Calculus Examples

$(x + 1) \frac{d y}{d x} + y = \ln (| x |)$ ; con $y (1) = 10$

Step 1

Check if the left side of the equation is the result of the derivative of the term $(x + 1) y$ .

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Step 1.1

Differentiate using the Product Rule which states that $\frac{d}{d x} [f (x) g (x)]$ is $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ where $f (x) = x + 1$ and $g (x) = y$ .

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

Step 1.2

Rewrite $\frac{d}{d x} [y]$ as $y'$ .

$(x + 1) y' + y \frac{d}{d x} [x + 1]$

Step 1.3

By the Sum Rule, the derivative of $x + 1$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [1]$ .

$(x + 1) y' + y (\frac{d}{d x} [x] + \frac{d}{d x} [1])$

Step 1.4

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

$(x + 1) y' + y (1 + \frac{d}{d x} [1])$

Step 1.5

Since $1$ is constant with respect to $x$ , the derivative of $1$ with respect to $x$ is $0$ .

$(x + 1) y' + y (1 + 0)$

Step 1.6

Add $1$ and $0$ .

$(x + 1) y' + y \cdot 1$

Step 1.7

Substitute $\frac{d y}{d x}$ for $y'$ .

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

Step 1.8

Reorder $y$ and $1$ .

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

Step 1.9

Multiply $y$ by $1$ .

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

Step 2

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

$\frac{d}{d x} [(x + 1) y] = \ln (| x |)$

Step 3

Set up an integral on each side.

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

Step 4

Integrate the left side.

$(x + 1) y = \int \ln (| x |) d x$

Step 5

Integrate the right side.

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Step 5.1

Integrate by parts using the formula $\int u d v = u v - \int v d u$ , where $u = \ln (| x |)$ and $d v = 1$ .

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

Step 5.2

Simplify.

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Step 5.2.1

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

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

Step 5.2.2

Cancel the common factor of $x$ .

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Step 5.2.2.1

Cancel the common factor.

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

Step 5.2.2.2

Rewrite the expression.

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

Step 5.3

Apply the constant rule.

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

Step 5.4

Simplify.

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

Step 6

Divide each term in $(x + 1) y = \ln (| x |) x - x + C$ by $x + 1$ and simplify.

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Step 6.1

Divide each term in $(x + 1) y = \ln (| x |) x - x + C$ by $x + 1$ .

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

Step 6.2

Simplify the left side.

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Step 6.2.1

Cancel the common factor of $x + 1$ .

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Step 6.2.1.1

Cancel the common factor.

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

Step 6.2.1.2

Divide $y$ by $1$ .

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

Step 6.3

Simplify the right side.

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Step 6.3.1

Move the negative in front of the fraction.

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

Step 6.3.2

Combine the numerators over the common denominator.

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

Step 6.3.3

Combine the numerators over the common denominator.

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

Step 6.3.4

Reorder factors in $\ln (| x |) x - x + C$ .

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

Step 7

Use the initial condition to find the value of $C$ by substituting $1$ for $x$ and $10$ for $y$ in $y = \frac{x \ln (| x |) - x + C}{x + 1}$ .

$10 = \frac{1 \ln (| 1 |) - 1 + C}{1 + 1}$

Step 8

Solve for $C$ .

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Step 8.1

Rewrite the equation as $\frac{1 \ln (| 1 |) - 1 + C}{1 + 1} = 10$ .

$\frac{1 \ln (| 1 |) - 1 + C}{1 + 1} = 10$

Step 8.2

Multiply both sides of the equation by $1 + 1$ .

$(1 + 1) \frac{1 \ln (| 1 |) - 1 + C}{1 + 1} = (1 + 1) \cdot 10$

Step 8.3

Simplify both sides of the equation.

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Step 8.3.1

Simplify the left side.

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Step 8.3.1.1

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

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Step 8.3.1.1.1

Simplify the numerator.

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Step 8.3.1.1.1.1

Multiply $\ln (| 1 |)$ by $1$ .

$(1 + 1) \frac{\ln (| 1 |) - 1 + C}{1 + 1} = (1 + 1) \cdot 10$

Step 8.3.1.1.1.2

The absolute value is the distance between a number and zero. The distance between $0$ and $1$ is $1$ .

$(1 + 1) \frac{\ln (1) - 1 + C}{1 + 1} = (1 + 1) \cdot 10$

Step 8.3.1.1.1.3

The natural logarithm of $1$ is $0$ .

$(1 + 1) \frac{0 - 1 + C}{1 + 1} = (1 + 1) \cdot 10$

Step 8.3.1.1.1.4

Subtract $1$ from $0$ .

$(1 + 1) \frac{- 1 + C}{1 + 1} = (1 + 1) \cdot 10$

Step 8.3.1.1.2

Reduce the expression by cancelling the common factors.

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Step 8.3.1.1.2.1

Add $1$ and $1$ .

$(1 + 1) \frac{- 1 + C}{2} = (1 + 1) \cdot 10$

Step 8.3.1.1.2.2

Add $1$ and $1$ .

$2 \frac{- 1 + C}{2} = (1 + 1) \cdot 10$

Step 8.3.1.1.2.3

Cancel the common factor of $2$ .

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Step 8.3.1.1.2.3.1

Cancel the common factor.

$2 \frac{- 1 + C}{2} = (1 + 1) \cdot 10$

Step 8.3.1.1.2.3.2

Rewrite the expression.

$- 1 + C = (1 + 1) \cdot 10$

Step 8.3.2

Simplify the right side.

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Step 8.3.2.1

Simplify $(1 + 1) \cdot 10$ .

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Step 8.3.2.1.1

Add $1$ and $1$ .

$- 1 + C = 2 \cdot 10$

Step 8.3.2.1.2

Multiply $2$ by $10$ .

$- 1 + C = 20$

Step 8.4

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

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Step 8.4.1

Add $1$ to both sides of the equation.

$C = 20 + 1$

Step 8.4.2

Add $20$ and $1$ .

$C = 21$

Step 9

Substitute $21$ for $C$ in $y = \frac{x \ln (| x |) - x + C}{x + 1}$ and simplify.

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Step 9.1

Substitute $21$ for $C$ .

$y = \frac{x \ln (| x |) - x + 21}{x + 1}$