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Variations of Parameters

  1. Oct 22, 2015 #1
    1. The problem statement, all variables and given/known data
    y'' + y = tan2x

    3. The attempt at a solution

    yh = c1sinx + c2cosx

    yp = -y1 ∫ (y2r/W) dx + y2 ∫ (y1r/W) dx

    r = tan2x

    y1 = sinx

    y2 = cosx

    W = 1

    I'm using integration by parts, but I've realized that the trig functions repeat (trivial).

    Is there another way to solve this equation or is by parts the way to go?
     
  2. jcsd
  3. Oct 22, 2015 #2

    ehild

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    Show your integrals, please. What have you tried?
     
  4. Oct 22, 2015 #3
    Thank you for the reply.

    I retried by parts and saw that trig identities simplify the integrals, so I don't have to do by parts a second time. I apologize for not displaying my integrals.

    My answer is y = c1sinx + c2cos +cosx

    if you still want to see my work I can display it, but it's a lot to type. Let me know.
     
  5. Oct 22, 2015 #4

    ehild

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    Is that really solution of the differential equation?
    I would like to see your integrals and what you have tried.
     
  6. Oct 22, 2015 #5
    W = 1
    r = tan2x
    y1 = sinx
    y2 = cosx

    yh = c1sinx + c2cosx

    yp = -y1 ∫ ((y2)(r)/W)dx + y2 ∫ ((y1)(r)/W)dx

    yp = -sinx ∫ ((cosx)(tan2x)dx) + cosx ∫ ((sinx)(tan2x)dx)

    yp = -sinx ∫ ((cosx)(sec2x +1)dx) + cosx ∫ ((sinx)(sec2x +1)dx)

    There's two by parts labeled 1 (left side) and 2 (right side)

    by parts 1

    u = cosx , du = sinx dx
    v = sec2x +1 , dv = tanx + x

    -sinx(cosx)(tanx +x) - ∫ (sinxtanx - sinx)dx
    -sinx(cosx)(tanx +x) + ∫ ((sin2x/cosx) - sinx)dx
    -sinx(cosx)(tanx +x) + ∫ (((cos2x + 1(/(cosx) + sinx)dx
    -sinx(cosx)(tanx +x) + ∫ (((cos2x + 1(/(cosx) - sinx)dx
    -sinx(cosx)(tanx +x) + ∫ (1/cosx)dx + ∫(cosx)dx - ∫(sinx)dx
    -sinxcosxtanx + (sinxcosx)x + ln cosx + sinx - cosx
    -sin2x + (sinxcosx)x +ln cosx + sinx - cosx

    by parts 2

    u = cosx , du = sinx dx
    v = sec2x +1 , dv = tanx + x

    + cosx(sinx)(tanx + x) -∫ (cosx)(tanx + x)dx
    + cosx(sinx)(tanx + x) -∫ (sinx)dx + ∫(cosx)dx
    + cosx(sinx)(tanx + x) +cosx + sinx
    + cosxsinxtanx + (cosxsinx)x + cosx + sinx
    + sin2x + (cosxsinx)x + cosx + sinx

    y = c1sinx + c2cosx -sin2x + (sinxcosx)x +ln cosx + sinx - cosx+ sin2x + (cosxsinx)x + cosx + sinx

    Final answer:

    y = c1sinx + c2cosx + (sin2xcos2x)x2 +ln cosx + 2sinx

    Thanks for having me type out this problem I found mistakes. How does this answer look?
     
  7. Oct 22, 2015 #6

    ehild

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    tan2x is not the same as sec2x +1. Check. But I would not use the secant. Would do it with tan(x)=sin(x)/cos(x) instead.
    I do not understand your way of integration by parts, either.
    And W=-1.
     
  8. Oct 22, 2015 #7

    Thank you, the secant should have been sec2x -1
     
  9. Oct 22, 2015 #8

    ehild

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  10. Oct 23, 2015 #9
    Hi Ehild,

    My apologies for thinking I could use sec2x - 1. I thought the derivation was tanx from sec2x. I'll use sinx/cosx.

    Thank you again for the help.
     
  11. Oct 23, 2015 #10
    My new answer is extremely long.

    y = c1sinx + c2cosx + sin2x - xcosx(xsinx - 1) - lntanx + secx - cosx(sinxtanx - 2) -xsinx(cosx + 1)
     
  12. Oct 23, 2015 #11

    ehild

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    Still not correct, I am afraid. Check the parentheses. I can not tell where your mistakes are if you do not show your work.
    You need the integrals ##\int(\cos(x)\tan^2(x)dx)## and ##\int(\sin(x)\tan^2(x)dx)## . How did you do them?Hint: when integrating by parts, integrate u'=sin(x).
     
    Last edited: Oct 23, 2015
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