/*! 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 37 Solve the differential equations... [FREE SOLUTION] | 91Ó°ÊÓ

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Solve the differential equations in Problems \(34-43 .\) Assume \(a, b,\) and \(k\) are nonzero constants. $$\frac{d Q}{d t}=b-Q$$

Short Answer

Expert verified
The solution is \( Q = b + Ce^{-t} \).

Step by step solution

01

Identify the Type of Differential Equation

The given differential equation is \( \frac{d Q}{d t} = b - Q \). This is a first-order linear differential equation since it involves the first derivative of \( Q \) and no higher derivatives.
02

Rearrange into Standard Form

Rearrange the equation into the standard linear form \( \frac{d Q}{d t} + Q = b \). This helps in identifying the integrating factor for solving the equation.
03

Find the Integrating Factor

The integrating factor \( \mu(t) \) for the equation \( \frac{d Q}{d t} + Q = b \) is \( e^{\int 1 \, dt} = e^t \). Use this factor to simplify the solution process.
04

Multiply by Integrating Factor

Multiply both sides of the equation by the integrating factor: \( e^t \frac{d Q}{d t} + e^t Q = b e^t \). This transforms the left side into the derivative of a product: \( \frac{d}{dt}(e^t Q) = b e^t \).
05

Integrate Both Sides

Integrate both sides with respect to \( t \). The left side becomes \( e^t Q \) and the right side becomes \( b \int e^t \, dt = b e^t + C \), where \( C \) is the constant of integration.
06

Solve for \( Q \)

Divide both sides by \( e^t \) to isolate \( Q \): \( Q = b + Ce^{-t} \). This is the general solution to the differential equation.

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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 is an equation that expresses the rate of change of a variable, dependent on itself and possibly other variables. It’s called "first-order" because it involves the first derivative of the unknown function. In our example, we have the equation \( \frac{d Q}{d t} = b - Q \), which is linear in the dependent variable \( Q \).

Feeling daunted by that? Don't be! The linear part simply means that we don't have any squares or higher powers of \( Q \) (or its derivatives) involved. This is why first-order linear differential equations are among the simplest and most approachable.

Understanding the structure helps us to know which method to use for solving it. Here, the linear form is rearranged to \( \frac{d Q}{d t} + Q = b \), which is crucial as it sets the stage for further steps like finding the right integrating factor.
Integrating Factor
The integrating factor is a tool to simplify a first-order linear differential equation, making it much easier to integrate and solve. Think of it as a magic multiplie that helps unravel the mystery of these equations. The idea is to find a function \( \mu(t) \), which when multiplied to the differential equation, allows us to write the equation as the derivative of a product.

In our example, the integrating factor is derived from the linear standard form: \( \frac{d Q}{d t} + Q = b \). We calculate \( \mu(t) = e^{\int 1 \, dt} = e^t \). By multiplying the entire differential equation by this integrating factor \( e^t \), you get an equation that’s easier to integrate across.

This integrating factor essentially converts a fragmentary structure into a cohesive product, allowing us to smoothly transition to the next step of integration.
General Solution
The general solution of a differential equation represents a family of all possible solutions that include arbitrary constants. For our example, the solution is given as \( Q = b + Ce^{-t} \). Each different value of the constant \( C \) represents a unique solution from the family if given specific initial conditions.

What does this mean in simple terms?
  • \( b \) is a constant part derived from the non-variable part of the equation.
  • The expression \( Ce^{-t} \) represents terms that can adjust depending on boundary conditions like initial values.
So, whenever you encounter a problem asking for the "general solution", it's all about expressing the solution in such a form that every solution is implicitly included. This prepares the equation to be tailored more closely to specific real-world situations.
Separation of Variables
Separation of variables wasn't directly used in solving this particular example, but it's a fundamental technique in dealing with differential equations. This method can sometimes be confused with linear methods, but it’s distinct. It involves rewriting a differential equation in a form where all terms depending only on the dependent variable are on one side, and all terms depending on the independent variable are on the other.

The chief idea is to isolate and play with the expressions, giving each variable its own side of the equation. Integrating both sides separately transports you to the general solution.

While our example utilized the integrating factor method, understanding separation of variables broadens your toolkit. Just remember: when the derivative involves terms that multiply each other, consider separation of variables as a possible strategy.

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Most popular questions from this chapter

Table 11.7 gives values for a logistic function \(P=f(t)\) (a) Estimate the maximum rate of change of \(P\) and estimate the value of \(t\) when it occurs. (b) If \(P\) represents the growth of a population, estimate the carrying capacity of the population. $$\begin{array}{c|c|c|c|c|c|c|c|c}\hline t & 0 & 10 & 20 & 30 & 40 & 50 & 60 & 70 \\ \hline P & 120 & 125 & 135 & 155 & 195 & 270 & 345 & 385 \\\\\hline\end{array}$$

Give an example of: A logistic differential equation for a quantity \(P\) such that the maximum rate of change of \(P\) occurs when \(P=75\)

Give the solution to the logistic differential equation with initial condition. $$\frac{d P}{d t}=0.8 P\left(1-\frac{P}{8500}\right) \text { with } P_{0}=500$$

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