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

Subtract $\frac{y}{x}$ from 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

Simplify.

Tap for more steps...

Step 1.4.1

Multiply $\frac{1}{- y + 1}$ by $\frac{- y + 1}{x}$ .

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

Step 1.4.2

Cancel the common factor of $- y + 1$ .

Tap for more steps...

Step 1.4.2.1

Cancel the common factor.

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

Step 1.4.2.2

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 $d u = - d y$ , so $- d u = d y$ . 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

Rewrite.

$\frac{- 1}{1}$

Step 2.2.1.1.2

Divide $- 1$ by $1$ .

$- 1$

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

Split the fraction into multiple fractions.

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

Step 2.2.3

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

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

Step 2.2.4

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

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

Step 2.2.5

Simplify.

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

Step 2.2.6

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

Add $\ln (| x |)$ to both sides of the equation.

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

Step 3.3

Divide each term in $- \ln (| - y + 1 |) = C + \ln (| x |)$ by $- 1$ and simplify.

Tap for more steps...

Step 3.3.1

Divide each term in $- \ln (| - y + 1 |) = C + \ln (| x |)$ by $- 1$ .

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

Step 3.3.2

Simplify the left side.

Tap for more steps...

Step 3.3.2.1

Dividing two negative values results in a positive value.

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

Step 3.3.2.2

Divide $\ln (| - y + 1 |)$ by $1$ .

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

Step 3.3.3

Simplify the right side.

Tap for more steps...

Step 3.3.3.1

Simplify each term.

Tap for more steps...

Step 3.3.3.1.1

Move the negative one from the denominator of $\frac{C}{- 1}$ .

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

Step 3.3.3.1.2

Rewrite $- 1 \cdot C$ as $- C$ .

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

Step 3.3.3.1.3

Move the negative one from the denominator of $\frac{\ln (| x |)}{- 1}$ .

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

Step 3.3.3.1.4

Rewrite $- 1 \cdot \ln (| x |)$ as $- \ln (| x |)$ .

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

Step 3.4

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

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

Step 3.5

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

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

Step 3.6

To multiply absolute values, multiply the terms inside each absolute value.

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

Step 3.7

Apply the distributive property.

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

Step 3.8

Multiply $x$ by $1$ .

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

Step 3.9

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

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

Step 3.10

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

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

Step 3.11

Solve for $y$ .

Tap for more steps...

Step 3.11.1

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

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

Step 3.11.2

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

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

Step 3.11.3

Subtract $x$ from both sides of the equation.

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

Step 3.11.4

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

Tap for more steps...

Step 3.11.4.1

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

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

Step 3.11.4.2

Simplify the left side.

Tap for more steps...

Step 3.11.4.2.1

Dividing two negative values results in a positive value.

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

Step 3.11.4.2.2

Cancel the common factor of $x$ .

Tap for more steps...

Step 3.11.4.2.2.1

Cancel the common factor.

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

Step 3.11.4.2.2.2

Divide $y$ by $1$ .

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

Step 3.11.4.3

Simplify the right side.

Tap for more steps...

Step 3.11.4.3.1

Simplify each term.

Tap for more steps...

Step 3.11.4.3.1.1

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

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

Step 3.11.4.3.1.2

Dividing two negative values results in a positive value.

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

Step 3.11.4.3.1.3

Cancel the common factor of $x$ .

Tap for more steps...

Step 3.11.4.3.1.3.1

Cancel the common factor.

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

Step 3.11.4.3.1.3.2

Rewrite the expression.

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

Step 4

Simplify the constant of integration.

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