Calculus Examples

$\frac{d y}{d x} = \frac{y (y - 2)}{x (y - 1)}$

Step 1

Separate the variables.

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Step 1.1

Regroup factors.

$\frac{d y}{d x} = \frac{y (y - 2)}{y - 1} \cdot \frac{1}{x}$

Step 1.2

Multiply both sides by $\frac{y - 1}{y (y - 2)}$ .

$\frac{y - 1}{y (y - 2)} \frac{d y}{d x} = \frac{y - 1}{y (y - 2)} (\frac{y (y - 2)}{y - 1} \cdot \frac{1}{x})$

Step 1.3

Simplify.

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Step 1.3.1

Multiply $\frac{y (y - 2)}{y - 1}$ by $\frac{1}{x}$ .

$\frac{y - 1}{y (y - 2)} \frac{d y}{d x} = \frac{y - 1}{y (y - 2)} \cdot \frac{y (y - 2)}{(y - 1) x}$

Step 1.3.2

Cancel the common factor of $y - 1$ .

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Step 1.3.2.1

Factor $y - 1$ out of $(y - 1) x$ .

$\frac{y - 1}{y (y - 2)} \frac{d y}{d x} = \frac{y - 1}{y (y - 2)} \cdot \frac{y (y - 2)}{(y - 1) (x)}$

Step 1.3.2.2

Cancel the common factor.

$\frac{y - 1}{y (y - 2)} \frac{d y}{d x} = \frac{y - 1}{y (y - 2)} \cdot \frac{y (y - 2)}{(y - 1) x}$

Step 1.3.2.3

Rewrite the expression.

$\frac{y - 1}{y (y - 2)} \frac{d y}{d x} = \frac{1}{y (y - 2)} \cdot \frac{y (y - 2)}{x}$

Step 1.3.3

Cancel the common factor of $y (y - 2)$ .

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Step 1.3.3.1

Cancel the common factor.

$\frac{y - 1}{y (y - 2)} \frac{d y}{d x} = \frac{1}{y (y - 2)} \cdot \frac{y (y - 2)}{x}$

Step 1.3.3.2

Rewrite the expression.

$\frac{y - 1}{y (y - 2)} \frac{d y}{d x} = \frac{1}{x}$

Step 1.4

Rewrite the equation.

$\frac{y - 1}{y (y - 2)} d y = \frac{1}{x} d x$

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

$\int \frac{y - 1}{y (y - 2)} d y = \int \frac{1}{x} d x$

Step 2.2

Integrate the left side.

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Step 2.2.1

Let $u = y (y - 2)$ . Then $d u = (2 y - 2) d y$ , so $\frac{1}{2} d u = (y - 1) d y$ . Rewrite using $u$ and $d$ $u$ .

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Step 2.2.1.1

Let $u = y (y - 2)$ . Find $\frac{d u}{d y}$ .

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Step 2.2.1.1.1

Differentiate $y (y - 2)$ .

$\frac{d}{d y} [y (y - 2)]$

Step 2.2.1.1.2

Differentiate using the Product Rule which states that $\frac{d}{d y} [f (y) g (y)]$ is $f (y) \frac{d}{d y} [g (y)] + g (y) \frac{d}{d y} [f (y)]$ where $f (y) = y$ and $g (y) = y - 2$ .

$y \frac{d}{d y} [y - 2] + (y - 2) \frac{d}{d y} [y]$

Step 2.2.1.1.3

Differentiate.

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Step 2.2.1.1.3.1

By the Sum Rule, the derivative of $y - 2$ with respect to $y$ is $\frac{d}{d y} [y] + \frac{d}{d y} [- 2]$ .

$y (\frac{d}{d y} [y] + \frac{d}{d y} [- 2]) + (y - 2) \frac{d}{d y} [y]$

Step 2.2.1.1.3.2

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

$y (1 + \frac{d}{d y} [- 2]) + (y - 2) \frac{d}{d y} [y]$

Step 2.2.1.1.3.3

Since $- 2$ is constant with respect to $y$ , the derivative of $- 2$ with respect to $y$ is $0$ .

$y (1 + 0) + (y - 2) \frac{d}{d y} [y]$

Step 2.2.1.1.3.4

Simplify the expression.

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Step 2.2.1.1.3.4.1

Add $1$ and $0$ .

$y \cdot 1 + (y - 2) \frac{d}{d y} [y]$

Step 2.2.1.1.3.4.2

Multiply $y$ by $1$ .

$y + (y - 2) \frac{d}{d y} [y]$

Step 2.2.1.1.3.5

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

$y + (y - 2) \cdot 1$

Step 2.2.1.1.3.6

Simplify by adding terms.

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Step 2.2.1.1.3.6.1

Multiply $y - 2$ by $1$ .

$y + y - 2$

Step 2.2.1.1.3.6.2

Add $y$ and $y$ .

$2 y - 2$

Step 2.2.1.2

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

$\int \frac{1}{u} \cdot \frac{1}{2} d u = \int \frac{1}{x} d x$

Step 2.2.2

Simplify.

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Step 2.2.2.1

Multiply $\frac{1}{u}$ by $\frac{1}{2}$ .

$\int \frac{1}{u \cdot 2} d u = \int \frac{1}{x} d x$

Step 2.2.2.2

Move $2$ to the left of $u$ .

$\int \frac{1}{2 u} d u = \int \frac{1}{x} d x$

Step 2.2.3

Since $\frac{1}{2}$ is constant with respect to $u$ , move $\frac{1}{2}$ out of the integral.

$\frac{1}{2} \int \frac{1}{u} d u = \int \frac{1}{x} d x$

Step 2.2.4

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

$\frac{1}{2} (\ln (| u |) + C_{1}) = \int \frac{1}{x} d x$

Step 2.2.5

Simplify.

$\frac{1}{2} \ln (| u |) + C_{1} = \int \frac{1}{x} d x$

Step 2.2.6

Replace all occurrences of $u$ with $y (y - 2)$ .

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

Step 2.3

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

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

Step 2.4

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

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