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Chapter 9: Problem 5

Find solutions to the differential equations in subject to the given initial condition. $$\frac{d p}{d q}=-0.1 p, \quad p=100 \text { when } q=5$$

Short Answer

Expert verified
The solution is \( p(q) = 100e^{0.5 - 0.1q} \).

Step by step solution

01

Identify the Type of Differential Equation

The given equation \( \frac{d p}{d q} = -0.1 p \) is a first-order linear differential equation. It suggests that the rate of change of \( p \) with respect to \( q \) is proportional to \( p \) itself.
02

Write the General Solution Equation

The general form for a first-order linear differential equation \( \frac{d p}{d q} = -k p \) is \( p(q) = Ce^{-kq} \). In this case, \( k = 0.1 \), so the general solution is \( p(q) = Ce^{-0.1q} \).
03

Apply the Initial Condition

An initial condition is given: \( p = 100 \) when \( q = 5 \). Substitute \( p(5) = 100 \) into the general solution: \( 100 = Ce^{-0.1 \times 5} \).
04

Solve for the Constant \( C \)

Solve \( 100 = Ce^{-0.5} \). Thus, \( C = 100e^{0.5} \).
05

Write the Specific Solution

Substitute \( C = 100e^{0.5} \) back into the general solution to get the specific solution: \( p(q) = 100e^{0.5}e^{-0.1q} \). Simplify this expression: \( p(q) = 100e^{0.5 - 0.1q} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

First-order linear differential equation
A first-order linear differential equation involves the derivative of a function and the function itself, combined in a linear manner. In simple terms, this type of equation describes how a quantity changes over time in a way that is proportional to its current value. For the equation \( \frac{d p}{d q} = -0.1 p \), this means that the rate at which \( p \) changes with respect to \( q \) is proportional to \( p \) itself. This is a characteristic feature of exponential growth or decay processes.
When identifying such equations:
  • The differential equation should be of the form \( \frac{dy}{dx} = ky \) or \( \frac{dy}{dx} = ky + b \), where \( k \) and \( b \) are constants.
  • The solution often captures natural processes, such as population growth or radioactive decay, which are inherently exponential in nature.
Understanding these equations helps us model real-world phenomena that exhibit exponential growth or decay characteristics.
Initial condition
An initial condition provides specific values for variables at a particular point, which allows us to find a particular solution to a differential equation. For example, in the problem \( p = 100 \) when \( q = 5 \) is the initial condition. By substituting these values into the general solution, we determine the constant that personalizes the solution to fit this initial scenario.
The process involves:
  • Substituting the values of \( p \) and \( q \) from the initial condition into the general solution equation.
  • Solving for the constant, often represented by \( C \).
This makes the general solution specific by providing a starting reference point. Initial conditions are crucial as they ensure the solution aligns perfectly with given or known data, thus making it unique to the specific problem scenario.
Exponential functions
Exponential functions are mathematical expressions where the variable appears in the exponent. They are fundamental in solving differential equations as seen in the exercise with the equation \( p(q) = Ce^{-0.1q} \).
Key characteristics of exponential functions include:
  • They grow or decay at a rate proportional to their current value, hence the term 'exponential growth' or 'exponential decay'.
  • The general form is \( y = Ce^{kx} \), where \( C \) is a constant and \( k \) determines the rate of growth (if positive) or decay (if negative).
  • In our example, \( p(q) = Ce^{-0.1q} \), signifies a decay process as indicated by the negative exponent.
Exponential functions model numerous real-life processes, such as population growth, radioactive decay, and even compound interest, showcasing their broad applicability in understanding natural and economic phenomena.

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Most popular questions from this chapter

Hydrocodone bitartrate is used as a cough suppressant. After the drug is fully absorbed, the quantity of drug in the body decreases at a rate proportional to the amount left in the body. The half-life of hydrocodone bitartrate in the body is 3.8 hours and the dose is 10 mg. (a) Write a differential equation for the quantity, \(Q,\) of hydrocodone bitartrate in the body at time \(t\), in hours since the drug was fully absorbed. (b) Solve the differential equation given in part (a). (c) Use the half-life to find the constant of proportionality, \(k\) (d) How much of the 10 -mg dose is still in the body after 12 hours?

In some chemical reactions, the rate at which the amount of a substance changes with time is proportional to the amount present. For example, this is the case as \(\delta\) glucono-lactone changes into gluconic acid. (a) Write a differential equation satisfied by \(y,\) the quantity of \(\delta\) -glucono-lactone present at time \(t\) (b) If 100 grams of \(\delta\) -glucono-lactone is reduced to 54.9 grams in one hour, how many grams will remain after 10 hours?

Check that \(y=t^{4}\) is a solution to the differential equation \(t \frac{d y}{d t}=4 y\)

(a) For \(d y / d x=x^{2}-y^{2},\) find the slope at the following points: \((1,0), \quad(0,1), \quad(1,1)\) \((2,1), \quad(1,2), \quad 2\) (2,2) (b) Sketch the slope ficld at these points.

A bank account that initially contains $$ 25,000\( earns interest at a continuous rate of \)4 \%\( per year. Withdrawals are made out of the account at a constant rate of $$ 2000 per year. Write a differential equation for the balance, \)B\( in the account as a function of the number of years, \)t$
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