Calculus Examples

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

Step 1

Separate the variables.

Tap for more steps...

Step 1.1

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

Tap for more steps...

Step 1.1.1

Subtract $\frac{1}{x}$ from both sides of the equation.

$\frac{d y}{d x} - \frac{y}{x} - \frac{1}{x} = 0$

Step 1.1.2

Move all terms not containing $\frac{d y}{d x}$ to the right side of the equation.

Tap for more steps...

Step 1.1.2.1

Add $\frac{y}{x}$ to both sides of the equation.

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

Step 1.1.2.2

Add $\frac{1}{x}$ to both sides of the equation.

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

Step 1.2

Combine the numerators over the common denominator.

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

Step 1.3

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

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

Step 1.4

Cancel the common factor of $y + 1$ .

Tap for more steps...

Step 1.4.1

Cancel the common factor.

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

Step 1.4.2

Rewrite the expression.

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

Step 1.5

Rewrite the equation.

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

Step 2

Integrate both sides.

Tap for more steps...

Step 2.1

Set up an integral on each side.

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

Step 2.2

Integrate the left side.

Tap for more steps...

Step 2.2.1

Let $u = y + 1$ . Then $d u = d y$ . Rewrite using $u$ and $d$ $u$ .

Tap for more steps...

Step 2.2.1.1

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

Tap for more steps...

Step 2.2.1.1.1

Differentiate $y + 1$ .

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

Step 2.2.1.1.2

By the Sum Rule, the derivative of $y + 1$ with respect to $y$ is $\frac{d}{d y} [y] + \frac{d}{d y} [1]$ .

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

Step 2.2.1.1.3

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

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

Step 2.2.1.1.4

Since $1$ is constant with respect to $y$ , the derivative of $1$ with respect to $y$ is $0$ .

$1 + 0$

Step 2.2.1.1.5

Add $1$ and $0$ .

$1$

Step 2.2.1.2

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

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

Step 2.2.2

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

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

Step 2.2.3

Replace all occurrences of $u$ with $y + 1$ .

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

Step 2.3

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

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

Step 2.4

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

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

Step 3

Solve for $y$ .

Tap for more steps...

Step 3.1

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

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

Step 3.2

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

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

Step 3.3

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

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

Step 3.4

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

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

Step 3.5

Solve for $y$ .

Tap for more steps...

Step 3.5.1

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

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

Step 3.5.2

Multiply both sides by $| x |$ .

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

Step 3.5.3

Simplify the left side.

Tap for more steps...

Step 3.5.3.1

Cancel the common factor of $| x |$ .

Tap for more steps...

Step 3.5.3.1.1

Cancel the common factor.

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

Step 3.5.3.1.2

Rewrite the expression.

$| y + 1 | = e^{C} | x |$

Step 3.5.4

Solve for $y$ .

Tap for more steps...

Step 3.5.4.1

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

$| y + 1 | = | x | e^{C}$

Step 3.5.4.2

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

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

Step 3.5.4.3

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

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

Step 3.5.4.4

Subtract $1$ from both sides of the equation.

$y = \pm e^{C} | x | - 1$

Step 4

Group the constant terms together.

Tap for more steps...

Step 4.1

Simplify the constant of integration.

$y = \pm C | x | - 1$

Step 4.2

Combine constants with the plus or minus.

$y = C | x | - 1$