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Math Help - how to solve this

  1. #1
    Senior Member Sambit's Avatar
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    Question how to solve this

    find the continuous solution of \frac{dy}{dx}+y=g(x), 0<x<\infty, y(0)=2 where g(x)=3, 0<x<\frac{\pi}{2}
    =\cos x, x\geq\frac{\pi}{2}

    for the first part, i got -ln(3-y)=x. but how can i solve it for the second part?
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  2. #2
    A Plied Mathematician
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    Solving for y in the first part of your solution yields

    y=3-e^{-x}.

    We get y(\pi/2)=3-e^{-\pi/2}. This is the initial condition for the next region's DE.

    So now you're solving y'(x)+y(x)=\cos(x), subject to y(\pi/2)=3-e^{-\pi/2}. Ideas?
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  3. #3
    Senior Member Sambit's Avatar
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    unfortunately, no. i have done solving algebraic equation subject to some constraint, but not differential equation. How to do this?
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  4. #4
    A Plied Mathematician
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    Hmm. I suggest you take a look at Chris's DE tutorial. You can find the technique you need there.
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  5. #5
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by Ackbeet View Post
    Solving for y in the first part of your solution yields

    y=3-e^{-x}.

    We get y(\pi/2)=3-e^{-\pi/2}. This is the initial condition for the next region's DE.

    So now you're solving y'(x)+y(x)=\cos(x), subject to y(\pi/2)=3-e^{-\pi/2}. Ideas?
    The solution You proposed is not properly adequate because the 'queue' of the y(x) due to the 'head' of g(x) doesn't vanish for x>3. The best approach to a problem of this type is the Laplace Transform...

    Kind regards

    \chi \sigma
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  6. #6
    Senior Member Sambit's Avatar
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    solving by integrating factor, i got ye^x=\frac{e^x(\sin x + \cos x)}{2} + C
    from the condition given in question, i am getting C=4.
    now what about "continuous solution"?
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  7. #7
    Senior Member Sambit's Avatar
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    solving by integrating factor, i got ye^x=\frac{e^x(\sin x + \cos x)}{2} + C
    from the condition given in question, i am getting C=4.
    now what about "continuous solution"?
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  8. #8
    A Plied Mathematician
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    Reply to chisigma:

    I'm not following you. What do you mean by "queue" and "head"? I would agree that the LT is a very convenient method of solution for this problem (non-homogeneous IVP's, especially with respect to piece-wise defined functions). However, as Sambit appears not to be familiar with DE's in general, I think the problem is more basic than that.

    I would argue that the way I've set the problem up, you will theoretically get a continuous solution on the desired interval. I mean, the initial condition for the region x >= 3 is as I've given, correct?
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  9. #9
    A Plied Mathematician
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    Reply to Sambit:

    Good so far. What next?
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  10. #10
    Senior Member Sambit's Avatar
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    as you said, y(\pi/2) should be equal to 3-e^{-\pi/2} but it does not hold here.
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  11. #11
    A Plied Mathematician
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    ...but it does not hold here...
    Why not? Are there no values of C which can make that happen? Incidentally, in post # 6, I was confused by your assertion that C = 4. How did you arrive at that? Is there a part of the problem statement that you haven't posted?
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  12. #12
    Senior Member Sambit's Avatar
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    in my question, i have y(0)=2. What does this mean? i think it means at x=0, y is 2. or am i wrong?
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  13. #13
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    Right, but you've already found the solution for the region 0 < x < pi/2. You're now finding a solution for pi/2 < x < infty. You'll have a piece-wise defined solution, just like you have a piece-wise defined RHS of the DE. Therefore, the condition y(0) = 2 does not apply to the piece you're trying to find now.
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  14. #14
    Senior Member Sambit's Avatar
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    so am i now to find a value of C such that y(\pi/2)=3-e^{-\pi/2}. holds?
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  15. #15
    Senior Member Sambit's Avatar
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    is C=6-2e^{-\pi/2}?
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