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Math Help - General Solution of DE

  1. #1
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    General Solution of DE

    Find the general solution of , where y is a function of t.

    Could someone help me out with this? I'm not sure how to go about solving for the general solution. All I've got is dy/dt = 1-2y.

    Thanks!
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  2. #2
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    It is a linear non-homogeneous equation.

    Solve for the integrating factor.
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  3. #3
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    Let  \mu = e^{ \int 2 dt} = e^{2t} . This will be our integrating factor.

    Multiplying  \mu through to the whole equation we get that

     e^{2t} y' + 2e^{2t} y = e^{2t}

    You can work it from here.
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  4. #4
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    Quote Originally Posted by cdlegendary View Post
    Could someone help me out with this? I'm not sure how to go about solving for the general solution. All I've got is dy/dt = 1-2y.
    good call, now, can you see your equation is separable?
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  5. #5
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    alright so this is what I ended up trying, but I don't think it's correct. any pointers on where I went wrong?

    e^{2t}y'+2e^{2t}y = e^{2t}

    (e^{2t}y)' = e^{2t}

    e^{2t}y = \int e^{2t} dt

    e^{2t}y = (e^{2t})/2 +c

    y= 1/2 +c
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  6. #6
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    Dy/dt=1-2y
    dy/(1-2y)=1 dt
    integrate both sides
    You should be able to get it from here
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  7. #7
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by cdlegendary View Post
    alright so this is what I ended up trying, but I don't think it's correct. any pointers on where I went wrong?

    e^{2t}y'+2e^{2t}y = e^{2t}

    (e^{2t}y)' = e^{2t}

    e^{2t}y = \int e^{2t} dt

    e^{2t}y = (e^{2t})/2 +c

    y= 1/2 +c
    Look at what Krizalid said. Since your equation is separable, how about:

    \int \frac{dy}{1-2y} = \int dt
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  8. #8
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    Quote Originally Posted by harish21 View Post
    Look at what Krizalid said. Since your equation is separable, how about:

    \int \frac{dy}{1-2y} = \int dt
    Alright, so now I've got:

    \int dy/1-2y = \int dt

    (-1/2)ln(1-2y) = t + c

    ln(1-2y) = -2t - 2c

    1-2y = e^{-2t-2c}

    y=(e^{-2t-2c}-1)/-2
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  9. #9
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by cdlegendary View Post
    Alright, so now I've got:

    \int dy/1-2y = \int dt

    (-1/2)ln(1-2y) = t + c

    ln(1-2y) = -2t - 2c

    1-2y = e^{-2t-2c}

    y=(e^{-2t-2c}-1)/-2
    That looks good.I would put it this way:

    1-2y = e^{-2t-2c} = e^{-2t} e^{-2c}

    y = \frac{1}{2}-\frac{e^{-2t} e^{-2c}}{2}

    y = \frac{1}{2} + {e^{-2t}. C}

    Note : C = \frac{-e^{-2c}}{2} is a constant
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  10. #10
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    y' +2y =1

    first find the characteristic roots

    y'+2y=0
    lambda=l

    l^2+2=0
    l=-2
    l=+-sqrt(2)i
    y=e^sqrt(2)i+e^-sqrt(2)i

    expand with eulers formula get sin and cosine = yg, then
    find particular solution
    y=A
    y'=0
    plug into the question
    2A=1
    A=1/2

    Now yp=1/2
    y=yg+yp

    Cheers
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