Calculus Examples

$\frac{d y}{d x} = \frac{3 y + 4}{7 x + 5}$

Step 1

Separate the variables.

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Step 1.1

Multiply both sides by $\frac{1}{3 y + 4}$ .

$\frac{1}{3 y + 4} \frac{d y}{d x} = \frac{1}{3 y + 4} \cdot \frac{3 y + 4}{7 x + 5}$

Step 1.2

Cancel the common factor of $3 y + 4$ .

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Step 1.2.1

Cancel the common factor.

$\frac{1}{3 y + 4} \frac{d y}{d x} = \frac{1}{3 y + 4} \cdot \frac{3 y + 4}{7 x + 5}$

Step 1.2.2

Rewrite the expression.

$\frac{1}{3 y + 4} \frac{d y}{d x} = \frac{1}{7 x + 5}$

Step 1.3

Rewrite the equation.

$\frac{1}{3 y + 4} d y = \frac{1}{7 x + 5} d x$

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

$\int \frac{1}{3 y + 4} d y = \int \frac{1}{7 x + 5} d x$

Step 2.2

Integrate the left side.

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Step 2.2.1

Let $u_{1} = 3 y + 4$ . Then $d u_{1} = 3 d y$ , so $\frac{1}{3} d u_{1} = d y$ . Rewrite using $u_{1}$ and $d$ $u_{1}$ .

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Step 2.2.1.1

Let $u_{1} = 3 y + 4$ . Find $\frac{d u_{1}}{d y}$ .

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Step 2.2.1.1.1

Differentiate $3 y + 4$ .

$\frac{d}{d y} [3 y + 4]$

Step 2.2.1.1.2

By the Sum Rule, the derivative of $3 y + 4$ with respect to $y$ is $\frac{d}{d y} [3 y] + \frac{d}{d y} [4]$ .

$\frac{d}{d y} [3 y] + \frac{d}{d y} [4]$

Step 2.2.1.1.3

Evaluate $\frac{d}{d y} [3 y]$ .

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Step 2.2.1.1.3.1

Since $3$ is constant with respect to $y$ , the derivative of $3 y$ with respect to $y$ is $3 \frac{d}{d y} [y]$ .

$3 \frac{d}{d y} [y] + \frac{d}{d y} [4]$

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 $n = 1$ .

$3 \cdot 1 + \frac{d}{d y} [4]$

Step 2.2.1.1.3.3

Multiply $3$ by $1$ .

$3 + \frac{d}{d y} [4]$

Step 2.2.1.1.4

Differentiate using the Constant Rule.

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Step 2.2.1.1.4.1

Since $4$ is constant with respect to $y$ , the derivative of $4$ with respect to $y$ is $0$ .

$3 + 0$

Step 2.2.1.1.4.2

Add $3$ and $0$ .

$3$

Step 2.2.1.2

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

$\int \frac{1}{u_{1}} \cdot \frac{1}{3} d u_{1} = \int \frac{1}{7 x + 5} d x$

Step 2.2.2

Simplify.

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Step 2.2.2.1

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

$\int \frac{1}{u_{1} \cdot 3} d u_{1} = \int \frac{1}{7 x + 5} d x$

Step 2.2.2.2

Move $3$ to the left of $u_{1}$ .

$\int \frac{1}{3 u_{1}} d u_{1} = \int \frac{1}{7 x + 5} d x$

Step 2.2.3

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

$\frac{1}{3} \int \frac{1}{u_{1}} d u_{1} = \int \frac{1}{7 x + 5} d x$

Step 2.2.4

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

$\frac{1}{3} (\ln (| u_{1} |) + C_{1}) = \int \frac{1}{7 x + 5} d x$

Step 2.2.5

Simplify.

$\frac{1}{3} \ln (| u_{1} |) + C_{1} = \int \frac{1}{7 x + 5} d x$

Step 2.2.6

Replace all occurrences of $u_{1}$ with $3 y + 4$ .

$\frac{1}{3} \ln (| 3 y + 4 |) + C_{1} = \int \frac{1}{7 x + 5} d x$

Step 2.3

Integrate the right side.

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Step 2.3.1

Let $u_{2} = 7 x + 5$ . Then $d u_{2} = 7 d x$ , so $\frac{1}{7} d u_{2} = d x$ . Rewrite using $u_{2}$ and $d$ $u_{2}$ .

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Step 2.3.1.1

Let $u_{2} = 7 x + 5$ . Find $\frac{d u_{2}}{d x}$ .

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Step 2.3.1.1.1

Differentiate $7 x + 5$ .

$\frac{d}{d x} [7 x + 5]$

Step 2.3.1.1.2

By the Sum Rule, the derivative of $7 x + 5$ with respect to $x$ is $\frac{d}{d x} [7 x] + \frac{d}{d x} [5]$ .

$\frac{d}{d x} [7 x] + \frac{d}{d x} [5]$

Step 2.3.1.1.3

Evaluate $\frac{d}{d x} [7 x]$ .

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Step 2.3.1.1.3.1

Since $7$ is constant with respect to $x$ , the derivative of $7 x$ with respect to $x$ is $7 \frac{d}{d x} [x]$ .

$7 \frac{d}{d x} [x] + \frac{d}{d x} [5]$

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 $n = 1$ .

$7 \cdot 1 + \frac{d}{d x} [5]$

Step 2.3.1.1.3.3

Multiply $7$ by $1$ .

$7 + \frac{d}{d x} [5]$

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 $0$ .

$7 + 0$

Step 2.3.1.1.4.2

Add $7$ and $0$ .

$7$

Step 2.3.1.2

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

$\frac{1}{3} \ln (| 3 y + 4 |) + C_{1} = \int \frac{1}{u_{2}} \cdot \frac{1}{7} d u_{2}$

Step 2.3.2

Simplify.

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Step 2.3.2.1

Multiply $\frac{1}{u_{2}}$ by $\frac{1}{7}$ .

$\frac{1}{3} \ln (| 3 y + 4 |) + C_{1} = \int \frac{1}{u_{2} \cdot 7} d u_{2}$

Step 2.3.2.2

Move $7$ to the left of $u_{2}$ .

$\frac{1}{3} \ln (| 3 y + 4 |) + C_{1} = \int \frac{1}{7 u_{2}} d u_{2}$

Step 2.3.3

Since $\frac{1}{7}$ is constant with respect to $u_{2}$ , move $\frac{1}{7}$ out of the integral.

$\frac{1}{3} \ln (| 3 y + 4 |) + C_{1} = \frac{1}{7} \int \frac{1}{u_{2}} d u_{2}$

Step 2.3.4

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

$\frac{1}{3} \ln (| 3 y + 4 |) + C_{1} = \frac{1}{7} (\ln (| u_{2} |) + C_{2})$

Step 2.3.5

Simplify.

$\frac{1}{3} \ln (| 3 y + 4 |) + C_{1} = \frac{1}{7} \ln (| u_{2} |) + C_{2}$

Step 2.3.6

Replace all occurrences of $u_{2}$ with $7 x + 5$ .

$\frac{1}{3} \ln (| 3 y + 4 |) + C_{1} = \frac{1}{7} \ln (| 7 x + 5 |) + C_{2}$

Step 2.4

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

$\frac{1}{3} \ln (| 3 y + 4 |) = \frac{1}{7} \ln (| 7 x + 5 |) + C$

Step 3

Solve for $y$ .

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Step 3.1

Multiply both sides of the equation by $3$ .

$3 (\frac{1}{3} \ln (| 3 y + 4 |)) = 3 (\frac{1}{7} \ln (| 7 x + 5 |) + C)$

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 $3 (\frac{1}{3} \ln (| 3 y + 4 |))$ .

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Step 3.2.1.1.1

Combine $\frac{1}{3}$ and $\ln (| 3 y + 4 |)$ .

$3 \frac{\ln (| 3 y + 4 |)}{3} = 3 (\frac{1}{7} \ln (| 7 x + 5 |) + C)$

Step 3.2.1.1.2

Cancel the common factor of $3$ .

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Step 3.2.1.1.2.1

Cancel the common factor.

$3 \frac{\ln (| 3 y + 4 |)}{3} = 3 (\frac{1}{7} \ln (| 7 x + 5 |) + C)$

Step 3.2.1.1.2.2

Rewrite the expression.

$\ln (| 3 y + 4 |) = 3 (\frac{1}{7} \ln (| 7 x + 5 |) + C)$

Step 3.2.2

Simplify the right side.

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Step 3.2.2.1

Simplify $3 (\frac{1}{7} \ln (| 7 x + 5 |) + C)$ .

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Step 3.2.2.1.1

Combine $\frac{1}{7}$ and $\ln (| 7 x + 5 |)$ .

$\ln (| 3 y + 4 |) = 3 (\frac{\ln (| 7 x + 5 |)}{7} + C)$

Step 3.2.2.1.2

To write $C$ as a fraction with a common denominator, multiply by $\frac{7}{7}$ .

$\ln (| 3 y + 4 |) = 3 (\frac{\ln (| 7 x + 5 |)}{7} + C \cdot \frac{7}{7})$

Step 3.2.2.1.3

Simplify terms.

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Step 3.2.2.1.3.1

Combine $C$ and $\frac{7}{7}$ .

$\ln (| 3 y + 4 |) = 3 (\frac{\ln (| 7 x + 5 |)}{7} + \frac{C \cdot 7}{7})$

Step 3.2.2.1.3.2

Combine the numerators over the common denominator.

$\ln (| 3 y + 4 |) = 3 \frac{\ln (| 7 x + 5 |) + C \cdot 7}{7}$

Step 3.2.2.1.4

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

$\ln (| 3 y + 4 |) = 3 \frac{\ln (| 7 x + 5 |) + 7 C}{7}$

Step 3.2.2.1.5

Combine $3$ and $\frac{\ln (| 7 x + 5 |) + 7 C}{7}$ .

$\ln (| 3 y + 4 |) = \frac{3 (\ln (| 7 x + 5 |) + 7 C)}{7}$

Step 3.3

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

$e^{\ln (| 3 y + 4 |)} = e^{\frac{3 (\ln (| 7 x + 5 |) + 7 C)}{7}}$

Step 3.4

Rewrite $\ln (| 3 y + 4 |) = \frac{3 (\ln (| 7 x + 5 |) + 7 C)}{7}$ 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 $b^{y} = x$ .

$e^{\frac{3 (\ln (| 7 x + 5 |) + 7 C)}{7}} = | 3 y + 4 |$

Step 3.5

Solve for $y$ .

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Step 3.5.1

Rewrite the equation as $| 3 y + 4 | = e^{\frac{3 (\ln (| 7 x + 5 |) + 7 C)}{7}}$ .

$| 3 y + 4 | = e^{\frac{3 (\ln (| 7 x + 5 |) + 7 C)}{7}}$

Step 3.5.2

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

$3 y + 4 = \pm e^{\frac{3 (\ln (| 7 x + 5 |) + 7 C)}{7}}$

Step 3.5.3

Subtract $4$ from both sides of the equation.

$3 y = \pm e^{\frac{3 (\ln (| 7 x + 5 |) + 7 C)}{7}} - 4$

Step 3.5.4

Divide each term in $3 y = \pm e^{\frac{3 (\ln (| 7 x + 5 |) + 7 C)}{7}} - 4$ by $3$ and simplify.

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Step 3.5.4.1

Divide each term in $3 y = \pm e^{\frac{3 (\ln (| 7 x + 5 |) + 7 C)}{7}} - 4$ by $3$ .

$\frac{3 y}{3} = \frac{\pm e^{\frac{3 (\ln (| 7 x + 5 |) + 7 C)}{7}}}{3} + \frac{- 4}{3}$

Step 3.5.4.2

Simplify the left side.

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Step 3.5.4.2.1

Cancel the common factor of $3$ .

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Step 3.5.4.2.1.1

Cancel the common factor.

$\frac{3 y}{3} = \frac{\pm e^{\frac{3 (\ln (| 7 x + 5 |) + 7 C)}{7}}}{3} + \frac{- 4}{3}$

Step 3.5.4.2.1.2

Divide $y$ by $1$ .

$y = \frac{\pm e^{\frac{3 (\ln (| 7 x + 5 |) + 7 C)}{7}}}{3} + \frac{- 4}{3}$

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.

$y = \frac{\pm e^{\frac{3 (\ln (| 7 x + 5 |) + 7 C)}{7}}}{3} - \frac{4}{3}$

Step 3.5.4.3.2

Combine the numerators over the common denominator.

$y = \frac{\pm e^{\frac{3 (\ln (| 7 x + 5 |) + 7 C)}{7}} - 4}{3}$

Step 4

Simplify the constant of integration.

$y = \frac{\pm e^{\frac{3 (\ln (| 7 x + 5 |) + C)}{7}} - 4}{3}$