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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

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  1. anonymous
    • 5 years ago
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    i am trying, give me some time, these problems are not short

  2. anonymous
    • 5 years ago
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    do you have the answer from the text book?

  3. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    then find \[y_p(x)=u_1(x)y_1(x) + u_2(x)y_2(x)\]

  5. anonymous
    • 5 years ago
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    then you would need wronskian which i am still working on, lol...

  6. anonymous
    • 5 years ago
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    \[w(y_1,y_2)=3e ^{-x}\] if anyone wants to correct any mistakes along the processes please feel free.. thanks

  7. anonymous
    • 5 years ago
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    \[u_1=-(7/6)e ^{2x}-(x/3)e ^{2x}\] this problem is really long, i am about to give up. =(

  8. anonymous
    • 5 years ago
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    i got y=4xe^x +x -(e^x)/3 +(7e^-2x)/3 G.S.

  9. anonymous
    • 5 years ago
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    start getting yc then the yp then y= yc+yp

  10. anonymous
    • 5 years ago
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    \[u_2 = 2x - (2x/9)e ^{3x} - (2/27)e ^{3x}\]

  11. anonymous
    • 5 years ago
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    thats why we ned to get the answer from the book itself...there are probably many solutions on this..

  12. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    \[y_p = [-(7/6)e ^{2x} - (x/3)e ^{2x} ][e ^{-2x}]+[2x-(2x/9)e ^{3x}-(2/27)e ^{3x}][e ^{x}]\]

  14. anonymous
    • 5 years ago
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    start w/ yc=c1e^x + c2e^-2x yp=Ae^x +Bxe^x+Cx

  15. anonymous
    • 5 years ago
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    ok then there must be other methods that will make this shorter perhaps laplace? or annihilator?

  16. anonymous
    • 5 years ago
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    i am done, i am happy i got that far til the y_p... haha.... sorry didn't help much!

  17. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    my method will still work, it just going to be long and most likely I already made a mistake

  19. anonymous
    • 5 years ago
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    yes you can do it from LAPLACE and get the same answer

  20. anonymous
    • 5 years ago
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    i could see lots lots lots of algebra involved with laplace!

  21. anonymous
    • 5 years ago
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    after subing back in, i get \[3Be^x -2Cx+C=6e^x-2x\] any ideas ? haha

  22. anonymous
    • 5 years ago
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    yes it might take sometimes in computing them

  23. anonymous
    • 5 years ago
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    3B=6 -2C=-2 but C=0?

  24. anonymous
    • 5 years ago
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    i got 2e^x +Be^x+C-2Cx=6e^x-2x

  25. anonymous
    • 5 years ago
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    thats why ive been stuck on it, unless the intial conditions allow c=1

  26. anonymous
    • 5 years ago
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    yes C=1

  27. anonymous
    • 5 years ago
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    so whats A equal to? 0?

  28. anonymous
    • 5 years ago
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    2e^x +Be^x+C-2Cx=6e^x-2x (2+B)e^x +C-2xC=6e^x-2x

  29. anonymous
    • 5 years ago
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    yes A=0

  30. anonymous
    • 5 years ago
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    did you get it nor rzawad?

  31. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    Yp=4xe^x +x Yc=C1e^x +C2e^-2x Y=Yc+Yp

  33. anonymous
    • 5 years ago
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    gives Y=4xe^x +x+C1e^x+C2e^-2x

  34. anonymous
    • 5 years ago
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    now you can take and subs the initial condition of the equation y(0)=2, y'(0)=0

  35. anonymous
    • 5 years ago
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    so \[c_1+c_2 = 2?\]

  36. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    C1=2-C2=2-7/3=-1/3 now subs this to your eq

  38. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    end......

  40. anonymous
    • 5 years ago
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    hope rzawad8 got the same solutions and answer..... lol

  41. anonymous
    • 5 years ago
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    rzawad. where did you get this problem from?

  42. anonymous
    • 5 years ago
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    \[y[x]=\frac{1}{2}+\frac{3 e^{-2 x}}{2}+x+2 e^x x \] \[y\text{''}[x]+y'[x]-2y[x]\text{=}6E{}^{\wedge}x-2x\text{//}\text{Simplify} \] \[\left\{6 e^x-2 x\right\}=6 e^x-2 x \]

  43. anonymous
    • 5 years ago
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    we worked hour+ for this problem and then look! Mathematica solved the D.E. in minute+ lol...

  44. anonymous
    • 5 years ago
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    wow thats real cool whats the answer from matematica did you ty to solved it from there?

  45. anonymous
    • 5 years ago
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    Even with a senior using Mathtematica it was more like and hour.

  46. anonymous
    • 5 years ago
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    an hour.

  47. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    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
    • 5 years ago
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    if anyone here is taking DE now you can email me and i might help you out with your HW etc

  50. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    wow i didnt knowthere is wolfram ....lol

  52. anonymous
    • 5 years ago
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    its from a past exam with no solutions dude soz,

  53. anonymous
    • 5 years ago
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    ok hope you got an idea of solving of this type of prob...thnx

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