MathSofiya 3 years ago \[y''+2y'+y=xe^{-x}\] \[r^2+2r+1=0\] \[r_1=r_2=1\] \[y_c=c_1e^{-x}+c_2e^{-x}\] for y_p would I have (before I derive it and solve for A) \[y_p=Ae^{-x}\] or \[y_p=Axe^{-x}\] ?

1. lgbasallote

2. MathSofiya

it's -1

3. MathSofiya

typo

4. MathSofiya

5. lgbasallote

it should be \[y_c = c_1 e^{-x} + c_2 x e^{-x}\] see your mistake?

6. MathSofiya

yeah r=-1

7. lgbasallote

i was actually referring to the x by c2

8. lgbasallote

if you have repeated roots you attach least power of x to the next arbitrary constant, remember?

9. MathSofiya

oh I'm sorry I see it now

10. lgbasallote

so do you know what y_p should be now?

11. MathSofiya

what does that have to do with y_p? isn't y_p based on what the RHS of the original equation is?

12. lgbasallote

are you using undetermined coefficients or variation of parameters>

13. lgbasallote

based on your question i presumed you were doing variation....which depends on the complementary solution....

14. MathSofiya

I don't quite know what either of those mean, but I trying to do it based on \[y_p=Ae^{bx}\]

15. MathSofiya

16. MathSofiya

I'm willing to learn =)

17. lgbasallote

can you describe to me the method you're familiar with? does it involve deriving the yp?

18. MathSofiya

according to my book I'm doing it based on THE METHOD OF UNDETERMINED COEFFICIENTS

19. lgbasallote

oh that

20. MathSofiya

yeah lol I guess THE METHOD OF VARIATION OF PARAMETERS is the next section....I haven't started that yet :P

21. lgbasallote

\[y_p = Ae^{bx}\] only works if you don't have repeated roots. for repeated roots, you use \[y_p = Ae^{bx} + Bxe^{bx}\]

22. lgbasallote

because like i said before, if you have repeated roots, you attach the least power of x to the succeeding constant. Since you have xe^-x on the right side, there must be an e^-x before it

23. lgbasallote

does that make sense?

24. MathSofiya

"least power of x to the succeeding constant" meaning \[y_c = c_1 e^{-x} + c_2 x e^{-x}\] referring to the x after the c_2

25. lgbasallote

yes. that. but in this case, we're doing y_p

26. MathSofiya

\[y_p=Ae^{-x}+Bxe^{-x}\]

27. MathSofiya

and then I just do the derivative twice to solve for A and B

28. lgbasallote

sounds right

29. MathSofiya

thank ya!

30. lgbasallote

welcome

31. MathSofiya

What did I do wrong? I get 0=x \[y_p=Ae^{-x}+Bxe^{-x}\] \[y_p'=-Ae^{-x}-Bxe^{-x}(x-1)\] \[y_p''=Ae^{-x}+Be^{-x}(x-2)\]

32. Algebraic!

y ' =-Ae^(-x) -(x-1)Be^(-x) you have an extra x in there..

33. MathSofiya

oh I see

34. MathSofiya

ok I'll try it again

35. MathSofiya

I still get a zero for x....I'll show my work, give me a second to type it :)

36. Algebraic!

Hold up.

37. Algebraic!

I think you need a C*x^2*e-x term as part of your solution...

38. MathSofiya

e-x?

39. MathSofiya

e^(-x)?

40. MathSofiya

Why a \[Cx^2e^{-x}\] ?

41. MathSofiya

I think I'm gonna start a new thread. This one is getting too long

42. Algebraic!

the problem is that your forcing function is the same as one of the solutions to the homogeneous equation.

43. Algebraic!

so without a term that doesn't end up being zero, you're not going to be able to ever get your DE to be 'forced'.

44. Algebraic!

There's a quantitative way to justify it, but I can't recall how it goes. The point is xe^-x is the forcing term and one of the solutions to the homogeneous. So you need another term in your yp if you're ever going to get an actual solution that satisfies the original non-homogeneous equation.

45. Algebraic!

Qualitatively speaking, you need a faster transient, because the forcing term is resonant with the system, so the whole thing is getting driven to x*e^-x harder than any of the terms you're using now can drive it.

46. Algebraic!

Hope that helps; probably not, but I tried:)

47. MathSofiya

Yeah It kinda makes sense.