Calculus Examples

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

Step 1

Rewrite the differential equation as $\frac{d y}{d x} + M (x) y = Q (x)$ .

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Step 1.1

Rewrite the equation as $M (x) \frac{d y}{d x} + P (x) y = Q (x)$ .

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Step 1.1.1

Split the fraction $\frac{y + \ln (x)}{x}$ into two fractions.

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

Step 1.1.2

Subtract $\frac{y}{x}$ from both sides of the equation.

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

Step 1.2

Factor $y$ out of $- \frac{y}{x}$ .

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

Step 1.3

Reorder $y$ and $- \frac{1}{x}$ .

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

Step 2

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 2.1

Set up the integration.

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

Step 2.2

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

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Step 2.2.1

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

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

Step 2.2.2

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

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

Step 2.2.3

Simplify.

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

Step 2.3

Remove the constant of integration.

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

Step 2.4

Use the logarithmic power rule.

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

Step 2.5

Exponentiation and log are inverse functions.

$x^{- 1}$

Step 2.6

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

$\frac{1}{x}$

Step 3

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

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Step 3.1

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

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

Step 3.2

Simplify each term.

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Step 3.2.1

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

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

Step 3.2.2

Rewrite using the commutative property of multiplication.

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

Step 3.2.3

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

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

Step 3.2.4

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

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Step 3.2.4.1

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

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

Step 3.2.4.2

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

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

Step 3.2.4.3

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

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

Step 3.2.4.4

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

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

Step 3.2.4.5

Add $1$ and $1$ .

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

Step 3.3

Multiply $\frac{1}{x} \cdot \frac{\ln (x)}{x}$ .

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Step 3.3.1

Multiply $\frac{1}{x}$ by $\frac{\ln (x)}{x}$ .

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

Step 3.3.2

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

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

Step 3.3.3

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

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

Step 3.3.4

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

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

Step 3.3.5

Add $1$ and $1$ .

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

Step 4

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

$\frac{d}{d x} [\frac{1}{x} y] = \frac{\ln (x)}{x^{2}}$

Step 5

Set up an integral on each side.

$\int \frac{d}{d x} [\frac{1}{x} y] d x = \int \frac{\ln (x)}{x^{2}} d x$

Step 6

Integrate the left side.

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

Step 7

Integrate the right side.

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Step 7.1

Apply basic rules of exponents.

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Step 7.1.1

Move $x^{2}$ out of the denominator by raising it to the $- 1$ power.

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

Step 7.1.2

Multiply the exponents in ${(x^{2})}^{- 1}$ .

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Step 7.1.2.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

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

Step 7.1.2.2

Multiply $2$ by $- 1$ .

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

Step 7.2

Integrate by parts using the formula $\int u d v = u v - \int v d u$ , where $u = \ln (x)$ and $d v = x^{- 2}$ .

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

Step 7.3

Simplify.

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Step 7.3.1

Combine $\ln (x)$ and $\frac{1}{x}$ .

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

Step 7.3.2

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

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

Step 7.3.3

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

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

Step 7.3.4

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

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

Step 7.3.5

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

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

Step 7.3.6

Add $1$ and $1$ .

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

Step 7.4

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

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

Step 7.5

Simplify the expression.

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Step 7.5.1

Simplify.

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Step 7.5.1.1

Multiply $- 1$ by $- 1$ .

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

Step 7.5.1.2

Multiply $\int \frac{1}{x^{2}} d x$ by $1$ .

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

Step 7.5.2

Apply basic rules of exponents.

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Step 7.5.2.1

Move $x^{2}$ out of the denominator by raising it to the $- 1$ power.

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

Step 7.5.2.2

Multiply the exponents in ${(x^{2})}^{- 1}$ .

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Step 7.5.2.2.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

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

Step 7.5.2.2.2

Multiply $2$ by $- 1$ .

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

Step 7.6

By the Power Rule, the integral of $x^{- 2}$ with respect to $x$ is $- x^{- 1}$ .

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

Step 7.7

Rewrite $- \frac{\ln (x)}{x} - x^{- 1} + C$ as $- \frac{\ln (x)}{x} - \frac{1}{x} + C$ .

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

Step 8

Solve for $y$ .

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Step 8.1

Move all terms containing variables to the left side of the equation.

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Step 8.1.1

Add $\frac{\ln (x)}{x}$ to both sides of the equation.

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

Step 8.1.2

Add $\frac{1}{x}$ to both sides of the equation.

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

Step 8.1.3

Subtract $C$ from both sides of the equation.

$\frac{1}{x} y + \frac{\ln (x)}{x} + \frac{1}{x} - C = 0$

Step 8.1.4

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

$\frac{y}{x} + \frac{\ln (x)}{x} + \frac{1}{x} - C = 0$

Step 8.2

Move all terms not containing $y$ to the right side of the equation.

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Step 8.2.1

Subtract $\frac{\ln (x)}{x}$ from both sides of the equation.

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

Step 8.2.2

Subtract $\frac{1}{x}$ from both sides of the equation.

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

Step 8.2.3

Add $C$ to both sides of the equation.

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

Step 8.3

Multiply both sides by $x$ .

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

Step 8.4

Simplify.

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Step 8.4.1

Simplify the left side.

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Step 8.4.1.1

Cancel the common factor of $x$ .

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Step 8.4.1.1.1

Cancel the common factor.

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

Step 8.4.1.1.2

Rewrite the expression.

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

Step 8.4.2

Simplify the right side.

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Step 8.4.2.1

Simplify $(- \frac{\ln (x)}{x} - \frac{1}{x} + C) x$ .

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Step 8.4.2.1.1

Apply the distributive property.

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

Step 8.4.2.1.2

Simplify.

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Step 8.4.2.1.2.1

Cancel the common factor of $x$ .

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Step 8.4.2.1.2.1.1

Move the leading negative in $- \frac{\ln (x)}{x}$ into the numerator.

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

Step 8.4.2.1.2.1.2

Cancel the common factor.

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

Step 8.4.2.1.2.1.3

Rewrite the expression.

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

Step 8.4.2.1.2.2

Cancel the common factor of $x$ .

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Step 8.4.2.1.2.2.1

Move the leading negative in $- \frac{1}{x}$ into the numerator.

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

Step 8.4.2.1.2.2.2

Cancel the common factor.

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

Step 8.4.2.1.2.2.3

Rewrite the expression.

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

Step 8.4.2.1.3

Move $- 1$ .

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

Step 8.4.2.1.4

Reorder $- \ln (x)$ and $C x$ .

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