Calculus Examples

$d x + (1 - x) d y = 0$

Step 1

Subtract $d x$ from both sides of the equation.

$(1 - x) d y = - d x$

Step 2

Multiply both sides by $\frac{1}{1 - x}$ .

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

Step 3

Simplify.

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Step 3.1

Cancel the common factor of $1 - x$ .

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Step 3.1.1

Cancel the common factor.

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

Step 3.1.2

Rewrite the expression.

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

Step 3.2

Combine $\frac{1}{1 - x}$ and $- 1$ .

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

Step 3.3

Move the negative in front of the fraction.

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

Step 4

Integrate both sides.

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Step 4.1

Set up an integral on each side.

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

Step 4.2

Apply the constant rule.

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

Step 4.3

Integrate the right side.

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Step 4.3.1

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

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

Step 4.3.2

Let $u = 1 - x$ . Then $d u = - d x$ , so $- d u = d x$ . Rewrite using $u$ and $d$ $u$ .

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Step 4.3.2.1

Let $u = 1 - x$ . Find $\frac{d u}{d x}$ .

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Step 4.3.2.1.1

Rewrite.

$\frac{- 1}{1}$

Step 4.3.2.1.2

Divide $- 1$ by $1$ .

$- 1$

Step 4.3.2.2

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

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

Step 4.3.3

Move the negative in front of the fraction.

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

Step 4.3.4

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

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

Step 4.3.5

Simplify.

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Step 4.3.5.1

Multiply $- 1$ by $- 1$ .

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

Step 4.3.5.2

Multiply $\int \frac{1}{u} d u$ by $1$ .

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

Step 4.3.6

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

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

Step 4.3.7

Replace all occurrences of $u$ with $1 - x$ .

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

Step 4.4

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

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