributive property.

$f (x, y) = \frac{y^{3}}{3} - 5 x - 5 (- \ln ({(x + 4)}^{4})) + C$

Step 15.1.4

Multiply $- 5 (- \ln ({(x + 4)}^{4}))$ .

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Step 15.1.4.1

Multiply $- 1$ by $- 5$ .

$f (x, y) = \frac{y^{3}}{3} - 5 x + 5 \ln ({(x + 4)}^{4}) + C$

Step 15.1.4.2

Simplify $5 \ln ({(x + 4)}^{4})$ by moving $5$ inside the logarithm.

$f (x, y) = \frac{y^{3}}{3} - 5 x + \ln ({({(x + 4)}^{4})}^{5}) + C$

Step 15.1.5

Multiply the exponents in ${({(x + 4)}^{4})}^{5}$ .

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Step 15.1.5.1

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

$f (x, y) = \frac{y^{3}}{3} - 5 x + \ln ({(x + 4)}^{4 \cdot 5}) + C$

Step 15.1.5.2

Multiply $4$ by $5$ .

$f (x, y) = \frac{y^{3}}{3} - 5 x + \ln ({(x + 4)}^{20}) + C$

Step 15.2

To write $\ln ({(x + 4)}^{20})$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$f (x, y) = - 5 x + \frac{y^{3}}{3} + \ln ({(x + 4)}^{20}) \cdot \frac{3}{3} + C$

Step 15.3

Combine $\ln ({(x + 4)}^{20})$ and $\frac{3}{3}$ .

$f (x, y) = - 5 x + \frac{y^{3}}{3} + \frac{\ln ({(x + 4)}^{20}) \cdot 3}{3} + C$

Step 15.4

Combine the numerators over the common denominator.

$f (x, y) = - 5 x + \frac{y^{3} + \ln ({(x + 4)}^{20}) \cdot 3}{3} + C$

Step 15.5

Simplify the numerator.

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Step 15.5.1

Multiply $\ln ({(x + 4)}^{20}) \cdot 3$ .

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Step 15.5.1.1

Reorder $\ln ({(x + 4)}^{20})$ and $3$ .

$f (x, y) = - 5 x + \frac{y^{3} + 3 \cdot \ln ({(x + 4)}^{20})}{3} + C$

Step 15.5.1.2

Simplify $3 \cdot \ln ({(x + 4)}^{20})$ by moving $3$ inside the logarithm.

$f (x, y) = - 5 x + \frac{y^{3} + \ln ({({(x + 4)}^{20})}^{3})}{3} + C$

Step 15.5.2

Multiply the exponents in ${({(x + 4)}^{20})}^{3}$ .

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Step 15.5.2.1

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

$f (x, y) = - 5 x + \frac{y^{3} + \ln ({(x + 4)}^{20 \cdot 3})}{3} + C$

Step 15.5.2.2

Multiply $20$ by $3$ .

$f (x, y) = - 5 x + \frac{y^{3} + \ln ({(x + 4)}^{60})}{3} + C$