(b) Find the solution of the following initial value problem y" – 4y" – 5y' = 9 + 5x, y(0) = 0, y' = 0 and y"(0) = 4 The general solution is y = Yn + Y, (The notation and symbols as used in the lecture notes) Yh = C1 + cz exp( )+cz exp(5z), where c1, c2 and cz are arbitrary constants The nonhomogeneous solution can be represented in the form Yp = Ap(x) + Bq(x) + Cr(x): where p(x), q(z) and r(x) are polynomial of degree n> 0. Find the values of the arbitrary constants A, B, C. A= B= C = { Express your answer in decimal where applicable} Using initial conditions, find the corresponding values of the arbitrary constants С1, сэ and cз. C1 = { Express your answer in decimal where applicable }. Thus the particular solution of the ordinary differential equation is given by y = (exp( +1))
(b) Find the solution of the following initial value problem y" – 4y" – 5y' = 9 + 5x, y(0) = 0, y' = 0 and y"(0) = 4 The general solution is y = Yn + Y, (The notation and symbols as used in the lecture notes) Yh = C1 + cz exp( )+cz exp(5z), where c1, c2 and cz are arbitrary constants The nonhomogeneous solution can be represented in the form Yp = Ap(x) + Bq(x) + Cr(x): where p(x), q(z) and r(x) are polynomial of degree n> 0. Find the values of the arbitrary constants A, B, C. A= B= C = { Express your answer in decimal where applicable} Using initial conditions, find the corresponding values of the arbitrary constants С1, сэ and cз. C1 = { Express your answer in decimal where applicable }. Thus the particular solution of the ordinary differential equation is given by y = (exp( +1))
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
Transcribed Image Text:(b) Find the solution of the following initial value problem
Finish
y" – 4y" – 5y' = 9+ 5x, y(0) = 0, y' = 0 and y"(0) = 4
Time
The general solution is y = Yr + y, (The notation and symbols as used in the lecture
notes)
Yh = C1 + c2 exp(
)+cz exp(5x), where c1, c2 and cz are arbitrary
constants
The nonhomogeneous solution can be represented in the form
Yp = Ap(x) + Bq(x)+ Cr(x): where p(x), q(x) and r(x) are polynomial of degree
n> 0.
Find the values of the arbitrary constants A, B, C. A =
‚B =
C =
{ Express your answer in decimal where applicable}
Using initial conditions, find the corresponding values of the arbitrary constants
C1, C2 and c3 .
C1 =
C2 =
{ Express your answer in decimal where applicable }.
Thus the particular solution of the ordinary differential equation is given by y = (exp(
+1))
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