Calculus Examples

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

Step 1

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

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

Step 2

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

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

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.

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

Step 3.1.2

Rewrite the expression.

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

Step 3.2

Rewrite using the commutative property of multiplication.

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

Step 3.3

Cancel the common factor of $y + 2$ .

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Step 3.3.1

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

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

Step 3.3.2

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

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

Step 3.3.3

Cancel the common factor.

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

Step 3.3.4

Rewrite the expression.

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

Step 3.4

Move the negative in front of the fraction.

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

Step 4

Integrate both sides.

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Step 4.1

Set up an integral on each side.

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

Step 4.2

Integrate the left side.

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Step 4.2.1

Let $u_{1} = y + 2$ . Then $d u_{1} = d y$ . Rewrite using $u_{1}$ and $d$ $u_{1}$ .

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Step 4.2.1.1

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

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Step 4.2.1.1.1

Differentiate $y + 2$ .

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

Step 4.2.1.1.2

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

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

Step 4.2.1.1.3

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

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

Step 4.2.1.1.4

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

$1 + 0$

Step 4.2.1.1.5

Add $1$ and $0$ .

$1$

Step 4.2.1.2

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

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

Step 4.2.2

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

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

Step 4.2.3

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

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

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.

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

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 $d u_{2}$ .

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

Step 4.3.3

Simplify.

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Step 4.3.3.1

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

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

Step 4.3.3.2

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

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

Step 4.3.4

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

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

Step 4.3.5

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

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

Step 4.3.6

Simplify.

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

Step 4.3.7

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

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

Step 4.4

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

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

Step 5

Solve for $y$ .

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Step 5.1

Simplify the right side.

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Step 5.1.1

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

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

Step 5.2

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

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

Step 5.3

To write $\ln (| y + 2 |)$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

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

Step 5.4

Simplify terms.

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Step 5.4.1

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

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

Step 5.4.2

Combine the numerators over the common denominator.

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

Step 5.5

Move $3$ to the left of $\ln (| y + 2 |)$ .

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

Step 5.6

Simplify the left side.

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Step 5.6.1

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

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Step 5.6.1.1

Simplify the numerator.

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Step 5.6.1.1.1

Simplify $3 \ln (| y + 2 |)$ by moving $3$ inside the logarithm.

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

Step 5.6.1.1.2

Use the product property of logarithms, $\log_{b} (x) + \log_{b} (y) = \log_{b} (x y)$ .

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

Step 5.6.1.2

Rewrite $\frac{\ln ({| y + 2 |}^{3} | 3 x + 2 |)}{3}$ as $\frac{1}{3} \ln ({| y + 2 |}^{3} | 3 x + 2 |)$ .

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

Step 5.6.1.3

Simplify $\frac{1}{3} \ln ({| y + 2 |}^{3} | 3 x + 2 |)$ by moving $\frac{1}{3}$ inside the logarithm.

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

Step 5.6.1.4

Apply the product rule to ${| y + 2 |}^{3} | 3 x + 2 |$ .

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

Step 5.6.1.5

Multiply the exponents in ${({| y + 2 |}^{3})}^{\frac{1}{3}}$ .

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Step 5.6.1.5.1

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

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

Step 5.6.1.5.2

Cancel the common factor of $3$ .

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Step 5.6.1.5.2.1

Cancel the common factor.

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

Step 5.6.1.5.2.2

Rewrite the expression.

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

Step 5.6.1.6

Simplify.

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

Step 5.6.1.7

Reorder factors in $\ln (| y + 2 | {| 3 x + 2 |}^{\frac{1}{3}})$ .

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

Step 5.7

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

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

Step 5.8

Rewrite $\ln ({| 3 x + 2 |}^{\frac{1}{3}} | y + 2 |) = C$ 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 $b^{y} = x$ .

$e^{C} = {| 3 x + 2 |}^{\frac{1}{3}} | y + 2 |$

Step 5.9

Solve for $y$ .

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Step 5.9.1

Rewrite the equation as ${| 3 x + 2 |}^{\frac{1}{3}} | y + 2 | = e^{C}$ .

${| 3 x + 2 |}^{\frac{1}{3}} | y + 2 | = e^{C}$

Step 5.9.2

Divide each term in ${| 3 x + 2 |}^{\frac{1}{3}} | y + 2 | = e^{C}$ by ${| 3 x + 2 |}^{\frac{1}{3}}$ and simplify.

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Step 5.9.2.1

Divide each term in ${| 3 x + 2 |}^{\frac{1}{3}} | y + 2 | = e^{C}$ by ${| 3 x + 2 |}^{\frac{1}{3}}$ .

$\frac{{| 3 x + 2 |}^{\frac{1}{3}} | y + 2 |}{{| 3 x + 2 |}^{\frac{1}{3}}} = \frac{e^{C}}{{| 3 x + 2 |}^{\frac{1}{3}}}$

Step 5.9.2.2

Simplify the left side.

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Step 5.9.2.2.1

Cancel the common factor.

$\frac{{| 3 x + 2 |}^{\frac{1}{3}} | y + 2 |}{{| 3 x + 2 |}^{\frac{1}{3}}} = \frac{e^{C}}{{| 3 x + 2 |}^{\frac{1}{3}}}$

Step 5.9.2.2.2

Divide $| y + 2 |$ by $1$ .

$| y + 2 | = \frac{e^{C}}{{| 3 x + 2 |}^{\frac{1}{3}}}$

Step 5.9.3

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

$y + 2 = \pm \frac{e^{C}}{{| 3 x + 2 |}^{\frac{1}{3}}}$

Step 5.9.4

Subtract $2$ from both sides of the equation.

$y = \pm \frac{e^{C}}{{| 3 x + 2 |}^{\frac{1}{3}}} - 2$

Step 6

Group the constant terms together.

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Step 6.1

Simplify the constant of integration.

$y = \pm \frac{C}{{| 3 x + 2 |}^{\frac{1}{3}}} - 2$

Step 6.2

Combine constants with the plus or minus.

$y = C \frac{1}{{| 3 x + 2 |}^{\frac{1}{3}}} - 2$