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:
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
Section: Chapter Questions
Problem 1RQ
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Question
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.
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)
Step by step
Solved in 3 steps with 3 images
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