Calculus Examples

$(y + 2) d x + (x - 2) d y = 0$

Step 1

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

$(x - 2) d y = - (y + 2) d x$

Step 2

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

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

Step 3

Simplify.

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Step 3.1

Cancel the common factor of $x - 2$ .

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Step 3.1.1

Cancel the common factor.

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

Step 3.1.2

Rewrite the expression.

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

Step 3.2

Rewrite using the commutative property of multiplication.

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

Step 3.3

Cancel the common factor of $y + 2$ .

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Step 3.3.1

Move the leading negative in $- \frac{1}{(x - 2) (y + 2)}$ into the numerator.

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

Step 3.3.2

Factor $y + 2$ out of $(x - 2) (y + 2)$ .

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

Step 3.3.3

Cancel the common factor.

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

Step 3.3.4

Rewrite the expression.

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

Step 3.4

Move the negative in front of the fraction.

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

Step 4

Integrate both sides.

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Step 4.1

Set up an integral on each side.

$\int \frac{1}{y + 2} d y = \int - \frac{1}{x - 2} d x$

Step 4.2

Integrate the left side.

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Step 4.2.1

Let $u_{1} = y + 2$ . Then $d u_{1} = d y$ . Rewrite using $u_{1}$ and $d$ $u_{1}$ .

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Step 4.2.1.1

Let $u_{1} = y + 2$ . Find $\frac{d u_{1}}{d y}$ .

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Step 4.2.1.1.1

Differentiate $y + 2$ .

$\frac{d}{d y} [y + 2]$

Step 4.2.1.1.2

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

$\frac{d}{d y} [y] + \frac{d}{d y} [2]$

Step 4.2.1.1.3

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

$1 + \frac{d}{d y} [2]$

Step 4.2.1.1.4

Since $2$ is constant with respect to $y$ , the derivative of $2$ with respect to $y$ is $0$ .

$1 + 0$

Step 4.2.1.1.5

Add $1$ and $0$ .

$1$

Step 4.2.1.2

Rewrite the problem using $u_{1}$ and $d u_{1}$ .

$\int \frac{1}{u_{1}} d u_{1} = \int - \frac{1}{x - 2} d x$

Step 4.2.2

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

$\ln (| u_{1} |) + C_{1} = \int - \frac{1}{x - 2} d x$

Step 4.2.3

Replace all occurrences of $u_{1}$ with $y + 2$ .

$\ln (| y + 2 |) + C_{1} = \int - \frac{1}{x - 2} d x$

Step 4.3

Integrate the right side.

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Step 4.3.1

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

$\ln (| y + 2 |) + C_{1} = - \int \frac{1}{x - 2} d x$

Step 4.3.2

Let $u_{2} = x - 2$ . Then $d u_{2} = d x$ . Rewrite using $u_{2}$ and $d$ $u_{2}$ .

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Step 4.3.2.1

Let $u_{2} = x - 2$ . Find $\frac{d u_{2}}{d x}$ .

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Step 4.3.2.1.1

Differentiate $x - 2$ .

$\frac{d}{d x} [x - 2]$

Step 4.3.2.1.2

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

$\frac{d}{d x} [x] + \frac{d}{d x} [- 2]$

Step 4.3.2.1.3

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

$1 + \frac{d}{d x} [- 2]$

Step 4.3.2.1.4

Since $- 2$ is constant with respect to $x$ , the derivative of $- 2$ with respect to $x$ is $0$ .

$1 + 0$

Step 4.3.2.1.5

Add $1$ and $0$ .

$1$

Step 4.3.2.2

Rewrite the problem using $u_{2}$ and $d u_{2}$ .

$\ln (| y + 2 |) + C_{1} = - \int \frac{1}{u_{2}} d u_{2}$

Step 4.3.3

The integral of $\frac{1}{u_{2}}$ with respect to $u_{2}$ is $\ln (| u_{2} |)$ .

$\ln (| y + 2 |) + C_{1} = - (\ln (| u_{2} |) + C_{2})$

Step 4.3.4

Simplify.

$\ln (| y + 2 |) + C_{1} = - \ln (| u_{2} |) + C_{2}$

Step 4.3.5

Replace all occurrences of $u_{2}$ with $x - 2$ .

$\ln (| y + 2 |) + C_{1} = - \ln (| x - 2 |) + C_{2}$

Step 4.4

Group the constant of integration on the right side as $C$ .

$\ln (| y + 2 |) = - \ln (| x - 2 |) + C$

Step 5

Solve for $y$ .

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Step 5.1

Move all the terms containing a logarithm to the left side of the equation.

$\ln (| y + 2 |) + \ln (| x - 2 |) = C$

Step 5.2

Use the product property of logarithms, $\log_{b} (x) + \log_{b} (y) = \log_{b} (x y)$ .

$\ln (| y + 2 | | x - 2 |) = C$

Step 5.3

To multiply absolute values, multiply the terms inside each absolute value.

$\ln (| (y + 2) (x - 2) |) = C$

Step 5.4

Expand $(y + 2) (x - 2)$ using the FOIL Method.

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Step 5.4.1

Apply the distributive property.

$\ln (| y (x - 2) + 2 (x - 2) |) = C$

Step 5.4.2

Apply the distributive property.

$\ln (| y x + y \cdot - 2 + 2 (x - 2) |) = C$

Step 5.4.3

Apply the distributive property.

$\ln (| y x + y \cdot - 2 + 2 x + 2 \cdot - 2 |) = C$

Step 5.5

Simplify each term.

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Step 5.5.1

Move $- 2$ to the left of $y$ .

$\ln (| y x - 2 \cdot y + 2 x + 2 \cdot - 2 |) = C$

Step 5.5.2

Multiply $2$ by $- 2$ .

$\ln (| y x - 2 y + 2 x - 4 |) = C$

Step 5.6

To solve for $y$ , rewrite the equation using properties of logarithms.

$e^{\ln (| y x - 2 y + 2 x - 4 |)} = e^{C}$

Step 5.7

Rewrite $\ln (| y x - 2 y + 2 x - 4 |) = C$ in exponential form using the definition of a logarithm. If $x$ and $b$ are positive real numbers and $b \neq 1$ , then $\log_{b} (x) = y$ is equivalent to $b^{y} = x$ .

$e^{C} = | y x - 2 y + 2 x - 4 |$

Step 5.8

Solve for $y$ .

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Step 5.8.1

Rewrite the equation as $| y x - 2 y + 2 x - 4 | = e^{C}$ .

$| y x - 2 y + 2 x - 4 | = e^{C}$

Step 5.8.2

Remove the absolute value term. This creates a $\pm$ on the right side of the equation because $| x | = \pm x$ .

$y x - 2 y + 2 x - 4 = \pm e^{C}$

Step 5.8.3

Move all terms not containing $y$ to the right side of the equation.

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Step 5.8.3.1

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

$y x - 2 y - 4 = \pm e^{C} - 2 x$

Step 5.8.3.2

Add $4$ to both sides of the equation.

$y x - 2 y = \pm e^{C} - 2 x + 4$

Step 5.8.4

Factor $y$ out of $y x - 2 y$ .

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Step 5.8.4.1

Factor $y$ out of $y x$ .

$y (x) - 2 y = \pm e^{C} - 2 x + 4$

Step 5.8.4.2

Factor $y$ out of $- 2 y$ .

$y (x) + y \cdot - 2 = \pm e^{C} - 2 x + 4$

Step 5.8.4.3

Factor $y$ out of $y (x) + y \cdot - 2$ .

$y (x - 2) = \pm e^{C} - 2 x + 4$

Step 5.8.5

Divide each term in $y (x - 2) = \pm e^{C} - 2 x + 4$ by $x - 2$ and simplify.

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Step 5.8.5.1

Divide each term in $y (x - 2) = \pm e^{C} - 2 x + 4$ by $x - 2$ .

$\frac{y (x - 2)}{x - 2} = \frac{\pm e^{C}}{x - 2} + \frac{- 2 x}{x - 2} + \frac{4}{x - 2}$

Step 5.8.5.2

Simplify the left side.

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Step 5.8.5.2.1

Cancel the common factor of $x - 2$ .

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Step 5.8.5.2.1.1

Cancel the common factor.

$\frac{y (x - 2)}{x - 2} = \frac{\pm e^{C}}{x - 2} + \frac{- 2 x}{x - 2} + \frac{4}{x - 2}$

Step 5.8.5.2.1.2

Divide $y$ by $1$ .

$y = \frac{\pm e^{C}}{x - 2} + \frac{- 2 x}{x - 2} + \frac{4}{x - 2}$

Step 5.8.5.3

Simplify the right side.

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Step 5.8.5.3.1

Move the negative in front of the fraction.

$y = \frac{\pm e^{C}}{x - 2} - \frac{2 x}{x - 2} + \frac{4}{x - 2}$

Step 5.8.5.3.2

Combine the numerators over the common denominator.

$y = \frac{\pm e^{C} - 2 x}{x - 2} + \frac{4}{x - 2}$

Step 5.8.5.3.3

Combine the numerators over the common denominator.

$y = \frac{\pm e^{C} - 2 x + 4}{x - 2}$

Step 6

Simplify the constant of integration.

$y = \frac{C - 2 x + 4}{x - 2}$