/*! 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 16 Solve the following initial valu... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Solve the following initial value problems. $$z^{\prime}(t)+\frac{z}{2}=6, z(-1)=0$$

Short Answer

Expert verified
Question: Solve the initial value problem (IVP) for z(t) given the ordinary differential equation (ODE) $$z^{\prime}(t) + \frac{z}{2} = 6$$ with the initial condition z(-1) = 0. Answer: The solution to the given initial value problem is $$z(t) = 12 - 12e^{\frac{-1}{2}}e^{-\frac{t}{2}}$$.

Step by step solution

01

Identify the equation type and coefficients

We have the first-order linear ODE $$z^{\prime}(t) + \frac{z}{2} = 6$$ The coefficient of z(t) is \(\frac{1}{2}\), and the constant term is 6.
02

Find the integrating factor

The integrating factor (IF) is given by the formula $$IF = e^{\int \frac{1}{2}dt}$$ Calculate the integral: $$IF = e^{\frac{t}{2}}$$
03

Multiply the equation by the integrating factor

Multiply both sides of the equation by the integrating factor: $$e^{\frac{t}{2}} z^{\prime}(t) + \frac{1}{2} e^{\frac{t}{2}} z(t) = 6e^{\frac{t}{2}}$$
04

Identify the left side as a derivative

Observe that the left side is the derivative of the product of the integrating factor and the unknown function: $$(e^{\frac{t}{2}} z(t))^{\prime} = 6e^{\frac{t}{2}}$$
05

Integrate both sides

Integrate both sides of the equation with respect to t: $$\int (e^{\frac{t}{2}} z(t))^{\prime} dt = \int 6e^{\frac{t}{2}} dt$$ Which gives us: $$e^{\frac{t}{2}} z(t) = 12e^{\frac{t}{2}} + C$$
06

Solve for z(t)

Solve for z(t) by dividing both sides by the integrating factor: $$z(t) = 12 + Ce^{-\frac{t}{2}}$$
07

Apply the initial condition

Apply the initial condition z(-1) = 0 to find the value of C: $$0 = 12 + Ce^{\frac{1}{2}}$$ $$C = -12e^{\frac{-1}{2}}$$
08

Write the final solution

Finally, substitute the value of C back into the expression for z(t) to obtain the specific solution: $$z(t) = 12 - 12e^{\frac{-1}{2}}e^{-\frac{t}{2}}$$

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 ODE
A first-order linear ordinary differential equation (ODE) has the general form \( y' + p(x)y = q(x) \) where \( y \) is the unknown function of \( x \) that we’re trying to solve for, \( y' \) is the derivative of \( y \) with respect to \( x \) also known as the rate of change of \( y \), \( p(x) \) is a known function, and \( q(x) \) is another known function.

The solution procedure typically involves finding an integrating factor, which is a function used to multiply both sides of the equation to aid in solving the ODE. The integrating factor generally transforms the left side of the equation into the derivative of a product. This simplifies the equation, allowing us to integrate and find \( y(x) \) explicitly. In the exercise provided, we are dealing with a first-order linear ODE, where \( p(t) = 1/2 \) and \( q(t) = 6 \).
Integrating Factor
The integrating factor is a powerful method employed to solve first-order linear ODEs. It is denoted as \( IF \) and is calculated using the coefficient of \( y \) from the ODE. The formula for the integrating factor is \( IF = e^{\int p(x)dx} \).

For our exercise, the integrating factor is determined from the coefficient \( \frac{1}{2} \) as \( IF = e^{\int \frac{1}{2} dt} = e^{\frac{t}{2}} \). When we multiply the entire ODE by this \( IF \) we make the left side of the differential equation a perfect derivative of the product of \( IF \) and the unknown function \( z(t) \), which tremendously simplifies the process of solving the equation.

Using the integrating factor transforms the problem into one that can be solved by direct integration, highlighting its usefulness in finding solutions to ODEs.
Separation of Variables
Separation of variables is yet another technique to solve differential equations, particularly useful when dealing with equations that can be factored into functions of \( y \) and functions of \( x \) independently. The idea is to rearrange the equation so that each variable and its derivatives are on opposite sides of the equation, and then integrate both sides.

This method, however, is only applicable when the differential equation can be manipulated into a separable form. While our given exercise does not specifically require this method, understanding separation of variables is crucial, as it is one of the simplest methods to solve ODEs when possible. In scenarios where equations are not readily separable or are non-linear, integration factors or other methods would need to be employed instead.

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

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?

Logistic equation for an epidemic When an infected person is introduced into a closed and otherwise healthy community, the number of people who contract the disease (in the absence of any intervention) may be modeled by the logistic equation $$\frac{d P}{d t}=k P\left(1-\frac{P}{A}\right), P(0)=P_{0}$$ where \(k\) is a positive infection rate, \(A\) is the number of people in the community, and \(P_{0}\) is the number of infected people at \(t=0\) The model also assumes no recovery. a. Find the solution of the initial value problem, for \(t \geq 0\), in terms of \(k, A,\) and \(P_{0}.\) b. Graph the solution in the case that \(k=0.025, A=300,\) and \(P_{0}=1.\) c. For a fixed value of \(k\) and \(A\), describe the long-term behavior of the solutions, for any \(P_{0}\) with \(0

Find the equilibrium solution of the following equations, make a sketch of the direction field, for \(t \geq 0,\) and determine whether the equilibrium solution is stable. The direction field needs to indicate only whether solutions are increasing or decreasing on either side of the equilibrium solution. $$y^{\prime}(t)-\frac{y}{4}-1=0$$

What is a carrying capacity? Mathematically, how does it appear on the graph of a population function?

Solve the following initial value problems and leave the solution in implicit form. Use graphing software to plot the solution. If the implicit solution describes more than one function, be sure to indicate which function corresponds to the solution of the initial value problem. $$z^{\prime}(x)=\frac{z^{2}+4}{x^{2}+16}, z(4)=2$$

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.