d'u d'u du - +2 dt = 3t² – 2e dt 3 dt +

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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Question

Consider a linear differential equation with t ≥ 0 given in photo 1 and an initial-value problem given under initial conditions given in photo 2.

question:

Solve the same initial-value problem using the Laplace transform.

d’u
d'u du
: = 3t² – 2e
+2-
+
dt
dt
dt
Transcribed Image Text:d’u d'u du : = 3t² – 2e +2- + dt dt dt
u(0) — и'(0) — и" (0) — 0
Transcribed Image Text:u(0) — и'(0) — и" (0) — 0
Expert Solution
Step 1

Solution of ordinary differential equation with constant coefficients:

Suppose we wish to solve the nth order ordinary differential equation with constant coefficients

dnudtn+a1dn-1udtn-1+...+anu=Ft,......(i)

where a1, ..., an are constants, subject to the initial conditions

u0=k0, u'0=k1, ..., un-10=kn-1.....(ii)

where k0, ..., kn-1 are constants.

On taking the Laplace transform of both sides of (i) and using (ii), we obtain an algebraic equation for the determination of Lut. The required solution is then obtained by finding the inverse Laplace transform of Lut. 

First Shifting Theorem

Statement: If L-1fs=Ft, then L-1fs-a=eatFt=eatL-1fs.

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