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

Cancel the common factor of $1 - x$ .

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Step 3.1.1

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 3.1.2

Rewrite the expression.

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

Step 3.2

Rewrite using the commutative property of multiplication.

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

Step 3.3

Cancel the common factor of $1 + y$ .

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Step 3.3.1

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 3.3.2

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 3.3.3

Cancel the common factor.

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

Step 3.3.4

Rewrite the expression.

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

Step 3.4

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

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

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Step 4.2.1.1

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

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Step 4.2.1.1.1

Differentiate $1 + y$ .

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

Step 4.2.1.1.2

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 4.2.1.1.3

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 4.2.1.1.4

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 4.2.1.1.5

Add $0$ and $1$ .

$1$

Step 4.2.1.2

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

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

Step 4.2.2

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

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

Step 4.2.3

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

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

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

Step 5.3

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

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

Step 5.4

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

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

Step 5.5

Solve for $y$ .

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Step 5.5.1

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

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

Step 5.5.2

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

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

Step 5.5.3

Simplify the left side.

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Step 5.5.3.1

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

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Step 5.5.3.1.1

Cancel the common factor.

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

Step 5.5.3.1.2

Rewrite the expression.

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

Step 5.5.4

Solve for $y$ .

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Step 5.5.4.1

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

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

Step 5.5.4.2

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

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

Step 5.5.4.3

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

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

Step 5.5.4.4

Subtract $1$ from both sides of the equation.

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

Step 6

Group the constant terms together.

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Step 6.1

Simplify the constant of integration.

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

Step 6.2

Combine constants with the plus or minus.

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