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anonymous

  • one year ago

Differential Equations SOS PLEASE

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  1. anonymous
    • one year ago
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    \[y'''+y'=\frac{ sinx }{ \cos^2x }\]

  2. anonymous
    • one year ago
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    \[K^3+K=0\]

  3. anonymous
    • one year ago
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    \[K(K+1)=0\]

  4. anonymous
    • one year ago
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    meaning \[K1=0, K2,3= +-i\]

  5. anonymous
    • one year ago
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    \[Y=C1+C2\cos(x)+C3\sin(x)+Yp\]

  6. anonymous
    • one year ago
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    now how do I find Yp?

  7. anonymous
    • one year ago
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    should I find the derivative ?

  8. ganeshie8
    • one year ago
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    familiar with variation of parameters ?

  9. anonymous
    • one year ago
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    yes, more or less

  10. anonymous
    • one year ago
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    @ganeshie8 ?

  11. anonymous
    • one year ago
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    I really need help with this one

  12. ganeshie8
    • one year ago
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    \[y'''+y'=\frac{ \sin x }{ \cos^2x }\] Let \( y'=v\), then the DE becomes \[v''+v=\frac{ \sin x }{ \cos^2x }\] which is a familiar second order ordinary eqn, you can find \(Y_p\) using wrokskian or any other tricks that you're familiar with

  13. anonymous
    • one year ago
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    hmmmm not sure I'm following

  14. anonymous
    • one year ago
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    I found the general solution

  15. anonymous
    • one year ago
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    All I need now is the private one

  16. anonymous
    • one year ago
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    Are you familiar with the method I'm using?

  17. ganeshie8
    • one year ago
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    http://gyazo.com/85b4d43f26ed36b347d32629ee0df5b0

  18. ganeshie8
    • one year ago
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    \[v''+v=\frac{ \sin x }{ \cos^2x }\] earlier you worked the general solution, general solution for above reduced DE is \(c_2\cos x+c_3\sin x\), yes ?

  19. anonymous
    • one year ago
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    +C1

  20. ganeshie8
    • one year ago
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    No, we're only looking at reduced DE for now.

  21. anonymous
    • one year ago
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    oh ok

  22. anonymous
    • one year ago
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    Can you please show how to do it using the formula you've attached?

  23. ganeshie8
    • one year ago
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    Okay, so from general solution we have \(y_1 = \cos x\) \(y_2 = \sin x\) Wronkskian\(W(y_1, y_2) = \begin{vmatrix} \cos x &\sin x\\\cos'x&\sin'x\end{vmatrix} = \cos^2x+\sin^2x = 1 \)

  24. ganeshie8
    • one year ago
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    simply plug it in the formula and evaluate the integral(s)

  25. anonymous
    • one year ago
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    but what is g(t)?

  26. anonymous
    • one year ago
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    whatit refers to?

  27. ganeshie8
    • one year ago
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    g(t) is whatever there on the right hand side

  28. ganeshie8
    • one year ago
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    \(g(\color{red}{x}) = \dfrac{\sin x}{\cos^2x}\)

  29. anonymous
    • one year ago
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    and how do I find C1 then?

  30. ganeshie8
    • one year ago
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    we will worry about that in the very end

  31. anonymous
    • one year ago
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    oh ok

  32. ganeshie8
    • one year ago
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    Keep in mind we're solving the "reduced" DE in \(v\)'s completely first, then we're gona substitute back \(v\)

  33. anonymous
    • one year ago
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    roger that

  34. anonymous
    • one year ago
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    I think I can handle with that by myself, but how then can I find C1?

  35. anonymous
    • one year ago
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    btw thanks a lot!

  36. ganeshie8
    • one year ago
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    \(\large v(x) = c_2\cos x + c_3\sin x+Y_p\) plug in \(v(x) = y'\) back

  37. ganeshie8
    • one year ago
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    and integrate

  38. anonymous
    • one year ago
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    thank you so much @ganeshie8 !

  39. anonymous
    • one year ago
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    I think I've got it! I'll try at home and see if it works out for me

  40. ikram002p
    • one year ago
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    good question, it was useful for me as well .

  41. ganeshie8
    • one year ago
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    you may refer to this solution generated by wolfram if you get stuck..

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  42. UsukiDoll
    • one year ago
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    Y_h portion is always easier than finding the Y_p

  43. UsukiDoll
    • one year ago
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    there are 3 different ways to solve 2nd order odes... method of undetermined coefficients, variation of parameters, and laplace transform. It's just that sometimes one method is easier to use than the other 2. >_<

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