Calculus Examples

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

Step 1

Separate the variables.

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Step 1.1

Add $y$ to both sides of the equation.

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

Step 1.2

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

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

Step 1.3

Cancel the common factor of $\frac{1}{y^{2}} + y$ .

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Step 1.3.1

Cancel the common factor.

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

Step 1.3.2

Rewrite the expression.

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

Step 1.4

Rewrite the equation.

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

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

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

Step 2.2

Integrate the left side.

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Step 2.2.1

Simplify.

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Step 2.2.1.1

Simplify the denominator.

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Step 2.2.1.1.1

To write $y$ as a fraction with a common denominator, multiply by $\frac{y^{2}}{y^{2}}$ .

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

Step 2.2.1.1.2

Combine the numerators over the common denominator.

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

Step 2.2.1.1.3

Simplify the numerator.

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Step 2.2.1.1.3.1

Rewrite $1$ as $1^{3}$ .

$\int \frac{1}{\frac{1^{3} + y \cdot y^{2}}{y^{2}}} d y = \int d x$

Step 2.2.1.1.3.2

Multiply $y$ by $y^{2}$ by adding the exponents.

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Step 2.2.1.1.3.2.1

Multiply $y$ by $y^{2}$ .

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Step 2.2.1.1.3.2.1.1

Raise $y$ to the power of $1$ .

$\int \frac{1}{\frac{1^{3} + y^{1} y^{2}}{y^{2}}} d y = \int d x$

Step 2.2.1.1.3.2.1.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$\int \frac{1}{\frac{1^{3} + y^{1 + 2}}{y^{2}}} d y = \int d x$

Step 2.2.1.1.3.2.2

Add $1$ and $2$ .

$\int \frac{1}{\frac{1^{3} + y^{3}}{y^{2}}} d y = \int d x$

Step 2.2.1.1.3.3

Since both terms are perfect cubes, factor using the sum of cubes formula, $a^{3} + b^{3} = (a + b) (a^{2} - a b + b^{2})$ where $a = 1$ and $b = y$ .

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

Step 2.2.1.1.3.4

Simplify.

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Step 2.2.1.1.3.4.1

One to any power is one.

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

Step 2.2.1.1.3.4.2

Rewrite $- 1 y$ as $- y$ .

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

Step 2.2.1.2

Multiply the numerator by the reciprocal of the denominator.

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

Step 2.2.1.3

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

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

Step 2.2.2

Let $u = (1 + y) (1 - y + y^{2})$ . Then $d u = 3 y^{2} d y$ , so $\frac{1}{3} d u = y^{2} d y$ . Rewrite using $u$ and $d$ $u$ .

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Step 2.2.2.1

Let $u = (1 + y) (1 - y + y^{2})$ . Find $\frac{d u}{d y}$ .

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Step 2.2.2.1.1

Differentiate $(1 + y) (1 - y + y^{2})$ .

$\frac{d}{d y} [(1 + y) (1 - y + y^{2})]$

Step 2.2.2.1.2

Differentiate using the Product Rule which states that $\frac{d}{d y} [f (y) g (y)]$ is $f (y) \frac{d}{d y} [g (y)] + g (y) \frac{d}{d y} [f (y)]$ where $f (y) = 1 + y$ and $g (y) = 1 - y + y^{2}$ .

$(1 + y) \frac{d}{d y} [1 - y + y^{2}] + (1 - y + y^{2}) \frac{d}{d y} [1 + y]$

Step 2.2.2.1.3

Differentiate.

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Step 2.2.2.1.3.1

By the Sum Rule, the derivative of $1 - y + y^{2}$ with respect to $y$ is $\frac{d}{d y} [1] + \frac{d}{d y} [- y] + \frac{d}{d y} [y^{2}]$ .

$(1 + y) (\frac{d}{d y} [1] + \frac{d}{d y} [- y] + \frac{d}{d y} [y^{2}]) + (1 - y + y^{2}) \frac{d}{d y} [1 + y]$

Step 2.2.2.1.3.2

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

$(1 + y) (0 + \frac{d}{d y} [- y] + \frac{d}{d y} [y^{2}]) + (1 - y + y^{2}) \frac{d}{d y} [1 + y]$

Step 2.2.2.1.3.3

Add $0$ and $\frac{d}{d y} [- y]$ .

$(1 + y) (\frac{d}{d y} [- y] + \frac{d}{d y} [y^{2}]) + (1 - y + y^{2}) \frac{d}{d y} [1 + y]$

Step 2.2.2.1.3.4

Since $- 1$ is constant with respect to $y$ , the derivative of $- y$ with respect to $y$ is $- \frac{d}{d y} [y]$ .

$(1 + y) (- \frac{d}{d y} [y] + \frac{d}{d y} [y^{2}]) + (1 - y + y^{2}) \frac{d}{d y} [1 + y]$

Step 2.2.2.1.3.5

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

$(1 + y) (- 1 \cdot 1 + \frac{d}{d y} [y^{2}]) + (1 - y + y^{2}) \frac{d}{d y} [1 + y]$

Step 2.2.2.1.3.6

Multiply $- 1$ by $1$ .

$(1 + y) (- 1 + \frac{d}{d y} [y^{2}]) + (1 - y + y^{2}) \frac{d}{d y} [1 + y]$

Step 2.2.2.1.3.7

Differentiate using the Power Rule which states that $\frac{d}{d y} [y^{n}]$ is $n y^{n - 1}$ where $n = 2$ .

$(1 + y) (- 1 + 2 y) + (1 - y + y^{2}) \frac{d}{d y} [1 + y]$

Step 2.2.2.1.3.8

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

$(1 + y) (- 1 + 2 y) + (1 - y + y^{2}) (\frac{d}{d y} [1] + \frac{d}{d y} [y])$

Step 2.2.2.1.3.9

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

$(1 + y) (- 1 + 2 y) + (1 - y + y^{2}) (0 + \frac{d}{d y} [y])$

Step 2.2.2.1.3.10

Add $0$ and $\frac{d}{d y} [y]$ .

$(1 + y) (- 1 + 2 y) + (1 - y + y^{2}) \frac{d}{d y} [y]$

Step 2.2.2.1.3.11

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

$(1 + y) (- 1 + 2 y) + (1 - y + y^{2}) \cdot 1$

Step 2.2.2.1.3.12

Multiply $1 - y + y^{2}$ by $1$ .

$(1 + y) (- 1 + 2 y) + 1 - y + y^{2}$

Step 2.2.2.1.4

Simplify.

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Step 2.2.2.1.4.1

Apply the distributive property.

$(1 + y) \cdot - 1 + (1 + y) (2 y) + 1 - y + y^{2}$

Step 2.2.2.1.4.2

Apply the distributive property.

$1 \cdot - 1 + y \cdot - 1 + (1 + y) (2 y) + 1 - y + y^{2}$

Step 2.2.2.1.4.3

Apply the distributive property.

$1 \cdot - 1 + y \cdot - 1 + 1 (2 y) + y (2 y) + 1 - y + y^{2}$

Step 2.2.2.1.4.4

Combine terms.

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Step 2.2.2.1.4.4.1

Multiply $- 1$ by $1$ .

$- 1 + y \cdot - 1 + 1 (2 y) + y (2 y) + 1 - y + y^{2}$

Step 2.2.2.1.4.4.2

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

$- 1 - 1 \cdot y + 1 (2 y) + y (2 y) + 1 - y + y^{2}$

Step 2.2.2.1.4.4.3

Rewrite $- 1 y$ as $- y$ .

$- 1 - y + 1 (2 y) + y (2 y) + 1 - y + y^{2}$

Step 2.2.2.1.4.4.4

Multiply $2$ by $1$ .

$- 1 - y + 2 y + y (2 y) + 1 - y + y^{2}$

Step 2.2.2.1.4.4.5

Raise $y$ to the power of $1$ .

$- 1 - y + 2 y + 2 (y^{1} y) + 1 - y + y^{2}$

Step 2.2.2.1.4.4.6

Raise $y$ to the power of $1$ .

$- 1 - y + 2 y + 2 (y^{1} y^{1}) + 1 - y + y^{2}$

Step 2.2.2.1.4.4.7

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$- 1 - y + 2 y + 2 y^{1 + 1} + 1 - y + y^{2}$

Step 2.2.2.1.4.4.8

Add $1$ and $1$ .

$- 1 - y + 2 y + 2 y^{2} + 1 - y + y^{2}$

Step 2.2.2.1.4.4.9

Add $- y$ and $2 y$ .

$- 1 + y + 2 y^{2} + 1 - y + y^{2}$

Step 2.2.2.1.4.4.10

Add $- 1$ and $1$ .

$y + 2 y^{2} + 0 - y + y^{2}$

Step 2.2.2.1.4.4.11

Add $y + 2 y^{2}$ and $0$ .

$y + 2 y^{2} - y + y^{2}$

Step 2.2.2.1.4.4.12

Subtract $y$ from $y$ .

$2 y^{2} + 0 + y^{2}$

Step 2.2.2.1.4.4.13

Add $2 y^{2}$ and $0$ .

$2 y^{2} + y^{2}$

Step 2.2.2.1.4.4.14

Add $2 y^{2}$ and $y^{2}$ .

$3 y^{2}$

Step 2.2.2.2

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

$\int \frac{1}{u} \cdot \frac{1}{3} d u = \int d x$

Step 2.2.3

Simplify.

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Step 2.2.3.1

Multiply $\frac{1}{u}$ by $\frac{1}{3}$ .

$\int \frac{1}{u \cdot 3} d u = \int d x$

Step 2.2.3.2

Move $3$ to the left of $u$ .

$\int \frac{1}{3 u} d u = \int d x$

Step 2.2.4

Since $\frac{1}{3}$ is constant with respect to $u$ , move $\frac{1}{3}$ out of the integral.

$\frac{1}{3} \int \frac{1}{u} d u = \int d x$

Step 2.2.5

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

$\frac{1}{3} (\ln (| u |) + C_{1}) = \int d x$

Step 2.2.6

Simplify.

$\frac{1}{3} \ln (| u |) + C_{1} = \int d x$

Step 2.2.7

Replace all occurrences of $u$ with $(1 + y) (1 - y + y^{2})$ .

$\frac{1}{3} \ln (| (1 + y) (1 - y + y^{2}) |) + C_{1} = \int d x$

Step 2.3

Apply the constant rule.

$\frac{1}{3} \ln (| (1 + y) (1 - y + y^{2}) |) + C_{1} = x + C_{2}$

Step 2.4

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

$\frac{1}{3} \ln (| (1 + y) (1 - y + y^{2}) |) = x + C$

Step 3

Solve for $y$ .

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Step 3.1

Multiply both sides of the equation by $3$ .

$3 (\frac{1}{3} \ln (| (1 + y) (1 - y + y^{2}) |)) = 3 (x + C)$

Step 3.2

Simplify both sides of the equation.

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Step 3.2.1

Simplify the left side.

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Step 3.2.1.1

Simplify $3 (\frac{1}{3} \ln (| (1 + y) (1 - y + y^{2}) |))$ .

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Step 3.2.1.1.1

Expand $(1 + y) (1 - y + y^{2})$ by multiplying each term in the first expression by each term in the second expression.

$3 (\frac{1}{3} \ln (| 1 \cdot 1 + 1 (- y) + 1 y^{2} + y \cdot 1 + y (- y) + y \cdot y^{2} |)) = 3 (x + C)$

Step 3.2.1.1.2

Simplify terms.

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Step 3.2.1.1.2.1

Simplify each term.

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Step 3.2.1.1.2.1.1

Multiply $1$ by $1$ .

$3 (\frac{1}{3} \ln (| 1 + 1 (- y) + 1 y^{2} + y \cdot 1 + y (- y) + y \cdot y^{2} |)) = 3 (x + C)$

Step 3.2.1.1.2.1.2

Multiply $- y$ by $1$ .

$3 (\frac{1}{3} \ln (| 1 - y + 1 y^{2} + y \cdot 1 + y (- y) + y \cdot y^{2} |)) = 3 (x + C)$

Step 3.2.1.1.2.1.3

Multiply $y^{2}$ by $1$ .

$3 (\frac{1}{3} \ln (| 1 - y + y^{2} + y \cdot 1 + y (- y) + y \cdot y^{2} |)) = 3 (x + C)$

Step 3.2.1.1.2.1.4

Multiply $y$ by $1$ .

$3 (\frac{1}{3} \ln (| 1 - y + y^{2} + y + y (- y) + y \cdot y^{2} |)) = 3 (x + C)$

Step 3.2.1.1.2.1.5

Rewrite using the commutative property of multiplication.

$3 (\frac{1}{3} \ln (| 1 - y + y^{2} + y - y \cdot y + y \cdot y^{2} |)) = 3 (x + C)$

Step 3.2.1.1.2.1.6

Multiply $y$ by $y$ by adding the exponents.

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Step 3.2.1.1.2.1.6.1

Move $y$ .

$3 (\frac{1}{3} \ln (| 1 - y + y^{2} + y - (y \cdot y) + y \cdot y^{2} |)) = 3 (x + C)$

Step 3.2.1.1.2.1.6.2

Multiply $y$ by $y$ .

$3 (\frac{1}{3} \ln (| 1 - y + y^{2} + y - y^{2} + y \cdot y^{2} |)) = 3 (x + C)$

Step 3.2.1.1.2.1.7

Multiply $y$ by $y^{2}$ by adding the exponents.

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Step 3.2.1.1.2.1.7.1

Multiply $y$ by $y^{2}$ .

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Step 3.2.1.1.2.1.7.1.1

Raise $y$ to the power of $1$ .

$3 (\frac{1}{3} \ln (| 1 - y + y^{2} + y - y^{2} + y^{1} y^{2} |)) = 3 (x + C)$

Step 3.2.1.1.2.1.7.1.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$3 (\frac{1}{3} \ln (| 1 - y + y^{2} + y - y^{2} + y^{1 + 2} |)) = 3 (x + C)$

Step 3.2.1.1.2.1.7.2

Add $1$ and $2$ .

$3 (\frac{1}{3} \ln (| 1 - y + y^{2} + y - y^{2} + y^{3} |)) = 3 (x + C)$

Step 3.2.1.1.2.2

Simplify terms.

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Step 3.2.1.1.2.2.1

Combine the opposite terms in $1 - y + y^{2} + y - y^{2} + y^{3}$ .

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Step 3.2.1.1.2.2.1.1

Add $- y$ and $y$ .

$3 (\frac{1}{3} \ln (| 1 + y^{2} + 0 - y^{2} + y^{3} |)) = 3 (x + C)$

Step 3.2.1.1.2.2.1.2

Add $1 + y^{2}$ and $0$ .

$3 (\frac{1}{3} \ln (| 1 + y^{2} - y^{2} + y^{3} |)) = 3 (x + C)$

Step 3.2.1.1.2.2.1.3

Subtract $y^{2}$ from $y^{2}$ .

$3 (\frac{1}{3} \ln (| 1 + 0 + y^{3} |)) = 3 (x + C)$

Step 3.2.1.1.2.2.1.4

Add $1$ and $0$ .

$3 (\frac{1}{3} \ln (| 1 + y^{3} |)) = 3 (x + C)$

Step 3.2.1.1.2.2.2

Combine $\frac{1}{3}$ and $\ln (| 1 + y^{3} |)$ .

$3 \frac{\ln (| 1 + y^{3} |)}{3} = 3 (x + C)$

Step 3.2.1.1.2.2.3

Cancel the common factor of $3$ .

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Step 3.2.1.1.2.2.3.1

Cancel the common factor.

$3 \frac{\ln (| 1 + y^{3} |)}{3} = 3 (x + C)$

Step 3.2.1.1.2.2.3.2

Rewrite the expression.

$\ln (| 1 + y^{3} |) = 3 (x + C)$

Step 3.2.2

Simplify the right side.

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Step 3.2.2.1

Apply the distributive property.

$\ln (| 1 + y^{3} |) = 3 x + 3 C$

Step 3.3

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

$e^{\ln (| 1 + y^{3} |)} = e^{3 x + 3 C}$

Step 3.4

Rewrite $\ln (| 1 + y^{3} |) = 3 x + 3 C$ 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 $b^{y} = x$ .

$e^{3 x + 3 C} = | 1 + y^{3} |$

Step 3.5

Solve for $y$ .

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Step 3.5.1

Rewrite the equation as $| 1 + y^{3} | = e^{3 x + 3 C}$ .

$| 1 + y^{3} | = e^{3 x + 3 C}$

Step 3.5.2

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

$1 + y^{3} = \pm e^{3 x + 3 C}$

Step 3.5.3

Subtract $1$ from both sides of the equation.

$y^{3} = \pm e^{3 x + 3 C} - 1$

Step 3.5.4

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$y = \sqrt[3]{\pm e^{3 x + 3 C} - 1}$

Step 4

Group the constant terms together.

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Step 4.1

Simplify the constant of integration.

$y = \sqrt[3]{\pm e^{3 x + C} - 1}$

Step 4.2

Rewrite $e^{3 x + C}$ as $e^{3 x} e^{C}$ .

$y = \sqrt[3]{\pm e^{3 x} e^{C} - 1}$

Step 4.3

Reorder $e^{3 x}$ and $e^{C}$ .

$y = \sqrt[3]{\pm e^{C} e^{3 x} - 1}$

Step 4.4

Combine constants with the plus or minus.

$y = \sqrt[3]{C e^{3 x} - 1}$