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hary

  • 2 years ago

du(x)/dx- tan(x)u(x)=-2sinx find u(x) inhomogeneous differential equation

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  1. SithsAndGiggles
    • 2 years ago
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    \[u'-(\tan x)u=-2\sin x\\ (\cos x)u'-(\sin x)u=-2\sin x\cos x\] What can you say about the LHS?

  2. zepdrix
    • 2 years ago
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    I saw you were stuck on this one earlier hary, what was the book answer again?

  3. hary
    • 2 years ago
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    u=cosx +C/cosx this was the book answer

  4. zepdrix
    • 2 years ago
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    Ah ok good. You must've made a small mistake near the end then. Because your integrating factor looked good.

  5. hary
    • 2 years ago
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    hmm im guessing the same too..but couldnt figure out where

  6. zepdrix
    • 2 years ago
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    \[\large u \cos x=-2 \int\limits \sin x \cos x dx\] \[\large m=\cos x\]\[\large -dm=\sin x dx\] Applying this substitution gives us,\[\large u \cos x=2 \int\limits m dm\] \[\large u \cos x=m^2+C\] \[\large u \cos x=\cos^2x+C\]

  7. hary
    • 2 years ago
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    owh god thats brilliant...did a blunder at my RHS integration

  8. myko
    • 2 years ago
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    lol, what a mistake. Time for me to go to sleep I think

  9. zepdrix
    • 2 years ago
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    heh :D

  10. hary
    • 2 years ago
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    @zepdix thanks alot!!!! was cracking my head on this for quite long...pheww thats a relief... thankx again...

  11. hary
    • 2 years ago
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    @myko thankx to you too...been a great help

  12. myko
    • 2 years ago
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    :)

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