Consider the equation ky′′ - ky′ − 180y = 0, with k real constant, k ≠ 0. If one solution to the equation is y1 (x) = e6x, then another solution to the linearly independent equation with y1 is:

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
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Consider the equation ky′′ - ky′ − 180y = 0, with k real constant, k ≠ 0. If one solution to the equation is y1 (x) = e6x, then another solution to the linearly independent equation with y1 is:

the answers are in the attached image.

171
a) y2(x) = el7z
b) y2(z) = e-5z
c) y2(z) = ekz
d) y2(x) = e
¯1lz
Transcribed Image Text:171 a) y2(x) = el7z b) y2(z) = e-5z c) y2(z) = ekz d) y2(x) = e ¯1lz
Expert Solution
Step 1

Given equation is, ky"-ky'-180y=0 

And one solution of the equation is y1(x)=e6x

We have to find another solution y2(x) 

 

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