Calculus Examples

$(2 x + 1) d y + y^{2} d x = 0$

Step 1

Subtract $y^{2} d x$ from both sides of the equation.

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

Step 2

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

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

Step 3

Simplify.

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Step 3.1

Cancel the common factor of $2 x + 1$ .

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Step 3.1.1

Factor $2 x + 1$ out of $(2 x + 1) y^{2}$ .

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

Step 3.1.2

Cancel the common factor.

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

Step 3.1.3

Rewrite the expression.

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

Step 3.2

Rewrite using the commutative property of multiplication.

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

Step 3.3

Cancel the common factor of $y^{2}$ .

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Step 3.3.1

Move the leading negative in $- \frac{1}{(2 x + 1) y^{2}}$ into the numerator.

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

Step 3.3.2

Factor $y^{2}$ out of $(2 x + 1) y^{2}$ .

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

Step 3.3.3

Cancel the common factor.

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

Step 3.3.4

Rewrite the expression.

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

Step 3.4

Move the negative in front of the fraction.

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

Step 4

Integrate both sides.

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Step 4.1

Set up an integral on each side.

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

Step 4.2

Integrate the left side.

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Step 4.2.1

Apply basic rules of exponents.

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Step 4.2.1.1

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

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

Step 4.2.1.2

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

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Step 4.2.1.2.1

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

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

Step 4.2.1.2.2

Multiply $2$ by $- 1$ .

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

Step 4.2.2

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

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

Step 4.2.3

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

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

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.

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

Step 4.3.2

Let $u = 2 x + 1$ . Then $d u = 2 d x$ , so $\frac{1}{2} d u = d x$ . Rewrite using $u$ and $d$ $u$ .

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Step 4.3.2.1

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

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Step 4.3.2.1.1

Differentiate $2 x + 1$ .

$\frac{d}{d x} [2 x + 1]$

Step 4.3.2.1.2

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

$\frac{d}{d x} [2 x] + \frac{d}{d x} [1]$

Step 4.3.2.1.3

Evaluate $\frac{d}{d x} [2 x]$ .

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Step 4.3.2.1.3.1

Since $2$ is constant with respect to $x$ , the derivative of $2 x$ with respect to $x$ is $2 \frac{d}{d x} [x]$ .

$2 \frac{d}{d x} [x] + \frac{d}{d x} [1]$

Step 4.3.2.1.3.2

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

$2 \cdot 1 + \frac{d}{d x} [1]$

Step 4.3.2.1.3.3

Multiply $2$ by $1$ .

$2 + \frac{d}{d x} [1]$

Step 4.3.2.1.4

Differentiate using the Constant Rule.

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Step 4.3.2.1.4.1

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

$2 + 0$

Step 4.3.2.1.4.2

Add $2$ and $0$ .

$2$

Step 4.3.2.2

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

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

Step 4.3.3

Simplify.

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Step 4.3.3.1

Multiply $\frac{1}{u}$ by $\frac{1}{2}$ .

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

Step 4.3.3.2

Move $2$ to the left of $u$ .

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

Step 4.3.4

Since $\frac{1}{2}$ is constant with respect to $u$ , move $\frac{1}{2}$ out of the integral.

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

Step 4.3.5

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

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

Step 4.3.6

Simplify.

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

Step 4.3.7

Replace all occurrences of $u$ with $2 x + 1$ .

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

Step 4.4

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

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

Step 5

Solve for $y$ .

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Step 5.1

Multiply $- \frac{1}{2} \ln (| 2 x + 1 |)$ .

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Step 5.1.1

Reorder $\ln (| 2 x + 1 |)$ and $\frac{1}{2}$ .

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

Step 5.1.2

Simplify $\frac{1}{2} \ln (| 2 x + 1 |)$ by moving $\frac{1}{2}$ inside the logarithm.

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

Step 5.2

Find the LCD of the terms in the equation.

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Step 5.2.1

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

$y, 1, 1$

Step 5.2.2

The LCM of one and any expression is the expression.

$y$

Step 5.3

Multiply each term in $- \frac{1}{y} = - \ln ({| 2 x + 1 |}^{\frac{1}{2}}) + C$ by $y$ to eliminate the fractions.

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Step 5.3.1

Multiply each term in $- \frac{1}{y} = - \ln ({| 2 x + 1 |}^{\frac{1}{2}}) + C$ by $y$ .

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

Step 5.3.2

Simplify the left side.

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Step 5.3.2.1

Cancel the common factor of $y$ .

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Step 5.3.2.1.1

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

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

Step 5.3.2.1.2

Cancel the common factor.

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

Step 5.3.2.1.3

Rewrite the expression.

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

Step 5.3.3

Simplify the right side.

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Step 5.3.3.1

Reorder factors in $- \ln ({| 2 x + 1 |}^{\frac{1}{2}}) y + C y$ .

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

Step 5.4

Solve the equation.

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Step 5.4.1

Rewrite the equation as $- y \ln ({| 2 x + 1 |}^{\frac{1}{2}}) + C y = - 1$ .

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

Step 5.4.2

Factor $y$ out of $- y \ln ({| 2 x + 1 |}^{\frac{1}{2}}) + C y$ .

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Step 5.4.2.1

Factor $y$ out of $- y \ln ({| 2 x + 1 |}^{\frac{1}{2}})$ .

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

Step 5.4.2.2

Factor $y$ out of $C y$ .

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

Step 5.4.2.3

Factor $y$ out of $y (- 1 \ln ({| 2 x + 1 |}^{\frac{1}{2}})) + y C$ .

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

Step 5.4.3

Rewrite $- 1 \ln ({| 2 x + 1 |}^{\frac{1}{2}})$ as $- \ln ({| 2 x + 1 |}^{\frac{1}{2}})$ .

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

Step 5.4.4

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

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Step 5.4.4.1

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

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

Step 5.4.4.2

Simplify the left side.

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Step 5.4.4.2.1

Divide $y$ by $1$ .

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

Step 5.4.4.3

Simplify the right side.

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Step 5.4.4.3.1

Move the negative in front of the fraction.

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

Step 5.4.4.3.2

Factor $- 1$ out of $- \ln ({| 2 x + 1 |}^{\frac{1}{2}})$ .

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

Step 5.4.4.3.3

Factor $- 1$ out of $C$ .

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

Step 5.4.4.3.4

Factor $- 1$ out of $- (\ln ({| 2 x + 1 |}^{\frac{1}{2}})) - 1 (- C)$ .

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

Step 5.4.4.3.5

Simplify the expression.

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Step 5.4.4.3.5.1

Rewrite $- (\ln ({| 2 x + 1 |}^{\frac{1}{2}}) - C)$ as $- 1 (\ln ({| 2 x + 1 |}^{\frac{1}{2}}) - C)$ .

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

Step 5.4.4.3.5.2

Move the negative in front of the fraction.

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

Step 5.4.4.3.5.3

Multiply $- 1$ by $- 1$ .

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

Step 5.4.4.3.5.4

Multiply $\frac{1}{\ln ({| 2 x + 1 |}^{\frac{1}{2}}) - C}$ by $1$ .

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

Step 6

Simplify the constant of integration.

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