Calculus Examples

$\frac{d y}{d x} = 2 \cdot \frac{1}{x + 7} \cdot y$

Step 1

Separate the variables.

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Step 1.1

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

$\frac{1}{y} \frac{d y}{d x} = \frac{1}{y} (2 \cdot \frac{1}{x + 7} \cdot y)$

Step 1.2

Simplify.

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Step 1.2.1

Rewrite using the commutative property of multiplication.

$\frac{1}{y} \frac{d y}{d x} = 2 \frac{1}{y} (\frac{1}{x + 7} y)$

Step 1.2.2

Combine $2$ and $\frac{1}{y}$ .

$\frac{1}{y} \frac{d y}{d x} = \frac{2}{y} (\frac{1}{x + 7} y)$

Step 1.2.3

Cancel the common factor of $y$ .

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Step 1.2.3.1

Factor $y$ out of $\frac{1}{x + 7} y$ .

$\frac{1}{y} \frac{d y}{d x} = \frac{2}{y} (y \frac{1}{x + 7})$

Step 1.2.3.2

Cancel the common factor.

$\frac{1}{y} \frac{d y}{d x} = \frac{2}{y} (y \frac{1}{x + 7})$

Step 1.2.3.3

Rewrite the expression.

$\frac{1}{y} \frac{d y}{d x} = 2 \frac{1}{x + 7}$

Step 1.2.4

Combine $2$ and $\frac{1}{x + 7}$ .

$\frac{1}{y} \frac{d y}{d x} = \frac{2}{x + 7}$

Step 1.3

Rewrite the equation.

$\frac{1}{y} d y = \frac{2}{x + 7} d x$

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

$\int \frac{1}{y} d y = \int \frac{2}{x + 7} d x$

Step 2.2

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

$\ln (| y |) + C_{1} = \int \frac{2}{x + 7} d x$

Step 2.3

Integrate the right side.

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Step 2.3.1

Since $2$ is constant with respect to $x$ , move $2$ out of the integral.

$\ln (| y |) + C_{1} = 2 \int \frac{1}{x + 7} d x$

Step 2.3.2

Let $u = x + 7$ . Then $d u = d x$ . Rewrite using $u$ and $d$ $u$ .

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Step 2.3.2.1

Let $u = x + 7$ . Find $\frac{d u}{d x}$ .

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Step 2.3.2.1.1

Differentiate $x + 7$ .

$\frac{d}{d x} [x + 7]$

Step 2.3.2.1.2

By the Sum Rule, the derivative of $x + 7$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [7]$ .

$\frac{d}{d x} [x] + \frac{d}{d x} [7]$

Step 2.3.2.1.3

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

$1 + \frac{d}{d x} [7]$

Step 2.3.2.1.4

Since $7$ is constant with respect to $x$ , the derivative of $7$ with respect to $x$ is $0$ .

$1 + 0$

Step 2.3.2.1.5

Add $1$ and $0$ .

$1$

Step 2.3.2.2

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

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

Step 2.3.3

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

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

Step 2.3.4

Simplify.

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

Step 2.3.5

Replace all occurrences of $u$ with $x + 7$ .

$\ln (| y |) + C_{1} = 2 \ln (| x + 7 |) + C_{2}$

Step 2.4

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

$\ln (| y |) = 2 \ln (| x + 7 |) + C$

Step 3

Solve for $y$ .

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Step 3.1

Move all the terms containing a logarithm to the left side of the equation.

$\ln (| y |) - 2 \ln (| x + 7 |) = C$

Step 3.2

Simplify the left side.

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Step 3.2.1

Simplify $\ln (| y |) - 2 \ln (| x + 7 |)$ .

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Step 3.2.1.1

Simplify each term.

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Step 3.2.1.1.1

Simplify $- 2 \ln (| x + 7 |)$ by moving $2$ inside the logarithm.

$\ln (| y |) - \ln ({| x + 7 |}^{2}) = C$

Step 3.2.1.1.2

Remove the absolute value in ${| x + 7 |}^{2}$ because exponentiations with even powers are always positive.

$\ln (| y |) - \ln ({(x + 7)}^{2}) = C$

Step 3.2.1.2

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

$\ln (\frac{| y |}{{(x + 7)}^{2}}) = C$

Step 3.3

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

$e^{\ln (\frac{| y |}{{(x + 7)}^{2}})} = e^{C}$

Step 3.4

Rewrite $\ln (\frac{| y |}{{(x + 7)}^{2}}) = C$ 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 $b^{y} = x$ .

$e^{C} = \frac{| y |}{{(x + 7)}^{2}}$

Step 3.5

Solve for $y$ .

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Step 3.5.1

Rewrite the equation as $\frac{| y |}{{(x + 7)}^{2}} = e^{C}$ .

$\frac{| y |}{{(x + 7)}^{2}} = e^{C}$

Step 3.5.2

Multiply both sides by ${(x + 7)}^{2}$ .

$\frac{| y |}{{(x + 7)}^{2}} {(x + 7)}^{2} = e^{C} {(x + 7)}^{2}$

Step 3.5.3

Simplify the left side.

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Step 3.5.3.1

Cancel the common factor of ${(x + 7)}^{2}$ .

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Step 3.5.3.1.1

Cancel the common factor.

$\frac{| y |}{{(x + 7)}^{2}} {(x + 7)}^{2} = e^{C} {(x + 7)}^{2}$

Step 3.5.3.1.2

Rewrite the expression.

$| y | = e^{C} {(x + 7)}^{2}$

Step 3.5.4

Solve for $y$ .

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Step 3.5.4.1

Simplify $e^{C} {(x + 7)}^{2}$ .

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Step 3.5.4.1.1

Rewrite ${(x + 7)}^{2}$ as $(x + 7) (x + 7)$ .

$| y | = e^{C} ((x + 7) (x + 7))$

Step 3.5.4.1.2

Expand $(x + 7) (x + 7)$ using the FOIL Method.

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Step 3.5.4.1.2.1

Apply the distributive property.

$| y | = e^{C} (x (x + 7) + 7 (x + 7))$

Step 3.5.4.1.2.2

Apply the distributive property.

$| y | = e^{C} (x \cdot x + x \cdot 7 + 7 (x + 7))$

Step 3.5.4.1.2.3

Apply the distributive property.

$| y | = e^{C} (x \cdot x + x \cdot 7 + 7 x + 7 \cdot 7)$

Step 3.5.4.1.3

Simplify and combine like terms.

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Step 3.5.4.1.3.1

Simplify each term.

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Step 3.5.4.1.3.1.1

Multiply $x$ by $x$ .

$| y | = e^{C} (x^{2} + x \cdot 7 + 7 x + 7 \cdot 7)$

Step 3.5.4.1.3.1.2

Move $7$ to the left of $x$ .

$| y | = e^{C} (x^{2} + 7 \cdot x + 7 x + 7 \cdot 7)$

Step 3.5.4.1.3.1.3

Multiply $7$ by $7$ .

$| y | = e^{C} (x^{2} + 7 x + 7 x + 49)$

Step 3.5.4.1.3.2

Add $7 x$ and $7 x$ .

$| y | = e^{C} (x^{2} + 14 x + 49)$

Step 3.5.4.1.4

Apply the distributive property.

$| y | = e^{C} x^{2} + e^{C} (14 x) + e^{C} \cdot 49$

Step 3.5.4.1.5

Simplify.

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Step 3.5.4.1.5.1

Rewrite using the commutative property of multiplication.

$| y | = e^{C} x^{2} + 14 e^{C} x + e^{C} \cdot 49$

Step 3.5.4.1.5.2

Move $49$ to the left of $e^{C}$ .

$| y | = e^{C} x^{2} + 14 e^{C} x + 49 \cdot e^{C}$

Step 3.5.4.1.6

Reorder factors in $e^{C} x^{2} + 14 e^{C} x + 49 e^{C}$ .

$| y | = x^{2} e^{C} + 14 x e^{C} + 49 e^{C}$

Step 3.5.4.2

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

$y = \pm (x^{2} e^{C} + 14 x e^{C} + 49 e^{C})$

Step 4

Simplify the constant of integration.

$y = \pm (x^{2} C + 14 x C + C)$