Calculus Examples

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

Step 1

The integrating factor is defined by the formula $e^{\int P (x) d x}$ , where $P (x) = - \frac{1}{x}$ .

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Step 1.1

Set up the integration.

$e^{\int - \frac{1}{x} d x}$

Step 1.2

Integrate $- \frac{1}{x}$ .

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Step 1.2.1

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

$e^{- \int \frac{1}{x} d x}$

Step 1.2.2

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

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

Step 1.2.3

Simplify.

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

Step 1.3

Remove the constant of integration.

$e^{- \ln (x)}$

Step 1.4

Use the logarithmic power rule.

$e^{\ln (x^{- 1})}$

Step 1.5

Exponentiation and log are inverse functions.

$x^{- 1}$

Step 1.6

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$\frac{1}{x}$

Step 2

Multiply each term by the integrating factor $\frac{1}{x}$ .

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Step 2.1

Multiply each term by $\frac{1}{x}$ .

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

Step 2.2

Simplify each term.

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Step 2.2.1

Combine $\frac{1}{x}$ and $\frac{d y}{d x}$ .

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

Step 2.2.2

Rewrite using the commutative property of multiplication.

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

Step 2.2.3

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

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

Step 2.2.4

Multiply $- \frac{1}{x} \cdot \frac{y}{x}$ .

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Step 2.2.4.1

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

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

Step 2.2.4.2

Raise $x$ to the power of $1$ .

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

Step 2.2.4.3

Raise $x$ to the power of $1$ .

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

Step 2.2.4.4

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$\frac{\frac{d y}{d x}}{x} - \frac{y}{x^{1 + 1}} = \frac{1}{x} \cdot 2$

Step 2.2.4.5

Add $1$ and $1$ .

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

Step 2.3

Combine $\frac{1}{x}$ and $2$ .

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

Step 3

Rewrite the left side as a result of differentiating a product.

$\frac{d}{d x} [\frac{1}{x} y] = \frac{2}{x}$

Step 4

Set up an integral on each side.

$\int \frac{d}{d x} [\frac{1}{x} y] d x = \int \frac{2}{x} d x$

Step 5

Integrate the left side.

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

Step 6

Integrate the right side.

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Step 6.1

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

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

Step 6.2

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

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

Step 6.3

Simplify.

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

Step 7

Solve for $y$ .

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Step 7.1

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

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

Step 7.2

Simplify each term.

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Step 7.2.1

Simplify $2 \ln (| x |)$ by moving $2$ inside the logarithm.

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

Step 7.2.2

Remove the absolute value in ${| x |}^{2}$ because exponentiations with even powers are always positive.

$\frac{y}{x} = \ln (x^{2}) + C$

Step 7.3

Multiply both sides by $x$ .

$\frac{y}{x} x = (\ln (x^{2}) + C) x$

Step 7.4

Simplify.

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Step 7.4.1

Simplify the left side.

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Step 7.4.1.1

Cancel the common factor of $x$ .

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Step 7.4.1.1.1

Cancel the common factor.

$\frac{y}{x} x = (\ln (x^{2}) + C) x$

Step 7.4.1.1.2

Rewrite the expression.

$y = (\ln (x^{2}) + C) x$

Step 7.4.2

Simplify the right side.

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Step 7.4.2.1

Simplify $(\ln (x^{2}) + C) x$ .

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Step 7.4.2.1.1

Apply the distributive property.

$y = \ln (x^{2}) x + C x$

Step 7.4.2.1.2

Simplify the expression.

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Step 7.4.2.1.2.1

Reorder factors in $\ln (x^{2}) x + C x$ .

$y = x \ln (x^{2}) + C x$

Step 7.4.2.1.2.2

Reorder $x \ln (x^{2})$ and $C x$ .

$y = C x + x \ln (x^{2})$