Calculus Examples

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

Step 1

Separate the variables.

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Step 1.1

Multiply both sides by $\frac{1}{3 y - 2 y + 5}$ .

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

Step 1.2

Cancel the common factor of $3 y - 2 y + 5$ .

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Step 1.2.1

Cancel the common factor.

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

Step 1.2.2

Rewrite the expression.

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

Step 1.3

Rewrite the equation.

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

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

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

Step 2.2

Integrate the left side.

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Step 2.2.1

Subtract $2 y$ from $3 y$ .

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

Step 2.2.2

Let $u_{1} = y + 5$ . Then $d u_{1} = d y$ . Rewrite using $u_{1}$ and $d$ $u_{1}$ .

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Step 2.2.2.1

Let $u_{1} = y + 5$ . Find $\frac{d u_{1}}{d y}$ .

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Step 2.2.2.1.1

Differentiate $y + 5$ .

$\frac{d}{d y} [y + 5]$

Step 2.2.2.1.2

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

$\frac{d}{d y} [y] + \frac{d}{d y} [5]$

Step 2.2.2.1.3

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

$1 + \frac{d}{d y} [5]$

Step 2.2.2.1.4

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

$1 + 0$

Step 2.2.2.1.5

Add $1$ and $0$ .

$1$

Step 2.2.2.2

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

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

Step 2.2.3

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

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

Step 2.2.4

Replace all occurrences of $u_{1}$ with $y + 5$ .

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

Step 2.3

Integrate the right side.

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Step 2.3.1

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

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Step 2.3.1.1

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

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Step 2.3.1.1.1

Differentiate $2 x + 5$ .

$\frac{d}{d x} [2 x + 5]$

Step 2.3.1.1.2

By the Sum Rule, the derivative of $2 x + 5$ with respect to $x$ is $\frac{d}{d x} [2 x] + \frac{d}{d x} [5]$ .

$\frac{d}{d x} [2 x] + \frac{d}{d x} [5]$

Step 2.3.1.1.3

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

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Step 2.3.1.1.3.1

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

$2 \frac{d}{d x} [x] + \frac{d}{d x} [5]$

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 $n = 1$ .

$2 \cdot 1 + \frac{d}{d x} [5]$

Step 2.3.1.1.3.3

Multiply $2$ by $1$ .

$2 + \frac{d}{d x} [5]$

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 $0$ .

$2 + 0$

Step 2.3.1.1.4.2

Add $2$ and $0$ .

$2$

Step 2.3.1.2

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

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

Step 2.3.2

Simplify.

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Step 2.3.2.1

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

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

Step 2.3.2.2

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

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

Step 2.3.3

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

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

Step 2.3.4

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

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

Step 2.3.5

Simplify.

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

Step 2.3.6

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

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

Step 2.4

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

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

Step 3

Solve for $y$ .

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Step 3.1

Simplify the right side.

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Step 3.1.1

Combine $\frac{1}{2}$ and $\ln (| 2 x + 5 |)$ .

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

Step 3.2

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

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

Step 3.3

To write $\ln (| y + 5 |)$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

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

Step 3.4

Simplify terms.

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Step 3.4.1

Combine $\ln (| y + 5 |)$ and $\frac{2}{2}$ .

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

Step 3.4.2

Combine the numerators over the common denominator.

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

Step 3.5

Move $2$ to the left of $\ln (| y + 5 |)$ .

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

Step 3.6

Simplify the left side.

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Step 3.6.1

Simplify $\frac{2 \ln (| y + 5 |) - \ln (| 2 x + 5 |)}{2}$ .

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Step 3.6.1.1

Simplify the numerator.

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Step 3.6.1.1.1

Simplify $2 \ln (| y + 5 |)$ by moving $2$ inside the logarithm.

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

Step 3.6.1.1.2

Remove the absolute value in ${| y + 5 |}^{2}$ because exponentiations with even powers are always positive.

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

Step 3.6.1.1.3

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

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

Step 3.6.1.2

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

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

Step 3.6.1.3

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

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

Step 3.6.1.4

Apply the product rule to $\frac{{(y + 5)}^{2}}{| 2 x + 5 |}$ .

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

Step 3.6.1.5

Simplify the numerator.

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Step 3.6.1.5.1

Multiply the exponents in ${({(y + 5)}^{2})}^{\frac{1}{2}}$ .

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Step 3.6.1.5.1.1

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

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

Step 3.6.1.5.1.2

Cancel the common factor of $2$ .

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Step 3.6.1.5.1.2.1

Cancel the common factor.

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

Step 3.6.1.5.1.2.2

Rewrite the expression.

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

Step 3.6.1.5.2

Simplify.

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

Step 3.7

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

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

Step 3.8

Rewrite $\ln (\frac{y + 5}{{| 2 x + 5 |}^{\frac{1}{2}}}) = C$ 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 $b^{y} = x$ .

$e^{C} = \frac{y + 5}{{| 2 x + 5 |}^{\frac{1}{2}}}$

Step 3.9

Solve for $y$ .

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Step 3.9.1

Rewrite the equation as $\frac{y + 5}{{| 2 x + 5 |}^{\frac{1}{2}}} = e^{C}$ .

$\frac{y + 5}{{| 2 x + 5 |}^{\frac{1}{2}}} = e^{C}$

Step 3.9.2

Multiply both sides by ${| 2 x + 5 |}^{\frac{1}{2}}$ .

$\frac{y + 5}{{| 2 x + 5 |}^{\frac{1}{2}}} {| 2 x + 5 |}^{\frac{1}{2}} = e^{C} {| 2 x + 5 |}^{\frac{1}{2}}$

Step 3.9.3

Simplify the left side.

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Step 3.9.3.1

Cancel the common factor of ${| 2 x + 5 |}^{\frac{1}{2}}$ .

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Step 3.9.3.1.1

Cancel the common factor.

$\frac{y + 5}{{| 2 x + 5 |}^{\frac{1}{2}}} {| 2 x + 5 |}^{\frac{1}{2}} = e^{C} {| 2 x + 5 |}^{\frac{1}{2}}$

Step 3.9.3.1.2

Rewrite the expression.

$y + 5 = e^{C} {| 2 x + 5 |}^{\frac{1}{2}}$

Step 3.9.4

Subtract $5$ from both sides of the equation.

$y = e^{C} {| 2 x + 5 |}^{\frac{1}{2}} - 5$

Step 4

Simplify the constant of integration.

$y = C {| 2 x + 5 |}^{\frac{1}{2}} - 5$