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t, t<2 y" + 4y = g(t), y(0) = - 2, y'(0) = 0, where g(t) = . 1, t>2

t, t<2 y" + 4y = g(t), y(0) = - 2, y'(0) = 0, where g(t) = . 1, t>2

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Solve for​ Y(s), the Laplace transform of the solution​ y(t) to the initial value problem below.

 

Y(s)=

 

(Answer in terms of e)

t, t<2 y" + 4y = g(t), y(0) = - 2, y'(0) = 0, where g(t) = 1, t>2
Transcribed Image Text:t, t<2 y" + 4y = g(t), y(0) = - 2, y'(0) = 0, where g(t) = 1, t>2
Expert Solution
Step 1

Given the differential equation:

     y''+4y=g(t), y(0)=-2, y'(0)=0 where gt=t,   t<21,  t>2

Find Y(s), the Laplace transform of the solution y(t).

Step 2

Write g(t) in terms of unit step function.

Consider the function 

               ft=f0t,    t<t1f1t,    t≥t1

where f0 and f1 are defined on [0,∞). f(t) can be written in terms of unit step function as:

         ft=f0t+ut-t1f1t-f0t.

Thus, 

        gt=t+ut-2(1-t)      =t+u(t-2)-tu(t-2) 

 

Step 3

The differential equation becomes

   y''+4y=t+u(t-2)-tu(t-2)

Now, take the Laplace transform on both sides.

   Ly''+4y=Lt+ut-2-tut-2s2Ys-sy(0)-y'(0)+4Y(s)=Lt+Lut-2-Ltut-2

 

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