Mathematics, 03.04.2020 05:33, darg3990rgp2t0r2

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

Answers: 2

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Answers

answered: rbines

C. $\displaystyle P = 10 - 6e^{-t}$

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets Parenthesis Exponents Multiplication Division Addition Subtraction Left to Right

Equality Properties

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

Algebra I

FunctionsFunction NotationExponential Rule [Multiplying]: $\displaystyle b^m \cdot b^n = b^{m + n}$

Algebra II

Log PropertiesNatural log ln(x) and eˣ

Calculus

Derivatives

Derivative Notation

Basic Power Rule:

f(x) = cxⁿ f’(x) = c·nxⁿ⁻¹

Slope Fields

Solving differentialsSeparation 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.

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

Step 3: Find General Solution Pt. 1

[Equality Property] Integrate both sides: $\displaystyle \int {\frac{1}{10 - P}} \, dP = \int \, dt$ [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.

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

Step 5: Find General Solution Pt. 2

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

This is our general solution to the differential equation.

Step 6: Find Particular Solution

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

answered: Guest

the answer is c hope it

step-by-step explanation:

answered: Guest

i think the correct option is the second option.

Which function is shown on the graph below

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Which of the following is the solution to the differential equation dP/dt+P=10 with the initial cond...