Question

Which of the following is the solution to the differential equation dP/dt+P=10 with the initial condition P(0)=4

A. P= -1+sqrt20t+25
B. P=5-e^-t
C. P=10-6e^-t
D. P=10-6e^t

Answer 1

Answer:

C.  $\displaystyle P = 10 - 6e^{-t}$

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

1. Brackets
2. Parenthesis
3. Exponents
4. Multiplication
5. Division
6. Addition
7. Subtraction

• Left to Right  

Equality Properties

• Multiplication Property of Equality
• Division Property of Equality
• Addition Property of Equality
• Subtraction Property of Equality

Algebra I

• Functions
• Function Notation
• Exponential Rule [Multiplying]:                                                                        $\displaystyle b^m \cdot b^n = b^{m + n}$

Algebra II

• Log Properties
• Natural log ln(x) and eˣ

Calculus

Derivatives

Derivative Notation

Basic Power Rule:

• f(x) = cxⁿ
• f’(x) = c·nxⁿ⁻¹

Slope Fields

• Solving differentials
• Separation of Variables

Antiderivatives - Integrals

Integration Constant C

Integration Rule [Reverse Power Rule]:                                                                    $\displaystyle \int {x^n} \, dx = \frac{x^{n + 1}}{n + 1} + C$

Integration Property [Multiplied Constant]:                                                             $\displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx$

U-Substitution

Logarithmic Integration

Step-by-step explanation:

Step 1: Define

$\displaystyle \frac{dP}{dt} + P = 10$

$\displaystyle P(0) = 4$

Step 2: Rewrite

Separation of Variables. Get differential equation to a form where we can integrate both sides.

1. [Subtraction Property of Equality] Isolate  $\displaystyle \frac{dP}{dt}$:                                               $\displaystyle \frac{dP}{dt} = 10 - P$
2. [Multiplication Property of Equality] Multiply dt on both sides:                     $\displaystyle dP = (10 - P) dt$
3. [Division Property of Equality] Isolate P terms together:                               $\displaystyle \frac{1}{10 - P}dP = dt$

Step 3: Find General Solution Pt. 1

1. [Equality Property] Integrate both sides:                                                        $\displaystyle \int {\frac{1}{10 - P}} \, dP = \int \, dt$
2. [Left Integral] Integrate [Integration Rule - Reverse Power Rule]:                 $\displaystyle \int {\frac{1}{10 - P}} \, dP = t + C$

Step 4: Identify Variables for U-Substitution

Set variables for u-sub.

1. Set:                                                                                                                     $\displaystyle u = 10 - P$
2. Differentiate [Basic Power Rule]:                                                                      $\displaystyle \frac{du}{dP} = -1$
3. [Multiplication Property of Equality] Rewrite:                                                  $\displaystyle du = -dP$

Step 5: Find General Solution Pt. 2

1. [Integral] Rewrite [Integration Property - Multiplied Constant]:                     $\displaystyle -\int {\frac{-1}{10 - P}} \, dP = t + C$
2. [Integral] U-Substitution:                                                                                  $\displaystyle -\int {\frac{1}{u}} \, du = t + C$
3. [Integral] Integrate [Logarithmic Integration]:                                                 $\displaystyle -ln|u| = t + C$
4. [Division Property of Equality] Divide -1 on both sides:                                 $\displaystyle ln|u| = -t + C$
5. [Equality Property] e both sides:                                                                     $\displaystyle e^{ln|u|} = e^{-t + C}$
6. Simplify [Exponential Rule - Multiplying]:                                                         $\displaystyle |u| = e^{-t} \cdot e^{C}$
7. Simplify:                                                                                                             $\displaystyle |u| = Ce^{-t}$
8. Back-Substitute:                                                                                               $\displaystyle |10 - P| = Ce^{-t}$
9. Rewrite:                                                                                                             $\displaystyle 10 - P = \pm Ce^{-t}$

This is our general solution to the differential equation.

Step 6: Find Particular Solution

1. Substitute in point [General Solution]:                                                            $\displaystyle 10 - 4 = \pm Ce^{-0}$
2. Simplify:                                                                                                             $\displaystyle 6 = \pm C$
3. Substitute in C [General Solution]:                                                                   $\displaystyle 10 - P = 6e^{-t}$
4. [Subtraction Property of Equality] Subtract 10 on both sides:                       $\displaystyle -P = 6e^{-t} - 10$
5. [Division Property of Equality] Divide -1 on both sides:                                 $\displaystyle P = 10 - 6e^{-t}$

This is our particular solution to the differential equation.

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Slope Fields

Book: College Calculus 10e

Answer 2

Combining the particular and the homogeneous solution, it is found that the solution to the differential equation is:

• C. $P(t) = 10 - 6e^{-t}$

The differential equation is:

$\frac{dP}{dt} + P = 10$

What is the homogeneous solution?

• It's the solution to the following differential equation:

$\frac{dP}{dt} + P = 0$

Applying separation of variables:

$\frac{dP}{dt} = -P$

$\frac{dP}{P} = -dt$

$\int \frac{dP}{P} = -\int dt$

$\ln{P} = -t + K$

$P_h(t) = Ke^{-t}$

What is the particular solution?

• It's the solution considering an input similar to the left side.

Since the left-side is a constant, the input also is, hence:

$P_p(t) = A$

$\frac{dP_p(t)}{dt} = 0$

Then:

$\frac{dP}{dt} + P = 10$

$A = 10$

Hence, the complete solution, combining both, is:

$P(t) = P_h(t) + P_p(t)$

$P(t) = Ke^{-t} + 10$

Since P(0) = 4:

$4 = K + 10$

$K = -6$

Hence, option C is correct.

You can learn more about particular and the homogeneous solutions at https://brainly.com/question/25308859