o solve ith the initial values y(0)=0 and solution get the solution, we assume nor B are solutions of the homogeneous equation this ssumption will work. Substituting y in the differential equation we obtain and B=-1/2. Next, we use the initial conditions and obtain and D=1. A=-1 y=A+Bt C=-1/2 At yFcos A+B particular complementary A=1 y-At+B C=1/2 y'(0)=-1 y(0)=-1 y'(0)=1 y'(0)=2 y"+y' - 2y = 2t Ye= Ce 2t + Det. we first find the . Since neither
o solve ith the initial values y(0)=0 and solution get the solution, we assume nor B are solutions of the homogeneous equation this ssumption will work. Substituting y in the differential equation we obtain and B=-1/2. Next, we use the initial conditions and obtain and D=1. A=-1 y=A+Bt C=-1/2 At yFcos A+B particular complementary A=1 y-At+B C=1/2 y'(0)=-1 y(0)=-1 y'(0)=1 y'(0)=2 y"+y' - 2y = 2t Ye= Ce 2t + Det. we first find the . Since neither
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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Transcribed Image Text:To solve
with the initial values y(0)=0 and
solution
To get the
solution, we assume
nor B are solutions of the homogeneous equation this
assumption will work. Substituting y in the differential equation we obtain
and B=-1/2. Next, we use the initial conditions and obtain
and D=1.
A=-1
y=A+Bt
C=-1/2
At
y=cos A+B
particular
complementary
A=1
y=At+B
C=1/2
y'(0)=-1
y(0)=-1
y'(0)=1
y'(0)=2
y"+y' - 2y = 2t
Ye= Ce-2t + Det.
we first find the
. Since neither
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