Calculus Examples

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

Step 1

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

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

Step 2

Simplify.

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Step 2.1

Cancel the common factor of $x$ .

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Step 2.1.1

Cancel the common factor.

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

Step 2.1.2

Rewrite the expression.

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

Step 2.2

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

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Step 2.2.1

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

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

Step 2.2.2

Cancel the common factor.

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

Step 2.2.3

Rewrite the expression.

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

Step 3

Integrate both sides.

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Step 3.1

Set up an integral on each side.

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

Step 3.2

Integrate the left side.

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Step 3.2.1

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

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Step 3.2.1.1

Let $u = 3 - 2 y$ . Find $\frac{d u}{d y}$ .

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Step 3.2.1.1.1

Differentiate $3 - 2 y$ .

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

Step 3.2.1.1.2

Differentiate.

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Step 3.2.1.1.2.1

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

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

Step 3.2.1.1.2.2

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

$0 + \frac{d}{d y} [- 2 y]$

Step 3.2.1.1.3

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

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Step 3.2.1.1.3.1

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

$0 - 2 \frac{d}{d y} [y]$

Step 3.2.1.1.3.2

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

$0 - 2 \cdot 1$

Step 3.2.1.1.3.3

Multiply $- 2$ by $1$ .

$0 - 2$

Step 3.2.1.1.4

Subtract $2$ from $0$ .

$- 2$

Step 3.2.1.2

Rewrite the problem using $u$ and $d u$ .

$\int \frac{1}{u} \cdot \frac{1}{- 2} d u = \int \frac{1}{x} d x$

Step 3.2.2

Simplify.

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Step 3.2.2.1

Move the negative in front of the fraction.

$\int \frac{1}{u} (- \frac{1}{2}) d u = \int \frac{1}{x} d x$

Step 3.2.2.2

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

$\int - \frac{1}{u \cdot 2} d u = \int \frac{1}{x} d x$

Step 3.2.2.3

Move $2$ to the left of $u$ .

$\int - \frac{1}{2 u} d u = \int \frac{1}{x} d x$

Step 3.2.3

Since $- 1$ is constant with respect to $u$ , move $- 1$ out of the integral.

$- \int \frac{1}{2 u} d u = \int \frac{1}{x} d x$

Step 3.2.4

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

$- (\frac{1}{2} \int \frac{1}{u} d u) = \int \frac{1}{x} d x$

Step 3.2.5

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

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

Step 3.2.6

Simplify.

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

Step 3.2.7

Replace all occurrences of $u$ with $3 - 2 y$ .

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

Step 3.3

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

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

Step 3.4

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

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

Step 4

Solve for $y$ .

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Step 4.1

Multiply both sides of the equation by $- 2$ .

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

Step 4.2

Simplify both sides of the equation.

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Step 4.2.1

Simplify the left side.

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Step 4.2.1.1

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

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Step 4.2.1.1.1

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

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

Step 4.2.1.1.2

Cancel the common factor of $2$ .

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Step 4.2.1.1.2.1

Move the leading negative in $- \frac{\ln (| 3 - 2 y |)}{2}$ into the numerator.

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

Step 4.2.1.1.2.2

Factor $2$ out of $- 2$ .

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

Step 4.2.1.1.2.3

Cancel the common factor.

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

Step 4.2.1.1.2.4

Rewrite the expression.

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

Step 4.2.1.1.3

Multiply.

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Step 4.2.1.1.3.1

Multiply $- 1$ by $- 1$ .

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

Step 4.2.1.1.3.2

Multiply $\ln (| 3 - 2 y |)$ by $1$ .

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

Step 4.2.2

Simplify the right side.

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Step 4.2.2.1

Apply the distributive property.

$\ln (| 3 - 2 y |) = - 2 \ln (| x |) - 2 C$

Step 4.3

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

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

Step 4.4

Simplify the left side.

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Step 4.4.1

Simplify $\ln (| 3 - 2 y |) + 2 \ln (| x |)$ .

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Step 4.4.1.1

Simplify each term.

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Step 4.4.1.1.1

Simplify $2 \ln (| x |)$ by moving $2$ inside the logarithm.

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

Step 4.4.1.1.2

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

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

Step 4.4.1.2

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

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

Step 4.4.1.3

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

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

Step 4.5

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

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

Step 4.6

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

$e^{- 2 C} = x^{2} | 3 - 2 y |$

Step 4.7

Solve for $y$ .

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Step 4.7.1

Rewrite the equation as $x^{2} | 3 - 2 y | = e^{- 2 C}$ .

$x^{2} | 3 - 2 y | = e^{- 2 C}$

Step 4.7.2

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

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Step 4.7.2.1

Divide each term in $x^{2} | 3 - 2 y | = e^{- 2 C}$ by $x^{2}$ .

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

Step 4.7.2.2

Simplify the left side.

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Step 4.7.2.2.1

Cancel the common factor of $x^{2}$ .

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Step 4.7.2.2.1.1

Cancel the common factor.

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

Step 4.7.2.2.1.2

Divide $| 3 - 2 y |$ by $1$ .

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

Step 4.7.3

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

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

Step 4.7.4

Subtract $3$ from both sides of the equation.

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

Step 4.7.5

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

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Step 4.7.5.1

Divide each term in $- 2 y = \pm \frac{e^{- 2 C}}{x^{2}} - 3$ by $- 2$ .

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

Step 4.7.5.2

Simplify the left side.

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Step 4.7.5.2.1

Cancel the common factor of $- 2$ .

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Step 4.7.5.2.1.1

Cancel the common factor.

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

Step 4.7.5.2.1.2

Divide $y$ by $1$ .

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

Step 4.7.5.3

Simplify the right side.

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Step 4.7.5.3.1

Simplify each term.

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Step 4.7.5.3.1.1

Simplify $\frac{\pm \frac{e^{- 2 C}}{x^{2}}}{- 2}$ .

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

Step 4.7.5.3.1.2

Dividing two negative values results in a positive value.

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

Step 5

Group the constant terms together.

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Step 5.1

Simplify the constant of integration.

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

Step 5.2

Combine constants with the plus or minus.

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