## anonymous 5 years ago how would i solve (x double dot) + x(x dot)+x=0

1. Owlfred

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2. anonymous

help

3. anonymous

I think you can solve it by reduction of order

4. anonymous

how?

5. anonymous

you deleted that? it wasnt the same

6. anonymous

yea, i had it wrong

7. anonymous

that one can be solved differently

8. anonymous

i thought reduction of order was for coefficients which are random polynomials of in t

9. anonymous

$x \prime \prime+xx \prime+x=0$

10. anonymous

for this you say let u=dx/dt than d^2x/dt^2=u du/dt

11. anonymous

than the equation will be u du/dt + xu+x=0

12. anonymous

ahh, that is clever, i will try that

13. anonymous

thanks a ton

14. anonymous

and I guess you can use an integrating factor to solve this

15. anonymous

now I have to go to the shop. If you have a problem just post it

16. anonymous

sweet, thanks a ton you are amazing

17. anonymous

that integral doesn't make sense, it is with respect to the wrong variable

18. anonymous

you have to change u back to x

19. anonymous

I did not solve it, what did you get?

20. anonymous

ah, i'm saying even setting up the integration factor

21. anonymous

dot means that we differentiate with respect to t

22. anonymous

x is a function of t, but the integral should be with respect to t

23. anonymous

yea, i set up $u \prime+ux+x=0$

24. anonymous

u du/dt + xu+x=0 du/dt +u/x=-x

25. anonymous

u'=u du/dt by the chain rule

26. anonymous

e^integral1/x is the integrating factor that is just x

27. anonymous

so multiplying with x gives xdu/dt+u=-x^2

28. anonymous

(xu)dot=-x^2

29. anonymous

xu=-(x^3)/3 +C

30. anonymous

u=-(x^2)/3 +c/x

31. anonymous

u=dx/dt

32. anonymous

dx/dt=-(x^2)/3 +c/x

33. anonymous

so t= -(x^3)/9 +Clnx

34. anonymous

but when you set up that equation du/dt +u/x=-x we should be integrating with respect to t

35. anonymous

you are integrating with respect to x

36. anonymous

for the integrating factor no

37. anonymous

it is only for making it a full derivative

38. anonymous

the integrating factor makes sense

39. anonymous

I think it is like this, but I would not bet on it :-)

40. anonymous

the actual integratioin though does not and why is the equation u du/dt rather than du/dt

41. anonymous

u=dx/dt u dot= x double dot

42. anonymous

this part is not 100% clear for me :) but I am sure that is right and comes from the chain rule

43. anonymous

I just posted a question about it, maybe some1 is wise

44. anonymous

(I'm just a 1st year maths student)

45. anonymous

well, thank you for your help, i will check back later. i really appreciate it andras

46. anonymous