/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 35 The differential equation \(d Q ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The differential equation \(d Q / d t=-0.15 Q+25\) represents the quantity of a drug in the body if the drug is metabolized at a continuous rate of \(15 \%\) per day and an IV line is delivering the drug at a constant rate of \(25 \mathrm{mg}\) per hour.

Short Answer

Expert verified
The solution to the differential equation is \( Q(t) = -166.67 + Ce^{0.15t} \), where \( C \) is determined by initial conditions.

Step by step solution

01

Understanding the Problem

We have a differential equation \( \frac{dQ}{dt} = -0.15Q + 25 \) which describes the rate of change of the quantity \( Q \) of a drug in the body. The drug is metabolized at a rate of 15% per day, and an IV line is delivering it at a rate of 25 mg per hour.
02

Setting Up the Equation

The equation \( \frac{dQ}{dt} = -0.15Q + 25 \) is a first-order linear differential equation. Our goal is to solve it to find \( Q(t) \), the quantity of the drug at any time \( t \).
03

Identifying the Type of Differential Equation

Recognize that this equation is of the form \( \frac{dQ}{dt} = aQ + b \) where \( a = -0.15 \) and \( b = 25 \). It's a first-order linear non-homogeneous differential equation.
04

Finding the Integrating Factor

To solve this, we can use an integrating factor. The integrating factor \( \mu(t) \) is given by \( e^{\int -0.15 \, dt} = e^{-0.15t} \).
05

Multiplying the Equation by the Integrating Factor

Multiply the entire differential equation by the integrating factor \( e^{-0.15t} \): \( e^{-0.15t} \frac{dQ}{dt} - 0.15e^{-0.15t}Q = 25e^{-0.15t} \).
06

Rewriting the Left-Hand Side

Notice that the left-hand side of the equation can be rewritten as the derivative of a product: \( \frac{d}{dt} (e^{-0.15t} Q) = 25e^{-0.15t} \).
07

Integrating Both Sides

Integrate both sides with respect to \( t \): \( e^{-0.15t} Q = \int 25e^{-0.15t} \, dt \).
08

Solving the Integral

The integral \( \int 25e^{-0.15t} \, dt \) is solved as \( \frac{-25}{0.15} e^{-0.15t} + C \), where \( C \) is the constant of integration.
09

Expressing \( Q(t) \)

Multiply through by \( e^{0.15t} \) to solve for \( Q(t) \): \( Q(t) = \frac{-25}{0.15} + Ce^{0.15t} \).
10

Simplifying the Expression

Notice that \( \frac{-25}{0.15} = -\frac{2500}{15} = -\frac{500}{3} \). Thus, \( Q(t) = -\frac{500}{3} + Ce^{0.15t} \).
11

Finalizing the Solution

The final solution is \( Q(t) = -166.67 + Ce^{0.15t} \), where \( C \) is determined by any initial condition provided.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

First-order Linear Differential Equations
In the realm of mathematics and physics, differential equations play an essential role in modeling real-world situations. A first-order linear differential equation, like the one in our original exercise, is of the form \( \frac{dy}{dt} = ay + b \). Here, \( y \) is the function we want to find, \( a \) and \( b \) are constants, and \( t \) represents time.
  • This type of equation is termed "first-order" because it involves the first derivative of the unknown function.
  • It's "linear" because it is linear in terms of \( y \) and its derivatives.
Recognizing this form allows us to apply a systematic method for finding the solution. In the exercise's equation \( \frac{dQ}{dt} = -0.15Q + 25 \), \( a = -0.15 \) and \( b = 25 \). Every term is either a constant or involves \( Q \) raised only to the first power, enforcing its linearity. This property of linearity makes these equations solvable using an integrating factor, a powerful tool we'll explore next.
Integrating Factor
The integrating factor is a technique used to simplify and solve first-order linear differential equations. When faced with \( \frac{dy}{dt} = ay + b \), the integrating factor \( \mu(t) \) is chosen to make the equation easier to solve.

To find the integrating factor, we use the expression \( e^{\int a \, dt} \):
  • In our example, this becomes \( e^{\int -0.15 \, dt} = e^{-0.15t} \).
This factor is then multiplied throughout the equation to transform it into one where the left-hand side is the derivative of a product of \( y \) and the integrating factor. This transformation allows us to rewrite the equation neatly and integrate both sides easily. You apply this step to transform the equation from a differential problem into a straightforward algebraic solution. It's a clever method that unlocks the door to solutions that might otherwise seem complex.
Non-homogeneous Equations
The differential equation \( \frac{dy}{dt} = ay + b \) is known as a non-homogeneous equation. This term differentiates it from homogeneous equations, where the right side would be zero.
  • In non-homogeneous equations, like our example \( \frac{dQ}{dt} = -0.15Q + 25 \), the constant addition of \( b \) represents an external influence or force.
  • Here, \( 25 \) signifies the constant rate of drug delivery.
Non-homogeneous equations model systems under the influence of outside forces or inputs, making them highly applicable to many real-world scenarios, such as circuits, populations, or even drug concentration in the bloodstream as seen here.

Such equations reveal how the system, represented by the left side, reacts over time due to these persistent external factors. Understanding this concept is crucial, as it helps in distinguishing the behavior resulting from the system's inherent dynamics versus external influences.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Analyze the phase plane of the differential equations for \(x, y \geq 0 .\) Show the nullclines and equilibrium points, and sketch the direction of the trajectories in each region. $$\begin{aligned}&\frac{d x}{d t}=x\left(1-y-\frac{x}{3}\right)\\\&\frac{d y}{d t}=y\left(1-\frac{y}{2}-x\right)\end{aligned}$$

Consider the system of differential equations $$ \frac{d x}{d t}=-y \quad \frac{d y}{d t}=-x $$ (a) Convert this system to a second order differential equation in \(y\) by differentiating the second equation with respect to \(t\) and substituting for \(x\) from the first equation. (b) Solve the equation you obtained for \(y\) as a function of \(t ;\) hence find \(x\) as a function of \(t\).

Find the general solution to the given differential equation. $$z^{\prime \prime}+2 z^{\prime}=0$$

Solve the boundary value problem. $$y^{\prime \prime}+5 y^{\prime}+6 y=0, \quad y(0)=1, y(1)=0$$

We analyze world oil production. \(^{25}\) When annual world oil production peaks and starts to decline. major economic restructuring will be needed. We investigate when this slowdown is projected to occur. We define \(P\) to be the total oil production worldwide since 1859 in billions of barrels. In \(1993,\) annual world oil production was 22.0 billion barrels and the total production was \(P=724\) billion barrels. In 2013 , annual production was 27.8 billion barrels and the total production was \(P=1235\) billion barrels. Let \(t\) be time in years since 1993 (a) Estimate the rate of production, \(d P / d t,\) for 1993 and 2013 (b) Estimate the relative growth rate, \((1 / P)(d P / d \imath)\) for 1993 and 2013 (c) Find an equation for the relative growth rate. \((1 / P)(d P / d t),\) as a function of \(P,\) assuming that the function is linear. (d) Assuming that \(P\) increases logistically and that all oil in the ground will ultimately be extracted, estimate the world oil reserves in 1859 to the nearest billion barrels. (e) Write and solve the logistic differential equation modeling \(P\)

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.