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Differental Equation

Discussion in 'General' started by Murdoc, 27 Apr 2005.

  1. Murdoc

    Murdoc Gas Mask..ZOMG

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    I need to find particular solution for the diferental equation:

    2y+dy/dx -2 = 2 x>=0 y(0)=1

    I have tried many things, Scorpsel on #bit-tech was helping me out but im till having trouble.

    I have got a bunch of these to do I have done several that are regarded as more difficult but this one is being a bitch...

    Any help would be great!!

    Thanks in advance

    'doc
     
  2. whypick1

    whypick1 The über-Pick

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    y = 2 - e^-2x

    Don't have time to go through all the steps, but basically you use an integrating factor of e^2x, so your equation becomes:

    d/dt(ye^2x) = 4e^2x

    From there, simple integration and division.
     
  3. Murdoc

    Murdoc Gas Mask..ZOMG

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    That is the correct answer, thankyou :D

    I'm a bit confused on what you mean by intergrating factor.

    'doc
     
  4. Murdoc

    Murdoc Gas Mask..ZOMG

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    I get to d/dt(ye^2x) = 4e^2x but then i get stuck :/

    'doc
     
    Last edited: 27 Apr 2005
  5. whypick1

    whypick1 The über-Pick

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    Now that I'm done with my classes for the day (lucky for you, I'm taking Diff Eq this quarter), I'll go in detail what I did:

    The general way to solve diff eqs with the form y' + v(t)y = s(t) is to use what is known as the integrating factor, usually called µ(t), which is simply e raised to the power of the integral of v(t). So now the equation becomes:

    d/dt(y*µ(t)) = s(t)µ(t)

    To picture how it works, take the derivative of ye^µ(t). Since y is a function of t, you have to use the chain rule, which would give you:

    y*µ'(t) + y'µ(t) = s(t)µ(t)

    However, we established already that µ'(t) = v(t):

    y*v(t)*µ(t) + y'µ(t) = s(t)µ(t)

    Dividing both sides by µ(t):

    y' + yv(t) = s(t)

    Our original equation.

    Anyway, back to solving. Now you have this:

    d/dt(yµ(t)) = s(t)µ(t)

    Integrating both sides, you get:

    yµ(t) = integral s(t)µ(t)

    The integral should be fairly easy to solve. After that, all you need to do is divide by µ(t) to get y(t). If there's an initial value problem, you'll need to solve for the constant that you got from integrating s(t)µ(t).
     
  6. yodasarmpit

    yodasarmpit No longer the other Brett.

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    You know, I'm glad I never have to do calculus ever again.
     
  7. bort

    bort What's a Dremel?

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    I knew there was a reason why I quit being an engineering major... I hate math
     
  8. Bandit_zoraK

    Bandit_zoraK What's a Dremel?

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    Use Laplace, duh.
     

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