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Math Help - Particular Integrals part 2

  1. #1
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    Smile Particular Integrals part 2

    Could some one check these for me please?

    1) dy/dx = (3/x)*y + 5x^2 (x>0)

    What is the integrating factor?

    Answer = x^3

    2) d^2y/dx^2 - 16y =4e^-4x has the complementary function y = ae^4x + be^-4x

    What is a suitable particular integral?

    Answer = y = ae-4x

    Think these are right but not sure. Thanks in advance!!!

    N.B read a = alpha and b = beta
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  2. #2
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    Quote Originally Posted by thermalwarrior View Post
    Could some one check these for me please?

    1) dy/dx = (3/x)*y + 5x^2 (x>0)

    What is the integrating factor?

    Answer = x^3 Mr F says: No. It's 1/x^3.

    2) d^2y/dx^2 - 16y =4e^-4x has the complementary function y = ae^4x + be^-4x

    What is a suitable particular integral?

    Answer = y = ae-4x Mr F says: No. This particular solution is part of the homogenous solution. It will therefore make the left hand side of the DE equal to zero ..... Try again .....

    Think these are right but not sure. Thanks in advance!!!

    N.B read a = alpha and b = beta
    ..
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  3. #3
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    these questions are multiple choice and 1/x3 is not an answer.

    for number number 2 i did get answer p = 1/8e-4x

    so a general answer would be y = ae-4x
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  4. #4
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    Quote Originally Posted by thermalwarrior View Post
    these questions are multiple choice and 1/x3 is not an answer.

    [snip]
    I guess they missed option F.

    Actually, I'm being sarcastic. The question as posted by you is wrong if 1/x^3 is not the integrating factor:

    The DE as posted can be re-written \frac{dy}{dx} - \frac{3}{x} = 5x^2.

    I.F. = e^{\int - \frac{3}{x} \, dx} = e^{- 3 \ln x} = e^{\ln x^{-3}} = x^{-3} = \frac{1}{x^3}.
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  5. #5
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    i see where i went wrong! the answer was there i mis read it. What about the second part? any advice on where i went wrong there???
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  6. #6
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    Quote Originally Posted by thermalwarrior View Post
    [snip]
    for number number 2 i did get answer p = 1/8e-4x

    so a general answer would be y = ae-4x
    Substitute y = ae-4x into d^2y/dx^2 - 16y. You get zero, not a multiple of e^-4x. y = ae-4x will not give a particular solution. Try again. Go to your notes - what do they say to do when the inhomogenous bit is part of the homogenous solution .....?
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  7. #7
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    well im lost i not sure where to go from the initial question
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