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anonymous

  • one year ago

Yet another Calculus question im having trouble with . Find the particular solution to y " = sin(x) given the general solution y = −sin(x) + Ax + B and the initial conditions y(pi/2) = 0 and y '(pi/2) = -2.

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  1. anonymous
    • one year ago
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    not even sure where to start on this one

  2. freckles
    • one year ago
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    you make a guess for the particular solution I know the derivative of sin(x) is cos(x) and the derivative of cos(x) is -sin(x) so I might try something like \[ y_p=A sin(x)+Bcos(x)\] for the particular solution

  3. freckles
    • one year ago
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    you can call A and B something else since they already being used

  4. anonymous
    • one year ago
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    I'm given the answer choices do you think they might help?

  5. UsukiDoll
    • one year ago
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    method of undetermined coefficients...

  6. UsukiDoll
    • one year ago
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    for a second order ode....

  7. anonymous
    • one year ago
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    i'v never been thought second order.... are they much different?

  8. freckles
    • one year ago
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    yea you will get something like -Asin(x)-Bcos(x)=sin(x) shich means B=0 and -A=1 so A=-1 so we didn't need the cos part of the particular solution

  9. UsukiDoll
    • one year ago
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    second order odes are easier than first order odes... that's what a ton of ode students say

  10. anonymous
    • one year ago
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    these area the answer choices by the way −sin(x) +1 + π −sin(x) − 2x + π −sin(x) − 2x − π −sin(x) − 2x + π + 1

  11. freckles
    • one year ago
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    it looks like it already gave you the solution

  12. freckles
    • one year ago
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    just apply the conditions

  13. anonymous
    • one year ago
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    what do you mean .... im not even sure what to solve for

  14. freckles
    • one year ago
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    \[y=-\sin(x)+Ax+B \\ y(\frac{\pi}{2})=0 \text{ insert this into the solution }\] you will also need to find y' given that and then use the other condition y'(pi/2)=-2

  15. freckles
    • one year ago
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    you will wind up with a system of equations

  16. anonymous
    • one year ago
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    so you mean \[y(\frac{ \pi }{ 2 }) = -1 + Api + B\]

  17. freckles
    • one year ago
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    not exactly \[y(\frac{\pi}{2})=-\sin(\frac{\pi}{2})+A(\frac{\pi}{2})+B \\ 0=-1+A(\frac{\pi}{2})+B\]

  18. UsukiDoll
    • one year ago
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    x will be pi/2 and then that would be = 0 just like freckies put it

  19. freckles
    • one year ago
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    now what is y'

  20. anonymous
    • one year ago
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    do we have to derive -sin(x) + Ax + B before plugging in pi/2?

  21. freckles
    • one year ago
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    yes otherwise you are differentiating constants on both sides and wind up with 0=0 which doesn't really help \[y=-\sin(x)+Ax+B \\ y'=-\cos(x)+A+0 \\ y'=-\cos(x)+A\]

  22. freckles
    • one year ago
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    now plug in the condition

  23. freckles
    • one year ago
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    for y'

  24. anonymous
    • one year ago
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    yay i found a !!!!!!

  25. freckles
    • one year ago
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    go back to the first equation and find B

  26. anonymous
    • one year ago
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    okay ......aahhh im so exited for some reason !!!

  27. freckles
    • one year ago
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    because this was easier than you thought imaginable ?

  28. anonymous
    • one year ago
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    possibly ... okay so im getting B = -pi +1 is that right?

  29. freckles
    • one year ago
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    hmmm... I might be geting pi+1

  30. freckles
    • one year ago
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    one sec

  31. anonymous
    • one year ago
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    no yea your right its -2 not 2 my bad

  32. anonymous
    • one year ago
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    okay so what now

  33. freckles
    • one year ago
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    \[y'(\frac{\pi}{2})=-\cos(\frac{\pi}{2})+A \\ -2=-\cos(\frac{\pi}{2})+A \\ -2 =0+A \\ -2=A \\ 0=-1+A(\frac{\pi}{2})+B \\ 0=-1-2(\frac{\pi}{2})+B \\ 0=-1-\pi+B \\ B=\pi+1\] just wanted to make sure

  34. freckles
    • one year ago
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    well as said you already had the solution the conditions just needed to be applied so you found A and B

  35. freckles
    • one year ago
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    just plug them back in

  36. freckles
    • one year ago
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    into the solution they gave you

  37. freckles
    • one year ago
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    \[y=-\sin(x)+Ax+B \\ \text{ we found } A=-2 \\ \text{ and } B=\pi+1 \\ \text{ plug \in and done }\]

  38. anonymous
    • one year ago
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    holy crap i found it !!! its D AAAAAHH this is ssoo cool !!!

  39. freckles
    • one year ago
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    lol you're funny

  40. anonymous
    • one year ago
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    im such a pro !!!!

  41. anonymous
    • one year ago
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    okay maybe nmot .... but still !!!

  42. anonymous
    • one year ago
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    thanks for the help , i really appreciate it!

  43. freckles
    • one year ago
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    np

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