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anonymous

  • 5 years ago

how to solve y''+y-2+0.....

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  1. myininaya
    • 5 years ago
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    ?you can't solve it because it isnt an equation

  2. anonymous
    • 5 years ago
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    sorry.....its y'' + y =2

  3. myininaya
    • 5 years ago
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    first lets solve y''+y=2 I think you can find e's exponent by doing r^2+1=0

  4. myininaya
    • 5 years ago
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    its been awhile since it is imaginary you might use cos and sin is that right?

  5. myininaya
    • 5 years ago
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    i mean first lets solve y''+y=0

  6. myininaya
    • 5 years ago
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    lol oops

  7. myininaya
    • 5 years ago
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    the solution of that will be part of the solution to y''+y=2

  8. anonymous
    • 5 years ago
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    yc= c1 exp[-x] + c2 exp[-x] is this right

  9. myininaya
    • 5 years ago
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    you can check it y'=-c1exp(-x)-c2exp(-x) and y''=c1exp(-x)+c2exp(-x) so no u have to use the sin and cos solution

  10. anonymous
    • 5 years ago
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    how to right in terms of sin nd cos?

  11. myininaya
    • 5 years ago
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    i think its y=c1sin[sqrt(2)t]-c2sin[sqrt(2)]

  12. myininaya
    • 5 years ago
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    i missed my t in that second part

  13. anonymous
    • 5 years ago
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    roots to characteristic equation are +i,-i

  14. myininaya
    • 5 years ago
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    you are right i don't know where I got r^2+2

  15. myininaya
    • 5 years ago
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    should had been r^2+1

  16. myininaya
    • 5 years ago
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    so we replace those sqrt of 2 with just 1*t instead of sqrt(2)*t

  17. myininaya
    • 5 years ago
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    we have for homogeneous solution part is y=c1cost-c2sint

  18. myininaya
    • 5 years ago
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    and we can check it

  19. myininaya
    • 5 years ago
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    so y'=-c1sint-c2cost and y''=-c1cost+c2cost

  20. myininaya
    • 5 years ago
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    gosh i hate typing lol

  21. anonymous
    • 5 years ago
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    but i have two conditions y(x=0)=0 & y(x=1)=0

  22. myininaya
    • 5 years ago
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    y''=-c1cost+c2sint

  23. myininaya
    • 5 years ago
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    we arent done yet we will get to the conditions

  24. myininaya
    • 5 years ago
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    -c1cost+c2sint+c1cost-c2sint=0 so we are good for y''+y=0

  25. myininaya
    • 5 years ago
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    now we need to do y''+y=2

  26. anonymous
    • 5 years ago
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    ok....

  27. anonymous
    • 5 years ago
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    how to get fr dat 2???

  28. myininaya
    • 5 years ago
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    http://www.sosmath.com/tables/diffeq/diffeq.html this is an awesome site

  29. myininaya
    • 5 years ago
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    are solution for the homogenous was not the general solution it was only a particular

  30. myininaya
    • 5 years ago
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    do you see where it says 2nd order

  31. anonymous
    • 5 years ago
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    yep..

  32. anonymous
    • 5 years ago
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    y=c1.cos(x) + c2.sin(x) is the general solution i no....bt hw 2 get particular solution

  33. myininaya
    • 5 years ago
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    ok we did find the general solution since alpha=0 and Belta=1,-1 since that is the coefficent i

  34. myininaya
    • 5 years ago
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    yeah i just found that out

  35. myininaya
    • 5 years ago
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    so once we find the nonhomogeneous solution then we done with the solution part until we get to using initial condition's. We have to make a guess at y

  36. anonymous
    • 5 years ago
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    y=0 for both x=1 & x=0 are the given initial conditions...these conditions yield 0=0!!!

  37. myininaya
    • 5 years ago
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    we arent done finding the solution we cant use the intial condition yet

  38. anonymous
    • 5 years ago
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    so how to find the particular solution....i no dat v hav found d general solution

  39. myininaya
    • 5 years ago
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    we have only found the general solution for the homogeneous equation now we need to find the particular solution for the non homogeneous part. we have to make a guess at y

  40. myininaya
    • 5 years ago
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    what if we let y=A

  41. myininaya
    • 5 years ago
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    A is a constant.

  42. myininaya
    • 5 years ago
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    is y=A a particular solution

  43. myininaya
    • 5 years ago
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    y'= what?

  44. myininaya
    • 5 years ago
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    and what does y''=?

  45. anonymous
    • 5 years ago
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    wat vil b d pert soltn for dis prblm??

  46. myininaya
    • 5 years ago
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    so in order for both sides to be the same A has to equal what?

  47. myininaya
    • 5 years ago
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    what is y' and y'' if y=A where A is a constant

  48. anonymous
    • 5 years ago
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    0

  49. myininaya
    • 5 years ago
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    thank you so what does A have to be?

  50. anonymous
    • 5 years ago
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    2??

  51. myininaya
    • 5 years ago
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    so what is your whole solution without using the initial conditions

  52. anonymous
    • 5 years ago
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    y= 2 + c1.cos x - c2.sin x

  53. myininaya
    • 5 years ago
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    yes now do you know how to find the solution with initial conditions?

  54. anonymous
    • 5 years ago
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    ohh...yeah thank you very much!!!!

  55. myininaya
    • 5 years ago
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    ok have fun. thanks for the memory refresher

  56. anonymous
    • 5 years ago
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    and one more question i have about finite element analysis....i am new so can u tell me ver 2 ask d question??

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