Calculus Examples

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

Step 1

Separate the variables.

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Step 1.1

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

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

Step 1.2

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

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Step 1.2.1

Cancel the common factor.

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

Step 1.2.2

Rewrite the expression.

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

Step 1.3

Rewrite the equation.

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

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

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

Step 2.2

Integrate the left side.

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Step 2.2.1

Let $u = y - 1$ . Then $d u = d y$ . Rewrite using $u$ and $d$ $u$ .

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Step 2.2.1.1

Let $u = y - 1$ . Find $\frac{d u}{d y}$ .

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Step 2.2.1.1.1

Differentiate $y - 1$ .

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

Step 2.2.1.1.2

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 2.2.1.1.3

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

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

Step 2.2.1.1.4

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

$1 + 0$

Step 2.2.1.1.5

Add $1$ and $0$ .

$1$

Step 2.2.1.2

Rewrite the problem using $u$ and $d u$ .

$\int \frac{1}{u^{2}} d u = \int d x$

Step 2.2.2

Apply basic rules of exponents.

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Step 2.2.2.1

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

$\int {(u^{2})}^{- 1} d u = \int d x$

Step 2.2.2.2

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

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Step 2.2.2.2.1

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

$\int u^{2 \cdot - 1} d u = \int d x$

Step 2.2.2.2.2

Multiply $2$ by $- 1$ .

$\int u^{- 2} d u = \int d x$

Step 2.2.3

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

$- u^{- 1} + C_{1} = \int d x$

Step 2.2.4

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

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

Step 2.2.5

Replace all occurrences of $u$ with $y - 1$ .

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

Step 2.3

Apply the constant rule.

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

Step 2.4

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

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

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.

$y - 1, 1, 1$

Step 3.1.2

Remove parentheses.

$y - 1, 1, 1$

Step 3.1.3

The LCM of one and any expression is the expression.

$y - 1$

Step 3.2

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

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Step 3.2.1

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

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

Step 3.2.2

Simplify the left side.

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Step 3.2.2.1

Cancel the common factor of $y - 1$ .

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Step 3.2.2.1.1

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

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

Step 3.2.2.1.2

Cancel the common factor.

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

Step 3.2.2.1.3

Rewrite the expression.

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

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

Apply the distributive property.

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

Step 3.2.3.1.2

Move $- 1$ to the left of $x$ .

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

Step 3.2.3.1.3

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

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

Step 3.2.3.1.4

Apply the distributive property.

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

Step 3.2.3.1.5

Move $- 1$ to the left of $K$ .

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

Step 3.2.3.1.6

Rewrite $- 1 K$ as $- K$ .

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

Step 3.3

Solve the equation.

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Step 3.3.1

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

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

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 $x$ to both sides of the equation.

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

Step 3.3.2.2

Add $K$ to both sides of the equation.

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

Step 3.3.3

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

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Step 3.3.3.1

Factor $y$ out of $x y$ .

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

Step 3.3.3.2

Factor $y$ out of $K y$ .

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

Step 3.3.3.3

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

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

Step 3.3.4

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

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Step 3.3.4.1

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

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

Step 3.3.4.2

Simplify the left side.

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Step 3.3.4.2.1

Cancel the common factor of $x + K$ .

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Step 3.3.4.2.1.1

Cancel the common factor.

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

Step 3.3.4.2.1.2

Divide $y$ by $1$ .

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

Step 3.3.4.3

Simplify the right side.

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Step 3.3.4.3.1

Move the negative in front of the fraction.

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

Step 3.3.4.3.2

Combine the numerators over the common denominator.

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

Step 3.3.4.3.3

Combine the numerators over the common denominator.

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

Step 3.3.4.3.4

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

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

Step 3.3.4.3.5

Factor $- 1$ out of $x$ .

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

Step 3.3.4.3.6

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

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

Step 3.3.4.3.7

Factor $- 1$ out of $K$ .

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

Step 3.3.4.3.8

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

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

Step 3.3.4.3.9

Move the negative in front of the fraction.

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

Step 4

Simplify the constant of integration.

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