Calculus Examples

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

Step 1

Separate the variables.

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Step 1.1

Divide each term in ${(1 + x)}^{2} \frac{d y}{d x} = {(1 + y)}^{2}$ by ${(1 + x)}^{2}$ and simplify.

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Step 1.1.1

Divide each term in ${(1 + x)}^{2} \frac{d y}{d x} = {(1 + y)}^{2}$ by ${(1 + x)}^{2}$ .

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

Step 1.1.2

Simplify the left side.

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Step 1.1.2.1

Cancel the common factor of ${(1 + x)}^{2}$ .

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Step 1.1.2.1.1

Cancel the common factor.

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

Step 1.1.2.1.2

Divide $\frac{d y}{d x}$ by $1$ .

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

Step 1.2

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

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

Step 1.3

Simplify.

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Step 1.3.1

Combine.

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

Step 1.3.2

Cancel the common factor of ${(1 + y)}^{2}$ .

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Step 1.3.2.1

Cancel the common factor.

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

Step 1.3.2.2

Rewrite the expression.

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

Step 1.4

Rewrite the equation.

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

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

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

Step 2.2

Integrate the left side.

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Step 2.2.1

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

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Step 2.2.1.1

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

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Step 2.2.1.1.1

Differentiate $1 + y$ .

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

Step 2.2.1.1.2

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 2.2.1.1.3

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 2.2.1.1.4

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 2.2.1.1.5

Add $0$ and $1$ .

$1$

Step 2.2.1.2

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

$\int \frac{1}{{u_{1}}^{2}} d u_{1} = \int \frac{1}{{(1 + x)}^{2}} d x$

Step 2.2.2

Apply basic rules of exponents.

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Step 2.2.2.1

Move ${u_{1}}^{2}$ out of the denominator by raising it to the $- 1$ power.

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

Step 2.2.2.2

Multiply the exponents in ${({u_{1}}^{2})}^{- 1}$ .

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Step 2.2.2.2.1

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

$\int {u_{1}}^{2 \cdot - 1} d u_{1} = \int \frac{1}{{(1 + x)}^{2}} d x$

Step 2.2.2.2.2

Multiply $2$ by $- 1$ .

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

Step 2.2.3

By the Power Rule, the integral of ${u_{1}}^{- 2}$ with respect to $u_{1}$ is $- {u_{1}}^{- 1}$ .

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

Step 2.2.4

Rewrite $- {u_{1}}^{- 1} + C_{1}$ as $- \frac{1}{u_{1}} + C_{1}$ .

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

Step 2.2.5

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

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

Step 2.3

Integrate the right side.

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Step 2.3.1

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

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Step 2.3.1.1

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

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Step 2.3.1.1.1

Differentiate $1 + x$ .

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

Step 2.3.1.1.2

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

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

Step 2.3.1.1.3

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

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

Step 2.3.1.1.4

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

$0 + 1$

Step 2.3.1.1.5

Add $0$ and $1$ .

$1$

Step 2.3.1.2

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

$- \frac{1}{1 + y} + C_{1} = \int \frac{1}{{u_{2}}^{2}} d u_{2}$

Step 2.3.2

Apply basic rules of exponents.

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Step 2.3.2.1

Move ${u_{2}}^{2}$ out of the denominator by raising it to the $- 1$ power.

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

Step 2.3.2.2

Multiply the exponents in ${({u_{2}}^{2})}^{- 1}$ .

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Step 2.3.2.2.1

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

$- \frac{1}{1 + y} + C_{1} = \int {u_{2}}^{2 \cdot - 1} d u_{2}$

Step 2.3.2.2.2

Multiply $2$ by $- 1$ .

$- \frac{1}{1 + y} + C_{1} = \int {u_{2}}^{- 2} d u_{2}$

Step 2.3.3

By the Power Rule, the integral of ${u_{2}}^{- 2}$ with respect to $u_{2}$ is $- {u_{2}}^{- 1}$ .

$- \frac{1}{1 + y} + C_{1} = - {u_{2}}^{- 1} + C_{2}$

Step 2.3.4

Rewrite $- {u_{2}}^{- 1} + C_{2}$ as $- \frac{1}{u_{2}} + C_{2}$ .

$- \frac{1}{1 + y} + C_{1} = - \frac{1}{u_{2}} + C_{2}$

Step 2.3.5

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

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

Step 2.4

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

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

Step 3

Solve for $y$ .

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Step 3.1

Find the LCD of the terms in the equation.

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Step 3.1.1

Finding the LCD of a list of values is the same as finding the LCM of the denominators of those values.

$1 + y, 1 + x, 1$

Step 3.1.2

The LCM is the smallest positive number that all of the numbers divide into evenly.

1. List the prime factors of each number.

2. Multiply each factor the greatest number of times it occurs in either number.

Step 3.1.3

The number $1$ is not a prime number because it only has one positive factor, which is itself.

Not prime

Step 3.1.4

The LCM of $1, 1, 1$ is the result of multiplying all prime factors the greatest number of times they occur in either number.

$1$

Step 3.1.5

The factor for $1 + y$ is $1 + y$ itself.

$(1 + y) = 1 + y$

$(1 + y)$ occurs $1$ time.

Step 3.1.6

The factor for $1 + x$ is $1 + x$ itself.

$(1 + x) = 1 + x$

$(1 + x)$ occurs $1$ time.

Step 3.1.7

The LCM of $1 + y, 1 + x$ is the result of multiplying all factors the greatest number of times they occur in either term.

$(1 + y) (1 + x)$

Step 3.2

Multiply each term in $- \frac{1}{1 + y} = - \frac{1}{1 + x} + K$ by $(1 + y) (1 + x)$ to eliminate the fractions.

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Step 3.2.1

Multiply each term in $- \frac{1}{1 + y} = - \frac{1}{1 + x} + K$ by $(1 + y) (1 + x)$ .

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

Step 3.2.2

Simplify the left side.

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Step 3.2.2.1

Cancel the common factor of $1 + y$ .

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Step 3.2.2.1.1

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

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

Step 3.2.2.1.2

Cancel the common factor.

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

Step 3.2.2.1.3

Rewrite the expression.

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

Step 3.2.2.2

Apply the distributive property.

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

Step 3.2.2.3

Multiply $- 1$ by $1$ .

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

Step 3.2.3

Simplify the right side.

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Step 3.2.3.1

Simplify each term.

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Step 3.2.3.1.1

Cancel the common factor of $1 + x$ .

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Step 3.2.3.1.1.1

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

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

Step 3.2.3.1.1.2

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

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

Step 3.2.3.1.1.3

Cancel the common factor.

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

Step 3.2.3.1.1.4

Rewrite the expression.

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

Step 3.2.3.1.2

Apply the distributive property.

$- 1 - x = - 1 \cdot 1 - y + K ((1 + y) (1 + x))$

Step 3.2.3.1.3

Multiply $- 1$ by $1$ .

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

Step 3.2.3.1.4

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

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Step 3.2.3.1.4.1

Apply the distributive property.

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

Step 3.2.3.1.4.2

Apply the distributive property.

$- 1 - x = - 1 - y + K (1 \cdot 1 + 1 x + y (1 + x))$

Step 3.2.3.1.4.3

Apply the distributive property.

$- 1 - x = - 1 - y + K (1 \cdot 1 + 1 x + y \cdot 1 + y x)$

Step 3.2.3.1.5

Simplify each term.

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Step 3.2.3.1.5.1

Multiply $1$ by $1$ .

$- 1 - x = - 1 - y + K (1 + 1 x + y \cdot 1 + y x)$

Step 3.2.3.1.5.2

Multiply $x$ by $1$ .

$- 1 - x = - 1 - y + K (1 + x + y \cdot 1 + y x)$

Step 3.2.3.1.5.3

Multiply $y$ by $1$ .

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

Step 3.2.3.1.6

Apply the distributive property.

$- 1 - x = - 1 - y + K \cdot 1 + K (x) + K (y) + K (y x)$

Step 3.2.3.1.7

Multiply $K$ by $1$ .

$- 1 - x = - 1 - y + K + K x + K y + K y x$

Step 3.3

Solve the equation.

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Step 3.3.1

Rewrite the equation as $- 1 - y + K + K x + K y + K y x = - 1 - x$ .

$- 1 - y + K + K x + K y + K y x = - 1 - x$

Step 3.3.2

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

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Step 3.3.2.1

Add $1$ to both sides of the equation.

$- y + K + K x + K y + K y x = - 1 - x + 1$

Step 3.3.2.2

Subtract $K$ from both sides of the equation.

$- y + K x + K y + K y x = - 1 - x + 1 - K$

Step 3.3.2.3

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

$- y + K y + K y x = - 1 - x + 1 - K - K x$

Step 3.3.2.4

Combine the opposite terms in $- 1 - x + 1 - K - K x$ .

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Step 3.3.2.4.1

Add $- 1$ and $1$ .

$- y + K y + K y x = - x + 0 - K - K x$

Step 3.3.2.4.2

Add $- x$ and $0$ .

$- y + K y + K y x = - x - K - K x$

Step 3.3.3

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

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Step 3.3.3.1

Factor $y$ out of $- y$ .

$y \cdot - 1 + K y + K y x = - x - K - K x$

Step 3.3.3.2

Factor $y$ out of $K y$ .

$y \cdot - 1 + y K + K y x = - x - K - K x$

Step 3.3.3.3

Factor $y$ out of $K y x$ .

$y \cdot - 1 + y K + y (K x) = - x - K - K x$

Step 3.3.3.4

Factor $y$ out of $y \cdot - 1 + y K$ .

$y \cdot (- 1 + K) + y (K x) = - x - K - K x$

Step 3.3.3.5

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

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

Step 3.3.4

Divide each term in $y (- 1 + K + K x) = - x - K - K x$ by $- 1 + K + K x$ and simplify.

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Step 3.3.4.1

Divide each term in $y (- 1 + K + K x) = - x - K - K x$ by $- 1 + K + K x$ .

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

Step 3.3.4.2

Simplify the left side.

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Step 3.3.4.2.1

Cancel the common factor of $- 1 + K + K x$ .

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Step 3.3.4.2.1.1

Cancel the common factor.

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

Step 3.3.4.2.1.2

Divide $y$ by $1$ .

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

Step 3.3.4.3

Simplify the right side.

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Step 3.3.4.3.1

Simplify each term.

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Step 3.3.4.3.1.1

Move the negative in front of the fraction.

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

Step 3.3.4.3.1.2

Move the negative in front of the fraction.

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

Step 3.3.4.3.1.3

Move the negative in front of the fraction.

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

Step 3.3.4.3.2

Simplify terms.

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Step 3.3.4.3.2.1

Combine the numerators over the common denominator.

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

Step 3.3.4.3.2.2

Combine the numerators over the common denominator.

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

Step 3.3.4.3.2.3

Factor $- 1$ out of $- x$ .

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

Step 3.3.4.3.2.4

Factor $- 1$ out of $- (x) - K$ .

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

Step 3.3.4.3.2.5

Factor $- 1$ out of $- K x$ .

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

Step 3.3.4.3.2.6

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

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

Step 3.3.4.3.2.7

Rewrite $- (x + K + K x)$ as $- 1 (x + K + K x)$ .

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

Step 3.3.4.3.2.8

Rewrite $- 1$ as $- 1 (1)$ .

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

Step 3.3.4.3.2.9

Factor $- 1$ out of $K$ .

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

Step 3.3.4.3.2.10

Factor $- 1$ out of $- 1 (1) - 1 (- K)$ .

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

Step 3.3.4.3.2.11

Factor $- 1$ out of $K x$ .

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

Step 3.3.4.3.2.12

Factor $- 1$ out of $- 1 (1 - K) - (- K x)$ .

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

Step 3.3.4.3.2.13

Cancel the common factor.

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

Step 3.3.4.3.2.14

Rewrite the expression.

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

Step 4

Simplify the constant of integration.

$y = \frac{x + K + K x}{K + K x}$