Calculus Examples

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

Step 1

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

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

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.

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

Step 2.1.2

Rewrite the expression.

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

Step 2.2

Cancel the common factor of $y + 3$ .

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Step 2.2.1

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

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

Step 2.2.2

Cancel the common factor.

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

Step 2.2.3

Rewrite the expression.

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

Step 3

Integrate both sides.

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Step 3.1

Set up an integral on each side.

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

Step 3.2

Integrate the left side.

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Step 3.2.1

Let $u = y + 3$ . Then $d u = d y$ . Rewrite using $u$ and $d$ $u$ .

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Step 3.2.1.1

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

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Step 3.2.1.1.1

Differentiate $y + 3$ .

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

Step 3.2.1.1.2

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

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

Step 3.2.1.1.3

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

$1 + \frac{d}{d y} [3]$

Step 3.2.1.1.4

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

$1 + 0$

Step 3.2.1.1.5

Add $1$ and $0$ .

$1$

Step 3.2.1.2

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

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

Step 3.2.2

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

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

Step 3.2.3

Replace all occurrences of $u$ with $y + 3$ .

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

Step 3.3

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

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

Step 3.4

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

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

Step 4

Solve for $y$ .

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Step 4.1

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

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

Step 4.2

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

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

Step 4.3

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

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

Step 4.4

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

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

Step 4.5

Solve for $y$ .

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Step 4.5.1

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

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

Step 4.5.2

Multiply both sides by $| x |$ .

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

Step 4.5.3

Simplify the left side.

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Step 4.5.3.1

Cancel the common factor of $| x |$ .

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Step 4.5.3.1.1

Cancel the common factor.

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

Step 4.5.3.1.2

Rewrite the expression.

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

Step 4.5.4

Solve for $y$ .

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Step 4.5.4.1

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

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

Step 4.5.4.2

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

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

Step 4.5.4.3

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

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

Step 4.5.4.4

Subtract $3$ from both sides of the equation.

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

Step 5

Group the constant terms together.

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Step 5.1

Simplify the constant of integration.

$y = \pm C | x | - 3$

Step 5.2

Combine constants with the plus or minus.

$y = C | x | - 3$