Answes given, y"-sy =1622 let's Solve it step by step 1. Homogeneaus Solution The characteristic equation Corresponding to the homogeneous part of the equation 93: 43-8=0 Solving this equation , we find the boots r = 2, −1+3√3, −1–1√ So the homogenous Solution 73: 2x Y₁(x) = Ge² + Ge* 68 (√3x) + Czetin (13x). 2. particulas Solution the particulas Solution, we'll use for the method of undetermined coefficients. Since a quadrate polynomial. We'll assume 16x2 18 the partedal Solution has = Ax +B+C. the form yp (x)
Answes given, y"-sy =1622 let's Solve it step by step 1. Homogeneaus Solution The characteristic equation Corresponding to the homogeneous part of the equation 93: 43-8=0 Solving this equation , we find the boots r = 2, −1+3√3, −1–1√ So the homogenous Solution 73: 2x Y₁(x) = Ge² + Ge* 68 (√3x) + Czetin (13x). 2. particulas Solution the particulas Solution, we'll use for the method of undetermined coefficients. Since a quadrate polynomial. We'll assume 16x2 18 the partedal Solution has = Ax +B+C. the form yp (x)
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:Answes
given, y"-sy
=1622
let's
Solve
it step by step
1. Homogeneaus Solution
The characteristic equation Corresponding
to the homogeneous part of the equation 93:
43-8=0
Solving this equation
, we
find the boots
r
= 2, −1+3√3, −1–1√
So the homogenous Solution
73:
2x
Y₁(x) = Ge² + Ge* 68 (√3x) + Czetin
(13x).
2.
particulas Solution
the particulas Solution, we'll
use
for
the method of undetermined coefficients. Since
a
quadrate polynomial. We'll assume
16x2
18
the partedal Solution
has
=
Ax +B+C.
the form yp (x)
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