Calculus Examples

$\frac{d y}{d x} = \frac{2}{5 - x}$

Step 1

Rewrite the equation.

$d y = \frac{2}{5 - x} d x$

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

$\int d y = \int \frac{2}{5 - x} d x$

Step 2.2

Apply the constant rule.

$y + C_{1} = \int \frac{2}{5 - x} d x$

Step 2.3

Integrate the right side.

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Step 2.3.1

Since $2$ is constant with respect to $x$ , move $2$ out of the integral.

$y + C_{1} = 2 \int \frac{1}{5 - x} d x$

Step 2.3.2

Let $u = 5 - x$ . Then $d u = - d x$ , so $- d u = d x$ . Rewrite using $u$ and $d$ $u$ .

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Step 2.3.2.1

Let $u = 5 - x$ . Find $\frac{d u}{d x}$ .

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Step 2.3.2.1.1

Rewrite.

$\frac{- 1}{1}$

Step 2.3.2.1.2

Divide $- 1$ by $1$ .

$- 1$

Step 2.3.2.2

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

$y + C_{1} = 2 \int \frac{- 1}{u} d u$

Step 2.3.3

Move the negative in front of the fraction.

$y + C_{1} = 2 \int - \frac{1}{u} d u$

Step 2.3.4

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

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

Step 2.3.5

Multiply $- 1$ by $2$ .

$y + C_{1} = - 2 \int \frac{1}{u} d u$

Step 2.3.6

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

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

Step 2.3.7

Simplify.

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

Step 2.3.8

Replace all occurrences of $u$ with $5 - x$ .

$y + C_{1} = - 2 \ln (| 5 - x |) + C_{2}$

Step 2.4

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

$y = - 2 \ln (| 5 - x |) + C$