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anonymous
 5 years ago
Could someone tell me the steps of solving this 2nd order differential equation?
(x^2)y''  x y' + y = 8(x^3)
Is it Euler's equations for the left side and method of undertermined coeffficients for the right side? Thanks!!!
anonymous
 5 years ago
Could someone tell me the steps of solving this 2nd order differential equation? (x^2)y''  x y' + y = 8(x^3) Is it Euler's equations for the left side and method of undertermined coeffficients for the right side? Thanks!!!

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the A.E. is D^2  D +1 =0 D= 1+ SQRT(3) i/ 2 yc= e^1/2 (C1 cos sqrt(3)/2 + C2 sin sqrt(3)/2)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0particular solution = 1/ (D^2  D +1) * 8(x^3)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sry an error in typing. yc= e^1/2 (C1 cos sqrt(3)/2 x + C2 sin sqrt(3)/2 x)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ah again yc= e^1/2 x (C1 cos sqrt(3)/2 x + C2 sin sqrt(3)/2 x)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0why did you write: D^2  D +1 =0 at the very beginning? Did you divide both sides by x squared?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0this is auxiliary equation

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yeah but you can't just ignore the x squared right? i know the quadratic formula but i am not sure if this is the case here

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0put z=logx and xD=theta , x^2D^2= theta (theta 1)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Your way may work but I don't think I've seen it in class...it's gotta be easier than we think

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you've been super helpful

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no prob... i wish i cud help u a lot more.. :(
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