Calculus Examples

$\frac{d p}{d t} + 20 p - 240 = 0$

Step 1

Separate the variables.

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Step 1.1

Move all terms not containing $\frac{d p}{d t}$ to the right side of the equation.

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Step 1.1.1

Subtract $20 p$ from both sides of the equation.

$\frac{d p}{d t} - 240 = - 20 p$

Step 1.1.2

Add $240$ to both sides of the equation.

$\frac{d p}{d t} = - 20 p + 240$

Step 1.2

Multiply both sides by $\frac{1}{- 20 p + 240}$ .

$\frac{1}{- 20 p + 240} \frac{d p}{d t} = \frac{1}{- 20 p + 240} (- 20 p + 240)$

Step 1.3

Simplify.

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Step 1.3.1

Factor $20$ out of $- 20 p + 240$ .

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Step 1.3.1.1

Factor $20$ out of $- 20 p$ .

$\frac{1}{- 20 p + 240} \frac{d p}{d t} = \frac{1}{20 (- p) + 240} (- 20 p + 240)$

Step 1.3.1.2

Factor $20$ out of $240$ .

$\frac{1}{- 20 p + 240} \frac{d p}{d t} = \frac{1}{20 (- p) + 20 (12)} (- 20 p + 240)$

Step 1.3.1.3

Factor $20$ out of $20 (- p) + 20 (12)$ .

$\frac{1}{- 20 p + 240} \frac{d p}{d t} = \frac{1}{20 (- p + 12)} (- 20 p + 240)$

Step 1.3.2

Multiply $\frac{1}{20 (- p + 12)}$ by $- 20 p + 240$ .

$\frac{1}{- 20 p + 240} \frac{d p}{d t} = \frac{- 20 p + 240}{20 (- p + 12)}$

Step 1.3.3

Cancel the common factor of $- 20 p + 240$ and $20$ .

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Step 1.3.3.1

Factor $20$ out of $- 20 p$ .

$\frac{1}{- 20 p + 240} \frac{d p}{d t} = \frac{20 (- p) + 240}{20 (- p + 12)}$

Step 1.3.3.2

Factor $20$ out of $240$ .

$\frac{1}{- 20 p + 240} \frac{d p}{d t} = \frac{20 (- p) + 20 \cdot 12}{20 (- p + 12)}$

Step 1.3.3.3

Factor $20$ out of $20 (- p) + 20 (12)$ .

$\frac{1}{- 20 p + 240} \frac{d p}{d t} = \frac{20 (- p + 12)}{20 (- p + 12)}$

Step 1.3.3.4

Cancel the common factors.

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Step 1.3.3.4.1

Cancel the common factor.

$\frac{1}{- 20 p + 240} \frac{d p}{d t} = \frac{20 (- p + 12)}{20 (- p + 12)}$

Step 1.3.3.4.2

Rewrite the expression.

$\frac{1}{- 20 p + 240} \frac{d p}{d t} = \frac{- p + 12}{- p + 12}$

Step 1.3.4

Cancel the common factor of $- p + 12$ .

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Step 1.3.4.1

Cancel the common factor.

$\frac{1}{- 20 p + 240} \frac{d p}{d t} = \frac{- p + 12}{- p + 12}$

Step 1.3.4.2

Rewrite the expression.

$\frac{1}{- 20 p + 240} \frac{d p}{d t} = 1$

Step 1.4

Rewrite the equation.

$\frac{1}{- 20 p + 240} d p = 1 d t$

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

$\int \frac{1}{- 20 p + 240} d p = \int d t$

Step 2.2

Integrate the left side.

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Step 2.2.1

Let $u = - 20 p + 240$ . Then $d u = - 20 d p$ , so $- \frac{1}{20} d u = d p$ . Rewrite using $u$ and $d$ $u$ .

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Step 2.2.1.1

Let $u = - 20 p + 240$ . Find $\frac{d u}{d p}$ .

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Step 2.2.1.1.1

Differentiate $- 20 p + 240$ .

$\frac{d}{d p} [- 20 p + 240]$

Step 2.2.1.1.2

By the Sum Rule, the derivative of $- 20 p + 240$ with respect to $p$ is $\frac{d}{d p} [- 20 p] + \frac{d}{d p} [240]$ .

$\frac{d}{d p} [- 20 p] + \frac{d}{d p} [240]$

Step 2.2.1.1.3

Evaluate $\frac{d}{d p} [- 20 p]$ .

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Step 2.2.1.1.3.1

Since $- 20$ is constant with respect to $p$ , the derivative of $- 20 p$ with respect to $p$ is $- 20 \frac{d}{d p} [p]$ .

$- 20 \frac{d}{d p} [p] + \frac{d}{d p} [240]$

Step 2.2.1.1.3.2

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

$- 20 \cdot 1 + \frac{d}{d p} [240]$

Step 2.2.1.1.3.3

Multiply $- 20$ by $1$ .

$- 20 + \frac{d}{d p} [240]$

Step 2.2.1.1.4

Differentiate using the Constant Rule.

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Step 2.2.1.1.4.1

Since $240$ is constant with respect to $p$ , the derivative of $240$ with respect to $p$ is $0$ .

$- 20 + 0$

Step 2.2.1.1.4.2

Add $- 20$ and $0$ .

$- 20$

Step 2.2.1.2

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

$\int \frac{1}{u} \cdot \frac{1}{- 20} d u = \int d t$

Step 2.2.2

Simplify.

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Step 2.2.2.1

Move the negative in front of the fraction.

$\int \frac{1}{u} (- \frac{1}{20}) d u = \int d t$

Step 2.2.2.2

Multiply $\frac{1}{u}$ by $\frac{1}{20}$ .

$\int - \frac{1}{u \cdot 20} d u = \int d t$

Step 2.2.2.3

Move $20$ to the left of $u$ .

$\int - \frac{1}{20 u} d u = \int d t$

Step 2.2.3

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

$- \int \frac{1}{20 u} d u = \int d t$

Step 2.2.4

Since $\frac{1}{20}$ is constant with respect to $u$ , move $\frac{1}{20}$ out of the integral.

$- (\frac{1}{20} \int \frac{1}{u} d u) = \int d t$

Step 2.2.5

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

$- \frac{1}{20} (\ln (| u |) + C_{1}) = \int d t$

Step 2.2.6

Simplify.

$- \frac{1}{20} \ln (| u |) + C_{1} = \int d t$

Step 2.2.7

Replace all occurrences of $u$ with $- 20 p + 240$ .

$- \frac{1}{20} \ln (| - 20 p + 240 |) + C_{1} = \int d t$

Step 2.3

Apply the constant rule.

$- \frac{1}{20} \ln (| - 20 p + 240 |) + C_{1} = t + C_{2}$

Step 2.4

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

$- \frac{1}{20} \ln (| - 20 p + 240 |) = t + C$

Step 3

Solve for $p$ .

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Step 3.1

Multiply both sides of the equation by $- 20$ .

$- 20 (- \frac{1}{20} \ln (| - 20 p + 240 |)) = - 20 (t + C)$

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 $- 20 (- \frac{1}{20} \ln (| - 20 p + 240 |))$ .

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Step 3.2.1.1.1

Combine $\ln (| - 20 p + 240 |)$ and $\frac{1}{20}$ .

$- 20 (- \frac{\ln (| - 20 p + 240 |)}{20}) = - 20 (t + C)$

Step 3.2.1.1.2

Cancel the common factor of $20$ .

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Step 3.2.1.1.2.1

Move the leading negative in $- \frac{\ln (| - 20 p + 240 |)}{20}$ into the numerator.

$- 20 \frac{- \ln (| - 20 p + 240 |)}{20} = - 20 (t + C)$

Step 3.2.1.1.2.2

Factor $20$ out of $- 20$ .

$20 (- 1) \frac{- \ln (| - 20 p + 240 |)}{20} = - 20 (t + C)$

Step 3.2.1.1.2.3

Cancel the common factor.

$20 \cdot - 1 \frac{- \ln (| - 20 p + 240 |)}{20} = - 20 (t + C)$

Step 3.2.1.1.2.4

Rewrite the expression.

$- - \ln (| - 20 p + 240 |) = - 20 (t + C)$

Step 3.2.1.1.3

Multiply.

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Step 3.2.1.1.3.1

Multiply $- 1$ by $- 1$ .

$1 \ln (| - 20 p + 240 |) = - 20 (t + C)$

Step 3.2.1.1.3.2

Multiply $\ln (| - 20 p + 240 |)$ by $1$ .

$\ln (| - 20 p + 240 |) = - 20 (t + C)$

Step 3.2.2

Simplify the right side.

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Step 3.2.2.1

Apply the distributive property.

$\ln (| - 20 p + 240 |) = - 20 t - 20 C$

Step 3.3

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

$e^{\ln (| - 20 p + 240 |)} = e^{- 20 t - 20 C}$

Step 3.4

Rewrite $\ln (| - 20 p + 240 |) = - 20 t - 20 C$ 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 $b^{y} = x$ .

$e^{- 20 t - 20 C} = | - 20 p + 240 |$

Step 3.5

Solve for $p$ .

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Step 3.5.1

Rewrite the equation as $| - 20 p + 240 | = e^{- 20 t - 20 C}$ .

$| - 20 p + 240 | = e^{- 20 t - 20 C}$

Step 3.5.2

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

$- 20 p + 240 = \pm e^{- 20 t - 20 C}$

Step 3.5.3

Subtract $240$ from both sides of the equation.

$- 20 p = \pm e^{- 20 t - 20 C} - 240$

Step 3.5.4

Divide each term in $- 20 p = \pm e^{- 20 t - 20 C} - 240$ by $- 20$ and simplify.

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Step 3.5.4.1

Divide each term in $- 20 p = \pm e^{- 20 t - 20 C} - 240$ by $- 20$ .

$\frac{- 20 p}{- 20} = \frac{\pm e^{- 20 t - 20 C}}{- 20} + \frac{- 240}{- 20}$

Step 3.5.4.2

Simplify the left side.

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Step 3.5.4.2.1

Cancel the common factor of $- 20$ .

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Step 3.5.4.2.1.1

Cancel the common factor.

$\frac{- 20 p}{- 20} = \frac{\pm e^{- 20 t - 20 C}}{- 20} + \frac{- 240}{- 20}$

Step 3.5.4.2.1.2

Divide $p$ by $1$ .

$p = \frac{\pm e^{- 20 t - 20 C}}{- 20} + \frac{- 240}{- 20}$

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 $\frac{\pm e^{- 20 t - 20 C}}{- 20}$ .

$p = \frac{\pm e^{- 20 t - 20 C}}{20} + \frac{- 240}{- 20}$

Step 3.5.4.3.1.2

Divide $- 240$ by $- 20$ .

$p = \frac{\pm e^{- 20 t - 20 C}}{20} + 12$

Step 3.5.4.3.2

To write $12$ as a fraction with a common denominator, multiply by $\frac{20}{20}$ .

$p = \frac{\pm e^{- 20 t - 20 C}}{20} + 12 \cdot \frac{20}{20}$

Step 3.5.4.3.3

Combine $12$ and $\frac{20}{20}$ .

$p = \frac{\pm e^{- 20 t - 20 C}}{20} + \frac{12 \cdot 20}{20}$

Step 3.5.4.3.4

Combine the numerators over the common denominator.

$p = \frac{\pm e^{- 20 t - 20 C} + 12 \cdot 20}{20}$

Step 3.5.4.3.5

Multiply $12$ by $20$ .

$p = \frac{\pm e^{- 20 t - 20 C} + 240}{20}$

Step 4

Group the constant terms together.

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Step 4.1

Simplify the constant of integration.

$p = \frac{\pm e^{- 20 t + C} + 240}{20}$

Step 4.2

Rewrite $e^{- 20 t + C}$ as $e^{- 20 t} e^{C}$ .

$p = \frac{\pm e^{- 20 t} e^{C} + 240}{20}$

Step 4.3

Reorder $e^{- 20 t}$ and $e^{C}$ .

$p = \frac{\pm e^{C} e^{- 20 t} + 240}{20}$

Step 4.4

Combine constants with the plus or minus.

$p = \frac{C e^{- 20 t} + 240}{20}$