Calculus Examples

$\frac{d x}{d t} = 8 - 3 x$ ; , $x (0) = 4$

Step 1

Separate the variables.

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Step 1.1

Multiply both sides by $\frac{1}{8 - 3 x}$ .

$\frac{1}{8 - 3 x} \frac{d x}{d t} = \frac{1}{8 - 3 x} (8 - 3 x)$

Step 1.2

Cancel the common factor of $8 - 3 x$ .

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Step 1.2.1

Cancel the common factor.

$\frac{1}{8 - 3 x} \frac{d x}{d t} = \frac{1}{8 - 3 x} (8 - 3 x)$

Step 1.2.2

Rewrite the expression.

$\frac{1}{8 - 3 x} \frac{d x}{d t} = 1$

Step 1.3

Rewrite the equation.

$\frac{1}{8 - 3 x} d x = 1 d t$

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

$\int \frac{1}{8 - 3 x} d x = \int d t$

Step 2.2

Integrate the left side.

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Step 2.2.1

Let $u = 8 - 3 x$ . Then $d u = - 3 d x$ , so $- \frac{1}{3} d u = d x$ . Rewrite using $u$ and $d$ $u$ .

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Step 2.2.1.1

Let $u = 8 - 3 x$ . Find $\frac{d u}{d x}$ .

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Step 2.2.1.1.1

Differentiate $8 - 3 x$ .

$\frac{d}{d x} [8 - 3 x]$

Step 2.2.1.1.2

Differentiate.

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Step 2.2.1.1.2.1

By the Sum Rule, the derivative of $8 - 3 x$ with respect to $x$ is $\frac{d}{d x} [8] + \frac{d}{d x} [- 3 x]$ .

$\frac{d}{d x} [8] + \frac{d}{d x} [- 3 x]$

Step 2.2.1.1.2.2

Since $8$ is constant with respect to $x$ , the derivative of $8$ with respect to $x$ is $0$ .

$0 + \frac{d}{d x} [- 3 x]$

Step 2.2.1.1.3

Evaluate $\frac{d}{d x} [- 3 x]$ .

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Step 2.2.1.1.3.1

Since $- 3$ is constant with respect to $x$ , the derivative of $- 3 x$ with respect to $x$ is $- 3 \frac{d}{d x} [x]$ .

$0 - 3 \frac{d}{d x} [x]$

Step 2.2.1.1.3.2

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

$0 - 3 \cdot 1$

Step 2.2.1.1.3.3

Multiply $- 3$ by $1$ .

$0 - 3$

Step 2.2.1.1.4

Subtract $3$ from $0$ .

$- 3$

Step 2.2.1.2

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

$\int \frac{1}{u} \cdot \frac{1}{- 3} d u = \int d t$

Step 2.2.2

Simplify.

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Step 2.2.2.1

Move the negative in front of the fraction.

$\int \frac{1}{u} (- \frac{1}{3}) d u = \int d t$

Step 2.2.2.2

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

$\int - \frac{1}{u \cdot 3} d u = \int d t$

Step 2.2.2.3

Move $3$ to the left of $u$ .

$\int - \frac{1}{3 u} d u = \int d t$

Step 2.2.3

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

$- \int \frac{1}{3 u} d u = \int d t$

Step 2.2.4

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

$- (\frac{1}{3} \int \frac{1}{u} d u) = \int d t$

Step 2.2.5

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

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

Step 2.2.6

Simplify.

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

Step 2.2.7

Replace all occurrences of $u$ with $8 - 3 x$ .

$- \frac{1}{3} \ln (| 8 - 3 x |) + C_{1} = \int d t$

Step 2.3

Apply the constant rule.

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

Step 2.4

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

$- \frac{1}{3} \ln (| 8 - 3 x |) = t + C$

Step 3

Solve for $x$ .

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Step 3.1

Multiply both sides of the equation by $- 3$ .

$- 3 (- \frac{1}{3} \ln (| 8 - 3 x |)) = - 3 (t + C)$

Step 3.2

Simplify both sides of the equation.

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Step 3.2.1

Simplify the left side.

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Step 3.2.1.1

Simplify $- 3 (- \frac{1}{3} \ln (| 8 - 3 x |))$ .

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Step 3.2.1.1.1

Combine $\ln (| 8 - 3 x |)$ and $\frac{1}{3}$ .

$- 3 (- \frac{\ln (| 8 - 3 x |)}{3}) = - 3 (t + C)$

Step 3.2.1.1.2

Cancel the common factor of $3$ .

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Step 3.2.1.1.2.1

Move the leading negative in $- \frac{\ln (| 8 - 3 x |)}{3}$ into the numerator.

$- 3 \frac{- \ln (| 8 - 3 x |)}{3} = - 3 (t + C)$

Step 3.2.1.1.2.2

Factor $3$ out of $- 3$ .

$3 (- 1) \frac{- \ln (| 8 - 3 x |)}{3} = - 3 (t + C)$

Step 3.2.1.1.2.3

Cancel the common factor.

$3 \cdot - 1 \frac{- \ln (| 8 - 3 x |)}{3} = - 3 (t + C)$

Step 3.2.1.1.2.4

Rewrite the expression.

$- - \ln (| 8 - 3 x |) = - 3 (t + C)$

Step 3.2.1.1.3

Multiply.

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Step 3.2.1.1.3.1

Multiply $- 1$ by $- 1$ .

$1 \ln (| 8 - 3 x |) = - 3 (t + C)$

Step 3.2.1.1.3.2

Multiply $\ln (| 8 - 3 x |)$ by $1$ .

$\ln (| 8 - 3 x |) = - 3 (t + C)$

Step 3.2.2

Simplify the right side.

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Step 3.2.2.1

Apply the distributive property.

$\ln (| 8 - 3 x |) = - 3 t - 3 C$

Step 3.3

To solve for $x$ , rewrite the equation using properties of logarithms.

$e^{\ln (| 8 - 3 x |)} = e^{- 3 t - 3 C}$

Step 3.4

Rewrite $\ln (| 8 - 3 x |) = - 3 t - 3 C$ in exponential form using the definition of a logarithm. If $x$ and $b$ are positive real numbers and $b \neq 1$ , then $\log_{b} (x) = y$ is equivalent to $b^{y} = x$ .

$e^{- 3 t - 3 C} = | 8 - 3 x |$

Step 3.5

Solve for $x$ .

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Step 3.5.1

Rewrite the equation as $| 8 - 3 x | = e^{- 3 t - 3 C}$ .

$| 8 - 3 x | = e^{- 3 t - 3 C}$

Step 3.5.2

Remove the absolute value term. This creates a $\pm$ on the right side of the equation because $| x | = \pm x$ .

$8 - 3 x = \pm e^{- 3 t - 3 C}$

Step 3.5.3

Subtract $8$ from both sides of the equation.

$- 3 x = \pm e^{- 3 t - 3 C} - 8$

Step 3.5.4

Divide each term in $- 3 x = \pm e^{- 3 t - 3 C} - 8$ by $- 3$ and simplify.

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Step 3.5.4.1

Divide each term in $- 3 x = \pm e^{- 3 t - 3 C} - 8$ by $- 3$ .

$\frac{- 3 x}{- 3} = \frac{\pm e^{- 3 t - 3 C}}{- 3} + \frac{- 8}{- 3}$

Step 3.5.4.2

Simplify the left side.

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Step 3.5.4.2.1

Cancel the common factor of $- 3$ .

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Step 3.5.4.2.1.1

Cancel the common factor.

$\frac{- 3 x}{- 3} = \frac{\pm e^{- 3 t - 3 C}}{- 3} + \frac{- 8}{- 3}$

Step 3.5.4.2.1.2

Divide $x$ by $1$ .

$x = \frac{\pm e^{- 3 t - 3 C}}{- 3} + \frac{- 8}{- 3}$

Step 3.5.4.3

Simplify the right side.

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Step 3.5.4.3.1

Simplify each term.

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Step 3.5.4.3.1.1

Simplify $\frac{\pm e^{- 3 t - 3 C}}{- 3}$ .

$x = \frac{\pm e^{- 3 t - 3 C}}{3} + \frac{- 8}{- 3}$

Step 3.5.4.3.1.2

Dividing two negative values results in a positive value.

$x = \frac{\pm e^{- 3 t - 3 C}}{3} + \frac{8}{3}$

Step 3.5.4.3.2

Combine the numerators over the common denominator.

$x = \frac{\pm e^{- 3 t - 3 C} + 8}{3}$

Step 4

Group the constant terms together.

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Step 4.1

Simplify the constant of integration.

$x = \frac{\pm e^{- 3 t + C} + 8}{3}$

Step 4.2

Rewrite $e^{- 3 t + C}$ as $e^{- 3 t} e^{C}$ .

$x = \frac{\pm e^{- 3 t} e^{C} + 8}{3}$

Step 4.3

Reorder $e^{- 3 t}$ and $e^{C}$ .

$x = \frac{\pm e^{C} e^{- 3 t} + 8}{3}$

Step 4.4

Combine constants with the plus or minus.

$x = \frac{C e^{- 3 t} + 8}{3}$

Step 5

Use the initial condition to find the value of $C$ by substituting $0$ for $t$ and $4$ for $x$ in $x = \frac{C e^{- 3 t} + 8}{3}$ .

$4 = \frac{C e^{- 3 \cdot 0} + 8}{3}$

Step 6

Solve for $C$ .

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Step 6.1

Rewrite the equation as $\frac{C e^{- 3 \cdot 0} + 8}{3} = 4$ .

$\frac{C e^{- 3 \cdot 0} + 8}{3} = 4$

Step 6.2

Multiply both sides by $3$ .

$\frac{C e^{- 3 \cdot 0} + 8}{3} \cdot 3 = 4 \cdot 3$

Step 6.3

Simplify.

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Step 6.3.1

Simplify the left side.

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Step 6.3.1.1

Simplify $\frac{C e^{- 3 \cdot 0} + 8}{3} \cdot 3$ .

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Step 6.3.1.1.1

Simplify the numerator.

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Step 6.3.1.1.1.1

Multiply $- 3$ by $0$ .

$\frac{C e^{0} + 8}{3} \cdot 3 = 4 \cdot 3$

Step 6.3.1.1.1.2

Anything raised to $0$ is $1$ .

$\frac{C \cdot 1 + 8}{3} \cdot 3 = 4 \cdot 3$

Step 6.3.1.1.1.3

Multiply $C$ by $1$ .

$\frac{C + 8}{3} \cdot 3 = 4 \cdot 3$

Step 6.3.1.1.2

Cancel the common factor of $3$ .

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Step 6.3.1.1.2.1

Cancel the common factor.

$\frac{C + 8}{3} \cdot 3 = 4 \cdot 3$

Step 6.3.1.1.2.2

Rewrite the expression.

$C + 8 = 4 \cdot 3$

Step 6.3.2

Simplify the right side.

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Step 6.3.2.1

Multiply $4$ by $3$ .

$C + 8 = 12$

Step 6.4

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

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Step 6.4.1

Subtract $8$ from both sides of the equation.

$C = 12 - 8$

Step 6.4.2

Subtract $8$ from $12$ .

$C = 4$

Step 7

Substitute $4$ for $C$ in $x = \frac{C e^{- 3 t} + 8}{3}$ and simplify.

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Step 7.1

Substitute $4$ for $C$ .

$x = \frac{4 e^{- 3 t} + 8}{3}$

Step 7.2

Factor $4$ out of $4 e^{- 3 t} + 8$ .

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Step 7.2.1

Factor $4$ out of $4 e^{- 3 t}$ .

$x = \frac{4 (e^{- 3 t}) + 8}{3}$

Step 7.2.2

Factor $4$ out of $8$ .

$x = \frac{4 e^{- 3 t} + 4 \cdot 2}{3}$

Step 7.2.3

Factor $4$ out of $4 e^{- 3 t} + 4 \cdot 2$ .

$x = \frac{4 (e^{- 3 t} + 2)}{3}$