Calculus Examples

$(y + 1) d x - (x + 1) d y = 0$

Step 1

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

$- (x + 1) d y = - (y + 1) d x$

Step 2

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

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

Step 3

Simplify.

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Step 3.1

Rewrite using the commutative property of multiplication.

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

Step 3.2

Cancel the common factor of $x + 1$ .

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Step 3.2.1

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

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

Step 3.2.2

Cancel the common factor.

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

Step 3.2.3

Rewrite the expression.

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

Step 3.3

Move the negative in front of the fraction.

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

Step 3.4

Rewrite using the commutative property of multiplication.

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

Step 3.5

Cancel the common factor of $y + 1$ .

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Step 3.5.1

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

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

Step 3.5.2

Factor $y + 1$ out of $(x + 1) (y + 1)$ .

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

Step 3.5.3

Cancel the common factor.

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

Step 3.5.4

Rewrite the expression.

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

Step 3.6

Move the negative in front of the fraction.

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

Step 4

Integrate both sides.

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Step 4.1

Set up an integral on each side.

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

Step 4.2

Integrate the left side.

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Step 4.2.1

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

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

Step 4.2.2

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

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Step 4.2.2.1

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

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Step 4.2.2.1.1

Differentiate $y + 1$ .

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

Step 4.2.2.1.2

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

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

Step 4.2.2.1.3

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

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

Step 4.2.2.1.4

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

$1 + 0$

Step 4.2.2.1.5

Add $1$ and $0$ .

$1$

Step 4.2.2.2

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

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

Step 4.2.3

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

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

Step 4.2.4

Simplify.

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

Step 4.2.5

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

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

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.

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

Step 4.3.2

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

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Step 4.3.2.1

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

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Step 4.3.2.1.1

Differentiate $x + 1$ .

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

Step 4.3.2.1.2

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

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

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 $n = 1$ .

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

Step 4.3.2.1.4

Since $1$ is constant with respect to $x$ , the derivative of $1$ 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 $d u_{2}$ .

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

Step 4.3.3

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

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

Step 4.3.4

Simplify.

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

Step 4.3.5

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

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

Step 4.4

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

$- \ln (| y + 1 |) = - \ln (| x + 1 |) + C$

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.

$- \ln (| y + 1 |) + \ln (| x + 1 |) = C$

Step 5.2

Subtract $\ln (| x + 1 |)$ from both sides of the equation.

$- \ln (| y + 1 |) = C - \ln (| x + 1 |)$

Step 5.3

Divide each term in $- \ln (| y + 1 |) = C - \ln (| x + 1 |)$ by $- 1$ and simplify.

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Step 5.3.1

Divide each term in $- \ln (| y + 1 |) = C - \ln (| x + 1 |)$ by $- 1$ .

$\frac{- \ln (| y + 1 |)}{- 1} = \frac{C}{- 1} + \frac{- \ln (| x + 1 |)}{- 1}$

Step 5.3.2

Simplify the left side.

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Step 5.3.2.1

Dividing two negative values results in a positive value.

$\frac{\ln (| y + 1 |)}{1} = \frac{C}{- 1} + \frac{- \ln (| x + 1 |)}{- 1}$

Step 5.3.2.2

Divide $\ln (| y + 1 |)$ by $1$ .

$\ln (| y + 1 |) = \frac{C}{- 1} + \frac{- \ln (| x + 1 |)}{- 1}$

Step 5.3.3

Simplify the right side.

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Step 5.3.3.1

Simplify each term.

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Step 5.3.3.1.1

Move the negative one from the denominator of $\frac{C}{- 1}$ .

$\ln (| y + 1 |) = - 1 \cdot C + \frac{- \ln (| x + 1 |)}{- 1}$

Step 5.3.3.1.2

Rewrite $- 1 \cdot C$ as $- C$ .

$\ln (| y + 1 |) = - C + \frac{- \ln (| x + 1 |)}{- 1}$

Step 5.3.3.1.3

Dividing two negative values results in a positive value.

$\ln (| y + 1 |) = - C + \frac{\ln (| x + 1 |)}{1}$

Step 5.3.3.1.4

Divide $\ln (| x + 1 |)$ by $1$ .

$\ln (| y + 1 |) = - C + \ln (| x + 1 |)$

Step 5.4

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

$\ln (| y + 1 |) - \ln (| x + 1 |) = - C$

Step 5.5

Use the quotient property of logarithms, $\log_{b} (x) - \log_{b} (y) = \log_{b} (\frac{x}{y})$ .

$\ln (\frac{| y + 1 |}{| x + 1 |}) = - C$

Step 5.6

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

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

Step 5.7

Rewrite $\ln (\frac{| y + 1 |}{| x + 1 |}) = - C$ 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 $b^{y} = x$ .

$e^{- C} = \frac{| y + 1 |}{| x + 1 |}$

Step 5.8

Solve for $y$ .

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Step 5.8.1

Rewrite the equation as $\frac{| y + 1 |}{| x + 1 |} = e^{- C}$ .

$\frac{| y + 1 |}{| x + 1 |} = e^{- C}$

Step 5.8.2

Multiply both sides by $| x + 1 |$ .

$\frac{| y + 1 |}{| x + 1 |} | x + 1 | = e^{- C} | x + 1 |$

Step 5.8.3

Simplify the left side.

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Step 5.8.3.1

Cancel the common factor of $| x + 1 |$ .

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Step 5.8.3.1.1

Cancel the common factor.

$\frac{| y + 1 |}{| x + 1 |} | x + 1 | = e^{- C} | x + 1 |$

Step 5.8.3.1.2

Rewrite the expression.

$| y + 1 | = e^{- C} | x + 1 |$

Step 5.8.4

Solve for $y$ .

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Step 5.8.4.1

Reorder factors in $e^{- C} | x + 1 |$ .

$| y + 1 | = | x + 1 | e^{- C}$

Step 5.8.4.2

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

$y + 1 = \pm | x + 1 | e^{- C}$

Step 5.8.4.3

Reorder factors in $\pm | x + 1 | e^{- C}$ .

$y + 1 = \pm e^{- C} | x + 1 |$

Step 5.8.4.4

Subtract $1$ from both sides of the equation.

$y = \pm e^{- C} | x + 1 | - 1$

Step 6

Group the constant terms together.

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Step 6.1

Simplify the constant of integration.

$y = \pm C | x + 1 | - 1$

Step 6.2

Combine constants with the plus or minus.

$y = C | x + 1 | - 1$