d^2f/dx^2+(ax+b)df/dx+cf=const Could someone know an easier way to solute this ode?

reduce to a quadratic and solve for roots by quadratic formula

What do you mean with "reduce to a quadratic " ? Do you mean, I should transform it in dies form d^2f/dx^2+bdf/dx+cf=const-(ax+b)df/dx .

if you allow the coefficients to be the terms for a quadratic, i.e. \[r ^{2}+(ax+b)r+c = 0\] then use the quadratic formula \[(-b \pm \sqrt{b ^{2}-4ac})/2a\] to find r such that it satisfies the ODE

but you can't do that because r is also a function of x. r=r(x)

no you treat it as a constant

what do you have so far for your work and/or proof?

I though about another approach, please, have a look on this site, http://www.ltcconline.net/greenl/courses/204/PowerLaplace/seriesSolutions1.htm

did you figure out the u substitution? if you did then it should follow...but i honestly don't think you need to use series for it

I figure it, but it seems to be really complicatet to have a serie solution.

Because of the non-constant coefficients and it become very ugly

can I send you a copy of my work?

Try uploading a pic to http://imgur.com/ and pasting the link to it here.

yeah i would like to see it

(We'll be adding image upload to OpenStudy soon, I hope.)

You could try using laplace transforms. It would turn the the 2nd order ODE into a 1st order ODE. Usually 1st order ODE are easier to solve. Look at the example here http://tutorial.math.lamar.edu/Classes/DE/IVPWithNonConstantCoefficient.aspx

i think that's what he was trying to do

here is a copy http://imgur.com/9y7hu

hope u can read it,

ok yeah now it is better to see... i think that it works, after simplifying then you can solve the ODE. but i think its already simplified...

but when it is already simplified, what is the result now?

I'm gonna use matlab to solve,

you can also check your answer on this website: http://www.wolframalpha.com/input/?i=y''+%2B+(ax%2Bb)y'%2Bcy+%3D+f The answer seems a bit complicated :\

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