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Chapter 9: Problem 9

Find the general solution of the following equations. $$u^{\prime}(t)+12 u=15$$

Short Answer

Expert verified
Question: Determine the general solution of the first-order linear ordinary differential equation: \(u^{\prime}(t) + 12 u = 15\). Answer: The general solution is \(u(t) = \frac{5}{4} + \frac{C}{e^{12t}}\).

Step by step solution

01

Identify the integrating factor

To solve the differential equation, let's first identify the integrating factor (IF). The form of the integrating factor for an equation of the form \(u^{\prime}(t) + p(t)u = q(t)\) is \(\mu(t) = e^{\int p(t) dt}\). In our problem, \(p(t) = 12\). Therefore, the integrating factor becomes: $$\mu(t) = e^{\int 12 dt} = e^{12t}$$
02

Multiply the original equation by the integrating factor

Now, we need to multiply both sides of the equation by the integrating factor. This yields: $$e^{12t}[u^{\prime}(t) + 12 u] = e^{12t}(15)$$ After multiplying, we have: $$e^{12t}u^{\prime}(t) + 12 e^{12t}u = 15 e^{12t}$$
03

Identify that the left hand side is a product rule and integrate both sides

Upon inspection, we can notice that the left hand side of the equation is the derivative of the product of the integrating factor and the function \(u\), that is: $$\frac{d}{dt}[e^{12t}u(t)] = 15 e^{12t}$$ Now, we can integrate both sides of this equation with respect to \(t\): $$\int\frac{d}{dt}[e^{12t}u(t)] dt = \int 15 e^{12t} dt$$ This gives us: $$e^{12t}u(t) = \frac{15}{12}e^{12t} + C$$
04

Divide by the integrating factor to find the general equation

Finally, we divide both sides of the equation by the integrating factor \(e^{12t}\): $$u(t) = \frac{15}{12} + \frac{C}{e^{12t}}$$ The general solution to the given differential equation is: $$u(t) = \frac{5}{4} + \frac{C}{e^{12t}}$$-

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Key Concepts

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

Integrating Factor Method
Understanding the integrating factor method is essential when tackling first-order linear differential equations. This technique turns an equation that is not easily solvable into one that is. We look for a function, called an integrating factor, which when multiplied by the differential equation, simplifies it into an exact equation that is much easier to solve.

The process starts with identifying an integrating factor that is typically denoted as \(\mu(t)\) and is derived by taking the exponential of the integral of the coefficient of \(u\) in the original equation, \(e^{\int p(t) dt}\). By multiplying the entire equation by this factor, we leverage the properties of the derivative of a product鈥攕pecifically, the product rule in calculus. Subsequently, we integrate the resulting equation. The goal of this method is to isolate the unknown function and determine its form, which then represents the general solution to the original differential equation.
First-Order Linear Differential Equations
Equations of the form \(u' + p(t)u = q(t)\), where \(u'\) is the derivative of \(u\) with respect to \(t\), \(p(t)\) and \(q(t)\) are functions of \(t\), are known as first-order linear differential equations. These equations are 'first-order' since they involve the first derivative, and 'linear' because the equation involves linear combinations of the function \(u\) and its derivatives.

The integrating factor method is particularly suited for first-order linear differential equations because it linearizes them into a form that can be integrated directly. The solution to these equations is often crucial in various scientific and engineering applications, such as modeling population growth, radioactive decay, or the discharge of a capacitor.
General Solution of Differential Equations
In the context of differential equations, the 'general solution' refers to the solution that contains all possible solutions of the differential equation. It is expressed in terms of constants that can take various values for different specific solutions.

For first-order linear differential equations, the general solution typically encompasses a particular integral that solves the non-homogeneous equation and a complementary function that solves the homogeneous part. The constants in the solution represent the initial conditions that define the specific behavior of the function. Such flexibility is critical since it allows for the adaptation of the solution to specific boundary conditions, thus providing a broad understanding of the equation's dynamics.
Integration in Calculus
Integration is a fundamental concept in calculus, which is essentially the reverse process of differentiation. It is used to compute areas, volumes, central points, and many useful things. But, it is not just for solving geometry problems; it plays a core role in solving differential equations.

Integrating both sides of a differential equation is often necessary to find the general solution. In the integrating factor method, you use integration to determine the integrating factor and then to solve for the unknown function after the equation has been multiplied by this factor. The integral sign \(\int\) is a key indicator of the operation, and the solutions often involve indefinite integrals, which include a constant of integration \(C\). This constant plays a crucial role in formulating the general solution.

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Most popular questions from this chapter

In this section, several models are presented and the solution of the associated differential equation is given. Later in the chapter, we present methods for solving these differential equations. The delivery of a drug (such as an antibiotic) through an intravenous line may be modeled by the differential equation \(m^{\prime}(t)+k m(t)=I,\) where \(m(t)\) is the mass of the drug in the blood at time \(t \geq 0, k\) is a constant that describes the rate at which the drug is absorbed, and \(I\) is the infusion rate. a. Show by substitution that if the initial mass of drug in the blood is zero \((m(0)=0),\) then the solution of the initial value problem is \(m(t)=\frac{I}{k}\left(1-e^{-k t}\right)\) b. Graph the solution for \(I=10 \mathrm{mg} / \mathrm{hr}\) and \(k=0.05 \mathrm{hr}^{-1}\) c. Evaluate \(\lim _{t \rightarrow \infty} m(t),\) the steady-state drug level, and verify the result using the graph in part (b).

Equilibrium solutions \(A\) differential equation of the form \(y^{\prime}(t)=f(y)\) is said to be autonomous (the function \(f\) depends only on y. The constant function \(y=y_{0}\) is an equilibrium solution of the equation provided \(f\left(y_{0}\right)=0\) (because then \(y^{\prime}(t)=0\) and the solution remains constant for all \(t\) ). Note that equilibrium solutions correspond to horizontal lines in the direction field. Note also that for autonomous equations, the direction field is independent of t. Carry out the following analysis on the given equations. a. Find the equilibrium solutions. b. Sketch the direction field, for \(t \geq 0\). c. Sketch the solution curve that corresponds to the initial condition \(y(0)=1\). $$y^{\prime}(t)=y(y-3)(y+2)$$

An endowment is an investment account in which the balance ideally remains constant and withdrawals are made on the interest earned by the account. Such an account may be modeled by the initial value problem \(B^{\prime}(t)=r B-m,\) for \(t \geq 0,\) with \(B(0)=B_{0} .\) The constant \(r>0\) reflects the annual interest rate, \(m>0\) is the annual rate of withdrawal, \(B_{0}\) is the initial balance in the account, and \(t\) is measured in years. a. Solve the initial value problem with \(r=0.05, m=\$ 1000 /\) year, and \(B_{0}=\$ 15,000 .\) Does the balance in the account increase or decrease? b. If \(r=0.05\) and \(B_{0}=\$ 50,000,\) what is the annual withdrawal rate \(m\) that ensures a constant balance in the account? What is the constant balance?

Solve the initial value problem and graph the solution. Let \(r\) be the natural growth rate, \(K\) the carrying capacity, and \(P_{\mathrm{o}}\) the initial population. $$r=0.2, K=300, P_{0}=50$$

U.S. population projections According to the U.S. Census Bureau, the nation's population (to the nearest million) was 296 million in 2005 and 321 million in \(2015 .\) The Bureau also projects a 2050 population of 398 million. To construct a logistic model, both the growth rate and the carrying capacity must be estimated. There are several ways to estimate these parameters. Here is one approach: a. Assume \(t=0\) corresponds to 2005 and that the population growth is exponential for the first ten years; that is, between 2005 and \(2015,\) the population is given by \(P(t)=P(0) e^{\pi}\) Estimate the growth rate \(r\) using this assumption. b. Write the solution of the logistic equation with the value of \(r\) found in part (a). Use the projected value \(P(45)=398\) million to find a value of the carrying capacity \(K\) c. According to the logistic model determined in parts (a) and (b), when will the U.S. population reach \(95 \%\) of its carrying capacity? d. Estimations of this kind must be made and interpreted carefully. Suppose the projected population for 2050 is 410 million rather than 398 million. What is the value of the carrying capacity in this case? e. Repeat part (d) assuming the projected population for 2050 is 380 million rather than 398 million. What is the value of the carrying capacity in this case? E. Comment on the sensitivity of the carrying capacity to the 35-year population projection.
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