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Math Help - Second Order DE (non homogenous)

  1. #1
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    Second Order DE (non homogenous)

    Hello!

    I having problems getting started with this problem:

    Using subn  t = e^x  and    z(t) = y(x) rewrite:

    \frac{d^2y}{dx^2} - \frac {dy}{dx} + e^{2x}y = xe^2x -1

    hence find all solutions.

    When I perform the substitution using  t^2 = e^{2x}   and   x = lnt, I get:

     \frac{d^2z}{dt^2} - \frac{dz}{dt} + t^2z = t^2lnt +1

    I can't find any complimentary functions and probably due to my undergrad engineering major, I tried a Laplace transform, but end up with a horrible function owing to the t^2z part that would leave a \frac{d^2s}{dt^2} on the LHS- I suspect this is not be the correct method as alluded to within the question - there must be a reason for the subn.....

    I can't use an order reduction or variation of parameters as I don't have an already known solution- unless I use a trial and error! If anyone can kick me in the right direction, I will be most greatful.... I can't seem to see through it right now!

    Thanks for looking!
    Last edited by padawan; August 13th 2010 at 04:14 AM.
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  2. #2
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    Should this actually be

    \frac{d^2 y}{dx^2} - \frac{dy}{dx} + e^{2x}y = xe^{2x} -1?


    If so...

    You plan to make the substitutions t = e^x and z(t) = y(x).

    This would mean \frac{dt}{dx} = e^x = t, t^2 = e^{2x}, x = \ln{(t)}.

    It would also mean

    \frac{dy}{dx} = \frac{dz}{dx}

     = \frac{dt}{dx}\frac{dz}{dt}

     = t\,\frac{dz}{dt}

    and

    \frac{d^2y}{dx^2} = \frac{d}{dx}\left(t\,\frac{dz}{dt}\right)

     = \frac{dt}{dx}\frac{dz}{dt} + t\,\frac{d}{dx}\left(\frac{dz}{dt}\right)

     = t\,\frac{dz}{dt} + t\left[\frac{dt}{dx}\,\frac{d}{dt}\left(\frac{dz}{dt}\rig  ht)\right]

     = t\,\frac{dz}{dt} + t\,\frac{dt}{dx}\,\frac{d^2z}{dt^2}

     = t\,\frac{dz}{dt} + t^2\,\frac{d^2z}{dt^2}.


    So finally we can substitute into the original DE...

    \frac{d^2 y}{dx^2} - \frac{dy}{dx} + e^{2x}y = xe^{2x} -1

    t\,\frac{dz}{dt} + t^2\,\frac{d^2z}{dt^2} - t\,\frac{dz}{dt} + t^2z = t^2\ln{(t)} - 1

    t^2\,\frac{d^2z}{dt^2} + t^2z = t^2\ln{(t)} - 1

    t^2\left(\frac{d^2z}{dt^2} + z\right) = t^2\ln{(t)} - 1

    \frac{d^2z}{dt^2} + z= \ln{(t)} - t^{-2}.


    You should be able to solve this now...
    Last edited by Prove It; August 13th 2010 at 04:18 AM.
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  3. #3
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    Yes it is.... I'll change the original, also missed off a'y'..... My latex errors!
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  4. #4
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    Oh dear, I just solved the original DE without the extra y, see my edit above...

    Well, I think you can see the technique anyway, will edit further...

    Edited again... You now have a second order DE that you should be able to solve...
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  5. #5
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    Thanks Prove it!

    No need to edit if you havent done so... leaves me something to do. !

    Big thanks for pointing the technique to me!
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  6. #6
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    I am getting to a difficult integral on the \int w_1/w

    The aux eqn is m^2 + 1 = 0

    giving roots of +i or -i and giving compliemntary function:

    z = A cos(t) + B sin(t)

    Evaluating the wronskian determinents:

    w_1 = det \begin{bmatrix}<br />
0 & sin(t) \\<br />
ln(t) - t^{-2} & cos(t)<br />
\end{bmatrix}

    = -sin(t) [ln(t) - t^{-2}]

    w_2 = det \begin{bmatrix}<br />
cos(t) & 0 \\<br />
-sin(t) & ln(t) - t^{-2} <br />
\end{bmatrix}

    = cos(t) [ln(t) - t^{-2}]

    w = det \begin{bmatrix}<br />
cos(t) & sin(t) \\<br />
-sin(t) & cos(t)<br />
\end{bmatrix}


    which drops to be 1

    this leaves

    u_1 = \int \frac {w_1}{w} = \int -sin(t) [ln(t) - t^{-2}] dt

    and similarly for w_2 with -sin(t) replaced by cos(t)

    It does not seem to possible to integrate t^{-2}sin(t).... have I made an error in my working or gone off track somewhere?
    Last edited by padawan; August 13th 2010 at 12:54 PM.
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  7. #7
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    I might add that

    \frac{d^2z}{dt^2} + z= \ln{(t)} - t^{-2}

    has the particular solution z_p = \ln t.
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  8. #8
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    Cheers danny.... I see that works via subn in and evaluating the DE.... was that by observation or just through some other way?
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  9. #9
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    Quote Originally Posted by padawan View Post
    Cheers danny.... I see that works via subn in and evaluating the DE.... was that by observation or just through some other way?
    Actually, when I saw your ODE

    \dfrac{d^2y}{dx^2} - \dfrac {dy}{dx} + e^{2x}y = xe^{2x} -1

    I saw

    \dfrac{d^2y}{dx^2} - \left(\dfrac {dy}{dx}-1\right) + e^{2x}\left(y - x\right) = 0

    and  y = x identically satisfies your DE.
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