Calculus Examples

$3 y - x \frac{d y}{d x} = 6$

Step 1

Separate the variables.

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Step 1.1

Solve for $\frac{d y}{d x}$ .

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Step 1.1.1

Subtract $3 y$ from both sides of the equation.

$- x \frac{d y}{d x} = 6 - 3 y$

Step 1.1.2

Divide each term in $- x \frac{d y}{d x} = 6 - 3 y$ by $- x$ and simplify.

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Step 1.1.2.1

Divide each term in $- x \frac{d y}{d x} = 6 - 3 y$ by $- x$ .

$\frac{- x \frac{d y}{d x}}{- x} = \frac{6}{- x} + \frac{- 3 y}{- x}$

Step 1.1.2.2

Simplify the left side.

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Step 1.1.2.2.1

Dividing two negative values results in a positive value.

$\frac{x \frac{d y}{d x}}{x} = \frac{6}{- x} + \frac{- 3 y}{- x}$

Step 1.1.2.2.2

Cancel the common factor of $x$ .

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Step 1.1.2.2.2.1

Cancel the common factor.

$\frac{x \frac{d y}{d x}}{x} = \frac{6}{- x} + \frac{- 3 y}{- x}$

Step 1.1.2.2.2.2

Divide $\frac{d y}{d x}$ by $1$ .

$\frac{d y}{d x} = \frac{6}{- x} + \frac{- 3 y}{- x}$

Step 1.1.2.3

Simplify the right side.

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Step 1.1.2.3.1

Simplify each term.

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Step 1.1.2.3.1.1

Move the negative in front of the fraction.

$\frac{d y}{d x} = - \frac{6}{x} + \frac{- 3 y}{- x}$

Step 1.1.2.3.1.2

Dividing two negative values results in a positive value.

$\frac{d y}{d x} = - \frac{6}{x} + \frac{3 y}{x}$

Step 1.2

Factor.

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Step 1.2.1

Combine the numerators over the common denominator.

$\frac{d y}{d x} = \frac{- 6 + 3 y}{x}$

Step 1.2.2

Factor $3$ out of $- 6 + 3 y$ .

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Step 1.2.2.1

Factor $3$ out of $- 6$ .

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

Step 1.2.2.2

Factor $3$ out of $3 \cdot - 2 + 3 y$ .

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

Step 1.2.2.3

Multiply $3$ by $- 2 + y$ .

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

Step 1.3

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

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

Step 1.4

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

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Step 1.4.1

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

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

Step 1.4.2

Cancel the common factor.

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

Step 1.4.3

Rewrite the expression.

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

Step 1.5

Rewrite the equation.

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

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

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

Step 2.2

Integrate the left side.

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Step 2.2.1

Let $u = - 2 + y$ . Then $d u = d y$ . Rewrite using $u$ and $d$ $u$ .

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Step 2.2.1.1

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

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Step 2.2.1.1.1

Differentiate $- 2 + y$ .

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

Step 2.2.1.1.2

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

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

Step 2.2.1.1.3

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

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

Step 2.2.1.1.4

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

$0 + 1$

Step 2.2.1.1.5

Add $0$ and $1$ .

$1$

Step 2.2.1.2

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

$\int \frac{1}{u} d u = \int \frac{3}{x} d x$

Step 2.2.2

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

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

Step 2.2.3

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

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

Step 2.3

Integrate the right side.

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Step 2.3.1

Since $3$ is constant with respect to $x$ , move $3$ out of the integral.

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

Step 2.3.2

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

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

Step 2.3.3

Simplify.

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

Step 2.4

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

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

Step 3

Solve for $y$ .

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Step 3.1

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

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

Step 3.2

Simplify the left side.

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Step 3.2.1

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

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Step 3.2.1.1

Simplify $- 3 \ln (| x |)$ by moving $3$ inside the logarithm.

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

Step 3.2.1.2

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

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

Step 3.3

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

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

Step 3.4

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

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

Step 3.5

Solve for $y$ .

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Step 3.5.1

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

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

Step 3.5.2

Multiply both sides by ${| x |}^{3}$ .

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

Step 3.5.3

Simplify the left side.

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Step 3.5.3.1

Cancel the common factor of ${| x |}^{3}$ .

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Step 3.5.3.1.1

Cancel the common factor.

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

Step 3.5.3.1.2

Rewrite the expression.

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

Step 3.5.4

Solve for $y$ .

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Step 3.5.4.1

Reorder factors in $e^{C} {| x |}^{3}$ .

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

Step 3.5.4.2

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

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

Step 3.5.4.3

Reorder factors in $\pm {| x |}^{3} e^{C}$ .

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

Step 3.5.4.4

Add $2$ to both sides of the equation.

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

Step 4

Group the constant terms together.

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Step 4.1

Simplify the constant of integration.

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

Step 4.2

Combine constants with the plus or minus.

$y = C {| x |}^{3} + 2$