/*! 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} Q37E Find the functionf(t) of the fo... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Find the functionf(t)of the form f(t)=ae3t+be2tsuch that f(0)=1andf'(0)=4.

Short Answer

Expert verified

The function of the form f(t)=ae3t+be2tsuch that f(0)=1and f'(0)=4is f(t)=2e3t-e2t.

Step by step solution

01

Consider the points and substitute these in the standard equation

Consider the function f(t)=ae3t+be2t. Putf(0)=1 inf(t)=ae3t+be2t

f0=ae30+be201=a+b

02

Consider the derivative of the polynomial

Take the derivative of the polynomialf(t)=ae3t+be2t and putf'(0)=4 into its derivative.

f'=t=3ae3t+2be2tf'0=3ae3×0+2be2×04=3a+2b

03

Solve the above equations.

Arrange the equations obtained in above steps into the matrix form.

1132ab=14

Upon solving the values ofa,b are obtained asa=2,b=-1

Substitute these values in the polynomial equation.

role="math" localid="1659340662448" f(t)=ae3t+be2tf(t)=2e3t-1e2t

The functionft of the formf(t)=ae3t+be2t such thatf(0)=1 andf'(0)=4 is f(t)=2e3t-e2t.

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!

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

Consider a solutionx1→of the linear systemAx→=b→. Justify the facts stated in parts (a) and (b):

a. Ifx→his a solution of the systemAx→=0→, thenx1→+xh→ is a solution of the systemA=x→=b→.

b. Ifx2→is another solution of the systemAx→=b→, thenx1→+xh→is a solution of the system Ax→+0→.

c. Now suppose A is a2×2matrix. A solution vectorx1→of the systemAx→+b→is shown in the accompanying figure. We are told that the solutions of the systemAx→=0→form the line shown in the sketch. Draw the line consisting of all solutions of the systemAx→=b→.

If you are puzzled by the generality of this problem, think about an example first:

A=(1 â¶Ä…â¶Ä…â¶Ä…23 â¶Ä…â¶Ä…â¶Ä…6),b→=[39]andx1→=[11]

Compute the products Ax in Exercises 16 through 19

using paper and pencil (if the products are defined).

17.


[123456].[78]

Kyle is getting some flowers for Olivia, his Valentine. Being of a precise analytical mind, he plans to spend exactly \(24 on a bunch of exactly two dozen flowers. At the flower market they have lilies (\)3 each), roses (\(2 each), and daisies (\)0.50 each). Kyle knows that Olivia loves lilies; what is he to do?

Use the concept of a continuous dynamical system.Solve the differential equationdxdt=−kx. Solvethe system dx→dt=Ax→whenAis diagonalizable overR,and sketch the phase portrait for 2×2 matricesA.

Solve the initial value problems posed in Exercises 1through 5. Graph the solution.

4 .dydt=0.8twithy(0)=-0.8

in exercises 1 through 10, find all solutions of the linear systems using elimination.Then check your solutions.

3.2x+4y=33x+6y=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.