Calculus Examples

$\frac{d y}{d x} = \frac{1}{3 (7 - x)}$ ; with $y (6) = 19$

Step 1

Rewrite the equation.

$d y = \frac{1}{3 (7 - x)} d x$

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

$\int d y = \int \frac{1}{3 (7 - x)} d x$

Step 2.2

Apply the constant rule.

$y + C_{1} = \int \frac{1}{3 (7 - x)} d x$

Step 2.3

Integrate the right side.

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Step 2.3.1

Since $\frac{1}{3}$ is constant with respect to $x$ , move $\frac{1}{3}$ out of the integral.

$y + C_{1} = \frac{1}{3} \int \frac{1}{7 - x} d x$

Step 2.3.2

Let $u = 7 - x$ . Then $d u = - d x$ , so $- d u = d x$ . Rewrite using $u$ and $d$ $u$ .

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Step 2.3.2.1

Let $u = 7 - x$ . Find $\frac{d u}{d x}$ .

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Step 2.3.2.1.1

Rewrite.

$\frac{- 1}{1}$

Step 2.3.2.1.2

Divide $- 1$ by $1$ .

$- 1$

Step 2.3.2.2

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

$y + C_{1} = \frac{1}{3} \int \frac{- 1}{u} d u$

Step 2.3.3

Move the negative in front of the fraction.

$y + C_{1} = \frac{1}{3} \int - \frac{1}{u} d u$

Step 2.3.4

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

$y + C_{1} = \frac{1}{3} (- \int \frac{1}{u} d u)$

Step 2.3.5

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

$y + C_{1} = \frac{1}{3} (- (\ln (| u |) + C_{2}))$

Step 2.3.6

Simplify.

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Step 2.3.6.1

Simplify.

$y + C_{1} = \frac{1}{3} (- \ln (| u |)) + C_{2}$

Step 2.3.6.2

Combine $\frac{1}{3}$ and $\ln (| u |)$ .

$y + C_{1} = - \frac{\ln (| u |)}{3} + C_{2}$

Step 2.3.7

Replace all occurrences of $u$ with $7 - x$ .

$y + C_{1} = - \frac{\ln (| 7 - x |)}{3} + C_{2}$

Step 2.3.8

Reorder terms.

$y + C_{1} = - \frac{1}{3} \ln (| 7 - x |) + C_{2}$

Step 2.4

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

$y = - \frac{1}{3} \ln (| 7 - x |) + C$

Step 3

Use the initial condition to find the value of $C$ by substituting $6$ for $x$ and $19$ for $y$ in $y = - \frac{1}{3} \ln (| 7 - x |) + C$ .

$19 = - \frac{1}{3} \ln (| 7 - 6 |) + C$

Step 4

Solve for $C$ .

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Step 4.1

Rewrite the equation as $- \frac{1}{3} \cdot \ln (| 7 - 6 |) + C = 19$ .

$- \frac{1}{3} \cdot \ln (| 7 - 6 |) + C = 19$

Step 4.2

Simplify $- \frac{1}{3} \cdot \ln (| 7 - 6 |) + C$ .

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Step 4.2.1

Simplify each term.

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Step 4.2.1.1

Subtract $6$ from $7$ .

$- \frac{1}{3} \cdot \ln (| 1 |) + C = 19$

Step 4.2.1.2

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

$- \frac{1}{3} \cdot \ln (1) + C = 19$

Step 4.2.1.3

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

$- \frac{1}{3} \cdot 0 + C = 19$

Step 4.2.1.4

Multiply $- \frac{1}{3} \cdot 0$ .

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Step 4.2.1.4.1

Multiply $0$ by $- 1$ .

$0 (\frac{1}{3}) + C = 19$

Step 4.2.1.4.2

Multiply $0$ by $\frac{1}{3}$ .

$0 + C = 19$

Step 4.2.2

Add $0$ and $C$ .

$C = 19$

Step 5

Substitute $19$ for $C$ in $y = - \frac{1}{3} \ln (| 7 - x |) + C$ and simplify.

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Step 5.1

Substitute $19$ for $C$ .

$y = - \frac{1}{3} \ln (| 7 - x |) + 19$

Step 5.2

Multiply $- \frac{1}{3} \ln (| 7 - x |)$ .

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Step 5.2.1

Reorder $\ln (| 7 - x |)$ and $\frac{1}{3}$ .

$y = - (\frac{1}{3} \ln (| 7 - x |)) + 19$

Step 5.2.2

Simplify $\frac{1}{3} \ln (| 7 - x |)$ by moving $\frac{1}{3}$ inside the logarithm.

$y = - \ln ({| 7 - x |}^{\frac{1}{3}}) + 19$