Calculus Examples

$\frac{d y}{d x} = \frac{2 y + 3}{4 x + 5}$

Step 1

Separate the variables.

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Step 1.1

Multiply both sides by $\frac{1}{2 y + 3}$ .

$\frac{1}{2 y + 3} \frac{d y}{d x} = \frac{1}{2 y + 3} \cdot \frac{2 y + 3}{4 x + 5}$

Step 1.2

Cancel the common factor of $2 y + 3$ .

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Step 1.2.1

Cancel the common factor.

$\frac{1}{2 y + 3} \frac{d y}{d x} = \frac{1}{2 y + 3} \cdot \frac{2 y + 3}{4 x + 5}$

Step 1.2.2

Rewrite the expression.

$\frac{1}{2 y + 3} \frac{d y}{d x} = \frac{1}{4 x + 5}$

Step 1.3

Rewrite the equation.

$\frac{1}{2 y + 3} d y = \frac{1}{4 x + 5} d x$

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

$\int \frac{1}{2 y + 3} d y = \int \frac{1}{4 x + 5} d x$

Step 2.2

Integrate the left side.

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Step 2.2.1

Let $u_{1} = 2 y + 3$ . Then $d u_{1} = 2 d y$ , so $\frac{1}{2} d u_{1} = d y$ . Rewrite using $u_{1}$ and $d$ $u_{1}$ .

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Step 2.2.1.1

Let $u_{1} = 2 y + 3$ . Find $\frac{d u_{1}}{d y}$ .

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Step 2.2.1.1.1

Differentiate $2 y + 3$ .

$\frac{d}{d y} [2 y + 3]$

Step 2.2.1.1.2

By the Sum Rule, the derivative of $2 y + 3$ with respect to $y$ is $\frac{d}{d y} [2 y] + \frac{d}{d y} [3]$ .

$\frac{d}{d y} [2 y] + \frac{d}{d y} [3]$

Step 2.2.1.1.3

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

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Step 2.2.1.1.3.1

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

$2 \frac{d}{d y} [y] + \frac{d}{d y} [3]$

Step 2.2.1.1.3.2

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

$2 \cdot 1 + \frac{d}{d y} [3]$

Step 2.2.1.1.3.3

Multiply $2$ by $1$ .

$2 + \frac{d}{d y} [3]$

Step 2.2.1.1.4

Differentiate using the Constant Rule.

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Step 2.2.1.1.4.1

Since $3$ is constant with respect to $y$ , the derivative of $3$ with respect to $y$ is $0$ .

$2 + 0$

Step 2.2.1.1.4.2

Add $2$ and $0$ .

$2$

Step 2.2.1.2

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

$\int \frac{1}{u_{1}} \cdot \frac{1}{2} d u_{1} = \int \frac{1}{4 x + 5} d x$

Step 2.2.2

Simplify.

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Step 2.2.2.1

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

$\int \frac{1}{u_{1} \cdot 2} d u_{1} = \int \frac{1}{4 x + 5} d x$

Step 2.2.2.2

Move $2$ to the left of $u_{1}$ .

$\int \frac{1}{2 u_{1}} d u_{1} = \int \frac{1}{4 x + 5} d x$

Step 2.2.3

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

$\frac{1}{2} \int \frac{1}{u_{1}} d u_{1} = \int \frac{1}{4 x + 5} d x$

Step 2.2.4

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

$\frac{1}{2} (\ln (| u_{1} |) + C_{1}) = \int \frac{1}{4 x + 5} d x$

Step 2.2.5

Simplify.

$\frac{1}{2} \ln (| u_{1} |) + C_{1} = \int \frac{1}{4 x + 5} d x$

Step 2.2.6

Replace all occurrences of $u_{1}$ with $2 y + 3$ .

$\frac{1}{2} \ln (| 2 y + 3 |) + C_{1} = \int \frac{1}{4 x + 5} d x$

Step 2.3

Integrate the right side.

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Step 2.3.1

Let $u_{2} = 4 x + 5$ . Then $d u_{2} = 4 d x$ , so $\frac{1}{4} d u_{2} = d x$ . Rewrite using $u_{2}$ and $d$ $u_{2}$ .

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Step 2.3.1.1

Let $u_{2} = 4 x + 5$ . Find $\frac{d u_{2}}{d x}$ .

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Step 2.3.1.1.1

Differentiate $4 x + 5$ .

$\frac{d}{d x} [4 x + 5]$

Step 2.3.1.1.2

By the Sum Rule, the derivative of $4 x + 5$ with respect to $x$ is $\frac{d}{d x} [4 x] + \frac{d}{d x} [5]$ .

$\frac{d}{d x} [4 x] + \frac{d}{d x} [5]$

Step 2.3.1.1.3

Evaluate $\frac{d}{d x} [4 x]$ .

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Step 2.3.1.1.3.1

Since $4$ is constant with respect to $x$ , the derivative of $4 x$ with respect to $x$ is $4 \frac{d}{d x} [x]$ .

$4 \frac{d}{d x} [x] + \frac{d}{d x} [5]$

Step 2.3.1.1.3.2

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

$4 \cdot 1 + \frac{d}{d x} [5]$

Step 2.3.1.1.3.3

Multiply $4$ by $1$ .

$4 + \frac{d}{d x} [5]$

Step 2.3.1.1.4

Differentiate using the Constant Rule.

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Step 2.3.1.1.4.1

Since $5$ is constant with respect to $x$ , the derivative of $5$ with respect to $x$ is $0$ .

$4 + 0$

Step 2.3.1.1.4.2

Add $4$ and $0$ .

$4$

Step 2.3.1.2

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

$\frac{1}{2} \ln (| 2 y + 3 |) + C_{1} = \int \frac{1}{u_{2}} \cdot \frac{1}{4} d u_{2}$

Step 2.3.2

Simplify.

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Step 2.3.2.1

Multiply $\frac{1}{u_{2}}$ by $\frac{1}{4}$ .

$\frac{1}{2} \ln (| 2 y + 3 |) + C_{1} = \int \frac{1}{u_{2} \cdot 4} d u_{2}$

Step 2.3.2.2

Move $4$ to the left of $u_{2}$ .

$\frac{1}{2} \ln (| 2 y + 3 |) + C_{1} = \int \frac{1}{4 u_{2}} d u_{2}$

Step 2.3.3

Since $\frac{1}{4}$ is constant with respect to $u_{2}$ , move $\frac{1}{4}$ out of the integral.

$\frac{1}{2} \ln (| 2 y + 3 |) + C_{1} = \frac{1}{4} \int \frac{1}{u_{2}} d u_{2}$

Step 2.3.4

The integral of $\frac{1}{u_{2}}$ with respect to $u_{2}$ is $\ln (| u_{2} |)$ .

$\frac{1}{2} \ln (| 2 y + 3 |) + C_{1} = \frac{1}{4} (\ln (| u_{2} |) + C_{2})$

Step 2.3.5

Simplify.

$\frac{1}{2} \ln (| 2 y + 3 |) + C_{1} = \frac{1}{4} \ln (| u_{2} |) + C_{2}$

Step 2.3.6

Replace all occurrences of $u_{2}$ with $4 x + 5$ .

$\frac{1}{2} \ln (| 2 y + 3 |) + C_{1} = \frac{1}{4} \ln (| 4 x + 5 |) + C_{2}$

Step 2.4

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

$\frac{1}{2} \ln (| 2 y + 3 |) = \frac{1}{4} \ln (| 4 x + 5 |) + C$

Step 3

Solve for $y$ .

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Step 3.1

Multiply both sides of the equation by $2$ .

$2 (\frac{1}{2} \ln (| 2 y + 3 |)) = 2 (\frac{1}{4} \ln (| 4 x + 5 |) + C)$

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

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Step 3.2.1.1.1

Combine $\frac{1}{2}$ and $\ln (| 2 y + 3 |)$ .

$2 \frac{\ln (| 2 y + 3 |)}{2} = 2 (\frac{1}{4} \ln (| 4 x + 5 |) + C)$

Step 3.2.1.1.2

Cancel the common factor of $2$ .

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Step 3.2.1.1.2.1

Cancel the common factor.

$2 \frac{\ln (| 2 y + 3 |)}{2} = 2 (\frac{1}{4} \ln (| 4 x + 5 |) + C)$

Step 3.2.1.1.2.2

Rewrite the expression.

$\ln (| 2 y + 3 |) = 2 (\frac{1}{4} \ln (| 4 x + 5 |) + C)$

Step 3.2.2

Simplify the right side.

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Step 3.2.2.1

Simplify $2 (\frac{1}{4} \ln (| 4 x + 5 |) + C)$ .

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Step 3.2.2.1.1

Combine $\frac{1}{4}$ and $\ln (| 4 x + 5 |)$ .

$\ln (| 2 y + 3 |) = 2 (\frac{\ln (| 4 x + 5 |)}{4} + C)$

Step 3.2.2.1.2

To write $C$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

$\ln (| 2 y + 3 |) = 2 (\frac{\ln (| 4 x + 5 |)}{4} + C \cdot \frac{4}{4})$

Step 3.2.2.1.3

Simplify terms.

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Step 3.2.2.1.3.1

Combine $C$ and $\frac{4}{4}$ .

$\ln (| 2 y + 3 |) = 2 (\frac{\ln (| 4 x + 5 |)}{4} + \frac{C \cdot 4}{4})$

Step 3.2.2.1.3.2

Combine the numerators over the common denominator.

$\ln (| 2 y + 3 |) = 2 \frac{\ln (| 4 x + 5 |) + C \cdot 4}{4}$

Step 3.2.2.1.3.3

Cancel the common factor of $2$ .

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Step 3.2.2.1.3.3.1

Factor $2$ out of $4$ .

$\ln (| 2 y + 3 |) = 2 \frac{\ln (| 4 x + 5 |) + C \cdot 4}{2 (2)}$

Step 3.2.2.1.3.3.2

Cancel the common factor.

$\ln (| 2 y + 3 |) = 2 \frac{\ln (| 4 x + 5 |) + C \cdot 4}{2 \cdot 2}$

Step 3.2.2.1.3.3.3

Rewrite the expression.

$\ln (| 2 y + 3 |) = \frac{\ln (| 4 x + 5 |) + C \cdot 4}{2}$

Step 3.2.2.1.4

Move $4$ to the left of $C$ .

$\ln (| 2 y + 3 |) = \frac{\ln (| 4 x + 5 |) + 4 C}{2}$

Step 3.3

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

$e^{\ln (| 2 y + 3 |)} = e^{\frac{\ln (| 4 x + 5 |) + 4 C}{2}}$

Step 3.4

Rewrite $\ln (| 2 y + 3 |) = \frac{\ln (| 4 x + 5 |) + 4 C}{2}$ 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 $b^{y} = x$ .

$e^{\frac{\ln (| 4 x + 5 |) + 4 C}{2}} = | 2 y + 3 |$

Step 3.5

Solve for $y$ .

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Step 3.5.1

Rewrite the equation as $| 2 y + 3 | = e^{\frac{\ln (| 4 x + 5 |) + 4 C}{2}}$ .

$| 2 y + 3 | = e^{\frac{\ln (| 4 x + 5 |) + 4 C}{2}}$

Step 3.5.2

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

$2 y + 3 = \pm e^{\frac{\ln (| 4 x + 5 |) + 4 C}{2}}$

Step 3.5.3

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

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Step 3.5.3.1

Subtract $3$ from both sides of the equation.

$2 y = \pm e^{\frac{\ln (| 4 x + 5 |) + 4 C}{2}} - 3$

Step 3.5.3.2

Simplify each term.

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Step 3.5.3.2.1

Split the fraction $\frac{\ln (| 4 x + 5 |) + 4 C}{2}$ into two fractions.

$2 y = \pm e^{\frac{\ln (| 4 x + 5 |)}{2} + \frac{4 C}{2}} - 3$

Step 3.5.3.2.2

Simplify each term.

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Step 3.5.3.2.2.1

Rewrite $\frac{\ln (| 4 x + 5 |)}{2}$ as $\frac{1}{2} \ln (| 4 x + 5 |)$ .

$2 y = \pm e^{\frac{1}{2} \ln (| 4 x + 5 |) + \frac{4 C}{2}} - 3$

Step 3.5.3.2.2.2

Simplify $\frac{1}{2} \ln (| 4 x + 5 |)$ by moving $\frac{1}{2}$ inside the logarithm.

$2 y = \pm e^{\ln ({| 4 x + 5 |}^{\frac{1}{2}}) + \frac{4 C}{2}} - 3$

Step 3.5.3.2.2.3

Cancel the common factor of $4$ and $2$ .

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Step 3.5.3.2.2.3.1

Factor $2$ out of $4 C$ .

$2 y = \pm e^{\ln ({| 4 x + 5 |}^{\frac{1}{2}}) + \frac{2 (2 C)}{2}} - 3$

Step 3.5.3.2.2.3.2

Cancel the common factors.

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Step 3.5.3.2.2.3.2.1

Factor $2$ out of $2$ .

$2 y = \pm e^{\ln ({| 4 x + 5 |}^{\frac{1}{2}}) + \frac{2 (2 C)}{2 (1)}} - 3$

Step 3.5.3.2.2.3.2.2

Cancel the common factor.

$2 y = \pm e^{\ln ({| 4 x + 5 |}^{\frac{1}{2}}) + \frac{2 (2 C)}{2 \cdot 1}} - 3$

Step 3.5.3.2.2.3.2.3

Rewrite the expression.

$2 y = \pm e^{\ln ({| 4 x + 5 |}^{\frac{1}{2}}) + \frac{2 C}{1}} - 3$

Step 3.5.3.2.2.3.2.4

Divide $2 C$ by $1$ .

$2 y = \pm e^{\ln ({| 4 x + 5 |}^{\frac{1}{2}}) + 2 C} - 3$

Step 3.5.4

Divide each term in $2 y = \pm e^{\ln ({| 4 x + 5 |}^{\frac{1}{2}}) + 2 C} - 3$ by $2$ and simplify.

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Step 3.5.4.1

Divide each term in $2 y = \pm e^{\ln ({| 4 x + 5 |}^{\frac{1}{2}}) + 2 C} - 3$ by $2$ .

$\frac{2 y}{2} = \frac{\pm e^{\ln ({| 4 x + 5 |}^{\frac{1}{2}}) + 2 C}}{2} + \frac{- 3}{2}$

Step 3.5.4.2

Simplify the left side.

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Step 3.5.4.2.1

Cancel the common factor of $2$ .

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Step 3.5.4.2.1.1

Cancel the common factor.

$\frac{2 y}{2} = \frac{\pm e^{\ln ({| 4 x + 5 |}^{\frac{1}{2}}) + 2 C}}{2} + \frac{- 3}{2}$

Step 3.5.4.2.1.2

Divide $y$ by $1$ .

$y = \frac{\pm e^{\ln ({| 4 x + 5 |}^{\frac{1}{2}}) + 2 C}}{2} + \frac{- 3}{2}$

Step 3.5.4.3

Simplify the right side.

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Step 3.5.4.3.1

Move the negative in front of the fraction.

$y = \frac{\pm e^{\ln ({| 4 x + 5 |}^{\frac{1}{2}}) + 2 C}}{2} - \frac{3}{2}$

Step 3.5.4.3.2

Combine the numerators over the common denominator.

$y = \frac{\pm e^{\ln ({| 4 x + 5 |}^{\frac{1}{2}}) + 2 C} - 3}{2}$

Step 4

Group the constant terms together.

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Step 4.1

Simplify the constant of integration.

$y = \frac{\pm e^{\ln ({| 4 x + 5 |}^{\frac{1}{2}}) + C} - 3}{2}$

Step 4.2

Rewrite $e^{\ln ({| 4 x + 5 |}^{\frac{1}{2}}) + C}$ as $e^{\ln ({| 4 x + 5 |}^{\frac{1}{2}})} e^{C}$ .

$y = \frac{\pm e^{\ln ({| 4 x + 5 |}^{\frac{1}{2}})} e^{C} - 3}{2}$

Step 4.3

Reorder $e^{\ln ({| 4 x + 5 |}^{\frac{1}{2}})}$ and $e^{C}$ .

$y = \frac{\pm e^{C} e^{\ln ({| 4 x + 5 |}^{\frac{1}{2}})} - 3}{2}$

Step 4.4

Combine constants with the plus or minus.

$y = \frac{C e^{\ln ({| 4 x + 5 |}^{\frac{1}{2}})} - 3}{2}$