Calculus Examples

$(3 x + 2) d y + (2 y - 5) d x = 0$

Step 1

Subtract $(2 y - 5) d x$ from both sides of the equation.

$(3 x + 2) d y = - (2 y - 5) d x$

Step 2

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

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

Step 3

Simplify.

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Step 3.1

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

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Step 3.1.1

Cancel the common factor.

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

Step 3.1.2

Rewrite the expression.

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

Step 3.2

Rewrite using the commutative property of multiplication.

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

Step 3.3

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

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Step 3.3.1

Move the leading negative in $- \frac{1}{(3 x + 2) (2 y - 5)}$ into the numerator.

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

Step 3.3.2

Factor $2 y - 5$ out of $(3 x + 2) (2 y - 5)$ .

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

Step 3.3.3

Cancel the common factor.

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

Step 3.3.4

Rewrite the expression.

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

Step 3.4

Move the negative in front of the fraction.

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

Step 4

Integrate both sides.

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Step 4.1

Set up an integral on each side.

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

Step 4.2

Integrate the left side.

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Step 4.2.1

Let $u_{1} = 2 y - 5$ . 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 4.2.1.1

Let $u_{1} = 2 y - 5$ . Find $\frac{d u_{1}}{d y}$ .

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Step 4.2.1.1.1

Differentiate $2 y - 5$ .

$\frac{d}{d y} [2 y - 5]$

Step 4.2.1.1.2

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

$\frac{d}{d y} [2 y] + \frac{d}{d y} [- 5]$

Step 4.2.1.1.3

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

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Step 4.2.1.1.3.1

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

$2 \frac{d}{d y} [y] + \frac{d}{d y} [- 5]$

Step 4.2.1.1.3.2

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

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

Step 4.2.1.1.3.3

Multiply $2$ by $1$ .

$2 + \frac{d}{d y} [- 5]$

Step 4.2.1.1.4

Differentiate using the Constant Rule.

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Step 4.2.1.1.4.1

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

$2 + 0$

Step 4.2.1.1.4.2

Add $2$ and $0$ .

$2$

Step 4.2.1.2

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

$\int \frac{1}{u_{1}} \cdot \frac{1}{2} d u_{1} = \int - \frac{1}{3 x + 2} d x$

Step 4.2.2

Simplify.

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Step 4.2.2.1

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

$\int \frac{1}{u_{1} \cdot 2} d u_{1} = \int - \frac{1}{3 x + 2} d x$

Step 4.2.2.2

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

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

Step 4.2.3

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

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

Step 4.2.4

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

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

Step 4.2.5

Simplify.

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

Step 4.2.6

Replace all occurrences of $u_{1}$ with $2 y - 5$ .

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

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.

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

Step 4.3.2

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

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Step 4.3.2.1

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

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Step 4.3.2.1.1

Differentiate $3 x + 2$ .

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

Step 4.3.2.1.2

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

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

Step 4.3.2.1.3

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

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Step 4.3.2.1.3.1

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

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

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 $n = 1$ .

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

Step 4.3.2.1.3.3

Multiply $3$ by $1$ .

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

Step 4.3.2.1.4

Differentiate using the Constant Rule.

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Step 4.3.2.1.4.1

Since $2$ is constant with respect to $x$ , the derivative of $2$ with respect to $x$ is $0$ .

$3 + 0$

Step 4.3.2.1.4.2

Add $3$ and $0$ .

$3$

Step 4.3.2.2

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

$\frac{1}{2} \ln (| 2 y - 5 |) + C_{1} = - \int \frac{1}{u_{2}} \cdot \frac{1}{3} d u_{2}$

Step 4.3.3

Simplify.

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Step 4.3.3.1

Multiply $\frac{1}{u_{2}}$ by $\frac{1}{3}$ .

$\frac{1}{2} \ln (| 2 y - 5 |) + C_{1} = - \int \frac{1}{u_{2} \cdot 3} d u_{2}$

Step 4.3.3.2

Move $3$ to the left of $u_{2}$ .

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

Step 4.3.4

Since $\frac{1}{3}$ is constant with respect to $u_{2}$ , move $\frac{1}{3}$ out of the integral.

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

Step 4.3.5

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

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

Step 4.3.6

Simplify.

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

Step 4.3.7

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

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

Step 4.4

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

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

Step 5

Solve for $y$ .

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Step 5.1

Multiply both sides of the equation by $2$ .

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

Step 5.2

Simplify both sides of the equation.

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Step 5.2.1

Simplify the left side.

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Step 5.2.1.1

Simplify $2 (\frac{1}{2} \ln (| 2 y - 5 |))$ .

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Step 5.2.1.1.1

Combine $\frac{1}{2}$ and $\ln (| 2 y - 5 |)$ .

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

Step 5.2.1.1.2

Cancel the common factor of $2$ .

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Step 5.2.1.1.2.1

Cancel the common factor.

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

Step 5.2.1.1.2.2

Rewrite the expression.

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

Step 5.2.2

Simplify the right side.

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Step 5.2.2.1

Simplify $2 (- \frac{1}{3} \ln (| 3 x + 2 |) + C)$ .

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Step 5.2.2.1.1

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

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

Step 5.2.2.1.2

To write $C$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

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

Step 5.2.2.1.3

Simplify terms.

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Step 5.2.2.1.3.1

Combine $C$ and $\frac{3}{3}$ .

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

Step 5.2.2.1.3.2

Combine the numerators over the common denominator.

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

Step 5.2.2.1.4

Move $3$ to the left of $C$ .

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

Step 5.2.2.1.5

Simplify terms.

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Step 5.2.2.1.5.1

Combine $2$ and $\frac{- \ln (| 3 x + 2 |) + 3 C}{3}$ .

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

Step 5.2.2.1.5.2

Factor $- 1$ out of $- \ln (| 3 x + 2 |)$ .

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

Step 5.2.2.1.5.3

Factor $- 1$ out of $3 C$ .

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

Step 5.2.2.1.5.4

Factor $- 1$ out of $- (\ln (| 3 x + 2 |)) - (- 3 C)$ .

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

Step 5.2.2.1.5.5

Simplify the expression.

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Step 5.2.2.1.5.5.1

Rewrite $- (\ln (| 3 x + 2 |) - 3 C)$ as $- 1 (\ln (| 3 x + 2 |) - 3 C)$ .

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

Step 5.2.2.1.5.5.2

Move the negative in front of the fraction.

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

Step 5.3

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

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

Step 5.4

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

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

Step 5.5

Solve for $y$ .

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Step 5.5.1

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

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

Step 5.5.2

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

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

Step 5.5.3

Add $5$ to both sides of the equation.

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

Step 5.5.4

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

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Step 5.5.4.1

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

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

Step 5.5.4.2

Simplify the left side.

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Step 5.5.4.2.1

Cancel the common factor of $2$ .

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Step 5.5.4.2.1.1

Cancel the common factor.

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

Step 5.5.4.2.1.2

Divide $y$ by $1$ .

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

Step 5.5.4.3

Simplify the right side.

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Step 5.5.4.3.1

Combine the numerators over the common denominator.

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

Step 6

Simplify the constant of integration.

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