## anonymous 5 years ago d^2f/dx^2+(ax+b)df/dx+cf=const Could someone know an easier way to solute this ode?

1. anonymous

2. anonymous

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 .

3. anonymous

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

4. anonymous

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

5. anonymous

no you treat it as a constant

6. anonymous

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

7. anonymous

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

8. anonymous

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

9. anonymous

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

10. anonymous

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

11. anonymous

can I send you a copy of my work?

13. anonymous

yeah i would like to see it

15. anonymous

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

16. anonymous

i think that's what he was trying to do

17. anonymous

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

18. anonymous

19. anonymous

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...

20. anonymous

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

21. anonymous

I'm gonna use matlab to solve,

22. anonymous

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 :\