Calculus Examples

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

Step 1

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

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

Step 2

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

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

Step 3

Simplify.

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Step 3.1

Rewrite using the commutative property of multiplication.

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

Step 3.2

Cancel the common factor of $1 - x$ .

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Step 3.2.1

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

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

Step 3.2.2

Cancel the common factor.

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

Step 3.2.3

Rewrite the expression.

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

Step 3.3

Move the negative in front of the fraction.

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

Step 3.4

Rewrite using the commutative property of multiplication.

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

Step 3.5

Cancel the common factor of $1 + y$ .

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Step 3.5.1

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

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

Step 3.5.2

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

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

Step 3.5.3

Cancel the common factor.

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

Step 3.5.4

Rewrite the expression.

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

Step 3.6

Move the negative in front of the fraction.

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

Step 4

Integrate both sides.

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Step 4.1

Set up an integral on each side.

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

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.

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

Step 4.2.2

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

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Step 4.2.2.1

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

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Step 4.2.2.1.1

Differentiate $1 + y$ .

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

Step 4.2.2.1.2

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

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

Step 4.2.2.1.3

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

$0 + \frac{d}{d y} [y]$

Step 4.2.2.1.4

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

$0 + 1$

Step 4.2.2.1.5

Add $0$ and $1$ .

$1$

Step 4.2.2.2

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

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

Step 4.2.3

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

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

Step 4.2.4

Simplify.

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

Step 4.2.5

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

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

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.

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

Step 4.3.2

Let $u_{2} = 1 - x$ . Then $d u_{2} = - d x$ , so $- d u_{2} = d x$ . Rewrite using $u_{2}$ and $d$ $u_{2}$ .

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Step 4.3.2.1

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

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Step 4.3.2.1.1

Rewrite.

$\frac{- 1}{1}$

Step 4.3.2.1.2

Divide $- 1$ by $1$ .

$- 1$

Step 4.3.2.2

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

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

Step 4.3.3

Move the negative in front of the fraction.

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

Step 4.3.4

Since $- 1$ is constant with respect to $u_{2}$ , move $- 1$ out of the integral.

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

Step 4.3.5

Simplify.

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Step 4.3.5.1

Multiply $- 1$ by $- 1$ .

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

Step 4.3.5.2

Multiply $\int \frac{1}{u_{2}} d u_{2}$ by $1$ .

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

Step 4.3.6

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

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

Step 4.3.7

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

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

Step 4.4

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

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

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.

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

Step 5.2

Add $\ln (| 1 - x |)$ to both sides of the equation.

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

Step 5.3

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

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Step 5.3.1

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

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

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.

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

Step 5.3.2.2

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

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

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 $\frac{C}{- 1}$ .

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

Step 5.3.3.1.2

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

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

Step 5.3.3.1.3

Move the negative one from the denominator of $\frac{\ln (| 1 - x |)}{- 1}$ .

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

Step 5.3.3.1.4

Rewrite $- 1 \cdot \ln (| 1 - x |)$ as $- \ln (| 1 - x |)$ .

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

Step 5.4

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

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

Step 5.5

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

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

Step 5.6

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

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

Step 5.7

Expand $(1 + y) (1 - x)$ using the FOIL Method.

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Step 5.7.1

Apply the distributive property.

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

Step 5.7.2

Apply the distributive property.

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

Step 5.7.3

Apply the distributive property.

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

Step 5.8

Simplify each term.

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Step 5.8.1

Multiply $1$ by $1$ .

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

Step 5.8.2

Multiply $- x$ by $1$ .

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

Step 5.8.3

Multiply $y$ by $1$ .

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

Step 5.8.4

Rewrite using the commutative property of multiplication.

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

Step 5.9

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

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

Step 5.10

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

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

Step 5.11

Solve for $y$ .

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Step 5.11.1

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

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

Step 5.11.2

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

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

Step 5.11.3

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

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Step 5.11.3.1

Subtract $1$ from both sides of the equation.

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

Step 5.11.3.2

Add $x$ to both sides of the equation.

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

Step 5.11.4

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

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Step 5.11.4.1

Factor $y$ out of $y^{1}$ .

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

Step 5.11.4.2

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

$y \cdot 1 + y (- 1 x) = \pm e^{- C} - 1 + x$

Step 5.11.4.3

Factor $y$ out of $y \cdot 1 + y (- 1 x)$ .

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

Step 5.11.5

Rewrite $- 1 x$ as $- x$ .

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

Step 5.11.6

Divide each term in $y (1 - x) = \pm e^{- C} - 1 + x$ by $1 - x$ and simplify.

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Step 5.11.6.1

Divide each term in $y (1 - x) = \pm e^{- C} - 1 + x$ by $1 - x$ .

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

Step 5.11.6.2

Simplify the left side.

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Step 5.11.6.2.1

Cancel the common factor of $1 - x$ .

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Step 5.11.6.2.1.1

Cancel the common factor.

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

Step 5.11.6.2.1.2

Divide $y$ by $1$ .

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

Step 5.11.6.3

Simplify the right side.

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Step 5.11.6.3.1

Move the negative in front of the fraction.

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

Step 5.11.6.3.2

Combine the numerators over the common denominator.

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

Step 5.11.6.3.3

Combine the numerators over the common denominator.

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

Step 6

Simplify the constant of integration.

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