## anonymous 5 years ago find solution to following differential equation with initial conditions y''+y'-2y=6e^x-2x y(0)=2, y'(0)=0

1. anonymous

i am trying, give me some time, these problems are not short

2. anonymous

do you have the answer from the text book?

3. anonymous

1st solve y"+y'-2y=0 Charateristic Eq. m^2 + m - 2 = 0 (m+2)(m-1)=0 m=-2, m=1 so $y_c = c_1 e ^{-2x} + c_2 e ^{?}$

4. anonymous

then find $y_p(x)=u_1(x)y_1(x) + u_2(x)y_2(x)$

5. anonymous

then you would need wronskian which i am still working on, lol...

6. anonymous

$w(y_1,y_2)=3e ^{-x}$ if anyone wants to correct any mistakes along the processes please feel free.. thanks

7. anonymous

$u_1=-(7/6)e ^{2x}-(x/3)e ^{2x}$ this problem is really long, i am about to give up. =(

8. anonymous

i got y=4xe^x +x -(e^x)/3 +(7e^-2x)/3 G.S.

9. anonymous

start getting yc then the yp then y= yc+yp

10. anonymous

$u_2 = 2x - (2x/9)e ^{3x} - (2/27)e ^{3x}$

11. anonymous

thats why we ned to get the answer from the book itself...there are probably many solutions on this..

12. anonymous

its not really that long!, im pretty sure we can use the method of undetermined coeffeiients but i cant seem to get a right answer to fit, i think my yp is wrong, variation of paremeters might be unessecary?

13. anonymous

$y_p = [-(7/6)e ^{2x} - (x/3)e ^{2x} ][e ^{-2x}]+[2x-(2x/9)e ^{3x}-(2/27)e ^{3x}][e ^{x}]$

14. anonymous

start w/ yc=c1e^x + c2e^-2x yp=Ae^x +Bxe^x+Cx

15. anonymous

ok then there must be other methods that will make this shorter perhaps laplace? or annihilator?

16. anonymous

i am done, i am happy i got that far til the y_p... haha.... sorry didn't help much!

17. anonymous

yp'=Ae^x +B(xe^x+e^x)+C yp"=Ae^x+B(xe^x+e^x+e^x) now sbstitute this to the orig equation y''+y'-2y=6e^x-2x

18. anonymous

my method will still work, it just going to be long and most likely I already made a mistake

19. anonymous

yes you can do it from LAPLACE and get the same answer

20. anonymous

i could see lots lots lots of algebra involved with laplace!

21. anonymous

after subing back in, i get $3Be^x -2Cx+C=6e^x-2x$ any ideas ? haha

22. anonymous

yes it might take sometimes in computing them

23. anonymous

3B=6 -2C=-2 but C=0?

24. anonymous

i got 2e^x +Be^x+C-2Cx=6e^x-2x

25. anonymous

thats why ive been stuck on it, unless the intial conditions allow c=1

26. anonymous

yes C=1

27. anonymous

so whats A equal to? 0?

28. anonymous

2e^x +Be^x+C-2Cx=6e^x-2x (2+B)e^x +C-2xC=6e^x-2x

29. anonymous

yes A=0

30. anonymous

did you get it nor rzawad?

31. anonymous

2e^x +Be^x+C-2Cx=6e^x-2x (2+B)e^x +C-2xC=6e^x-2x 2+B=6 -> B=4 and -2xC=-2x -->C=1 now subs to Yp

32. anonymous

Yp=4xe^x +x Yc=C1e^x +C2e^-2x Y=Yc+Yp

33. anonymous

gives Y=4xe^x +x+C1e^x+C2e^-2x

34. anonymous

now you can take and subs the initial condition of the equation y(0)=2, y'(0)=0

35. anonymous

so $c_1+c_2 = 2?$

36. anonymous

Y(0)=C1+C2=2 and take the Y'=4(xe^x +e^x) +1 +C1e^x+C2e^-2x take Y'(o)=0=4(1)+1+C1-2C2 0=5+C1-2C2 sub C1=2-C2 0=5+2-C2-2C2 C2=7/3

37. anonymous

C1=2-C2=2-7/3=-1/3 now subs this to your eq

38. anonymous

Y=4xe^x +x+C1e^x+C2e^-2x subst here gives y=4xe^x +x -(e^x)/3 +(7e^-2x)/3 G.S.

39. anonymous

end......

40. anonymous

41. anonymous

rzawad. where did you get this problem from?

42. anonymous

$y[x]=\frac{1}{2}+\frac{3 e^{-2 x}}{2}+x+2 e^x x$ $y\text{''}[x]+y'[x]-2y[x]\text{=}6E{}^{\wedge}x-2x\text{//}\text{Simplify}$ $\left\{6 e^x-2 x\right\}=6 e^x-2 x$

43. anonymous

we worked hour+ for this problem and then look! Mathematica solved the D.E. in minute+ lol...

44. anonymous

wow thats real cool whats the answer from matematica did you ty to solved it from there?

45. anonymous

Even with a senior using Mathtematica it was more like and hour.

46. anonymous

an hour.

47. anonymous

well i just solved this prob in 15 mins lol..and hoping to get the answer from the text book from rzawad..lol

48. anonymous

last sem i had this Advanced engineering math even though we had a DE subject last sem. we had to cover this type of problems

49. anonymous

if anyone here is taking DE now you can email me and i might help you out with your HW etc

50. anonymous

http://www.wolframalpha.com/input/?i=y%27%27%2By%27-2y%3D6e%5Ex-2x+%2C+y%280%29%3D2+%2C+y%27%280%29%3D0 wolfram will give the same answer as robtobey's

51. anonymous

wow i didnt knowthere is wolfram ....lol

52. anonymous

its from a past exam with no solutions dude soz,

53. anonymous

ok hope you got an idea of solving of this type of prob...thnx