Calculus Examples

$\frac{d y}{d x} = 8 + \frac{1}{x}$ , $y (1) = 9$

Step 1

Rewrite the equation.

$d y = (8 + \frac{1}{x}) d x$

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

$\int d y = \int 8 + \frac{1}{x} d x$

Step 2.2

Apply the constant rule.

$y + C_{1} = \int 8 + \frac{1}{x} d x$

Step 2.3

Integrate the right side.

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Step 2.3.1

Split the single integral into multiple integrals.

$y + C_{1} = \int 8 d x + \int \frac{1}{x} d x$

Step 2.3.2

Apply the constant rule.

$y + C_{1} = 8 x + C_{2} + \int \frac{1}{x} d x$

Step 2.3.3

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

$y + C_{1} = 8 x + C_{2} + \ln (| x |) + C_{3}$

Step 2.3.4

Simplify.

$y + C_{1} = 8 x + \ln (| x |) + C_{4}$

Step 2.4

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

$y = 8 x + \ln (| x |) + C$

Step 3

Use the initial condition to find the value of $C$ by substituting $1$ for $x$ and $9$ for $y$ in $y = 8 x + \ln (| x |) + C$ .

$9 = 8 \cdot 1 + \ln (| 1 |) + C$

Step 4

Solve for $C$ .

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Step 4.1

Rewrite the equation as $8 \cdot 1 + \ln (| 1 |) + C = 9$ .

$8 \cdot 1 + \ln (| 1 |) + C = 9$

Step 4.2

Simplify $8 \cdot 1 + \ln (| 1 |) + C$ .

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Step 4.2.1

Simplify each term.

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Step 4.2.1.1

Multiply $8$ by $1$ .

$8 + \ln (| 1 |) + C = 9$

Step 4.2.1.2

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

$8 + \ln (1) + C = 9$

Step 4.2.1.3

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

$8 + 0 + C = 9$

Step 4.2.2

Add $8$ and $0$ .

$8 + C = 9$

Step 4.3

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

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Step 4.3.1

Subtract $8$ from both sides of the equation.

$C = 9 - 8$

Step 4.3.2

Subtract $8$ from $9$ .

$C = 1$

Step 5

Substitute $1$ for $C$ in $y = 8 x + \ln (| x |) + C$ and simplify.

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Step 5.1

Substitute $1$ for $C$ .

$y = 8 x + \ln (| x |) + 1$