5) Solve the differential equation2 u" + 3u – 2u = te-" , where u(0) = 0, u'(0) = -2.%3!

Name: 5) Solve the differential equation2 u" + 3u – 2u = te-" , where u(0)
Uploaded: 2024-03-20T07:06:41.000Z
Channel: Bartleby
Description: 5) Solve the differential equation2 u" + 3u – 2u = te-" , where u(0) = 0, u'(0) = -2.%3!

Answered: 2 u + 3ư - 2u =te , where u(0) = 0,…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

See similar textbooks

Related questions

Concept explainers

Topic Video

Question

5) Solve the differential equation
2 u" + 3u – 2u = te-" , where u(0) = 0, u'(0) = -2.
%3!

Expert Solution

Step 1

The objective is to solve the differential equation $2 u'' + 3 u' - 2 u = t e^{- 2 t}$ where $u (0) = 0, u' (0) = - 2$

Now let's find the solution for the homogenous part,

$2 u'' + 3 u' - 2 u = 0$

Take $m = \frac{d}{d t}$ then,

$2 m^{2} + 3 m - 2 = 0 2 m^{2} - 4 m + m - 2 = 0 2 m (m - 2) + (m - 2) = 0 (2 m + 1) (m - 2) = 0$

So the solution for the homogenous equation is, $y_{c} = c_{1} e^{- t / 2} + c_{2} e^{2 t}$ .

Step by step

Solved in 2 steps

SEE SOLUTION Check out a sample Q&A here

Knowledge Booster

$Discrete Probability Distributions$

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, advanced-math and related others by exploring similar questions and additional content below.

Similar questions