Results 1 to 9 of 9

Thread: Second Order DE (non homogenous)

  1. #1
    Junior Member
    Joined
    Aug 2010
    Posts
    53

    Second Order DE (non homogenous)

    Hello!

    I having problems getting started with this problem:

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

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

    hence find all solutions.

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

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

    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 $\displaystyle t^2z$ part that would leave a $\displaystyle \frac{d^2s}{dt^2}$ 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; Aug 13th 2010 at 04:14 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    12,880
    Thanks
    1946
    Should this actually be

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


    If so...

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

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

    It would also mean

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

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

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

    and

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

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

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

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

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


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

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

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

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

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

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


    You should be able to solve this now...
    Last edited by Prove It; Aug 13th 2010 at 04:18 AM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Aug 2010
    Posts
    53
    Yes it is.... I'll change the original, also missed off a'y'..... My latex errors!
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    12,880
    Thanks
    1946
    Oh dear, I just solved the original DE without the extra $\displaystyle y$, 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...
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Aug 2010
    Posts
    53
    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!
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member
    Joined
    Aug 2010
    Posts
    53
    I am getting to a difficult integral on the $\displaystyle \int w_1/w$

    The aux eqn is $\displaystyle m^2 + 1 = 0 $

    giving roots of $\displaystyle +i or -i $ and giving compliemntary function:

    $\displaystyle z = A cos(t) + B sin(t)$

    Evaluating the wronskian determinents:

    $\displaystyle w_1 = det \begin{bmatrix}
    0 & sin(t) \\
    ln(t) - t^{-2} & cos(t)
    \end{bmatrix}$

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

    $\displaystyle w_2 = det \begin{bmatrix}
    cos(t) & 0 \\
    -sin(t) & ln(t) - t^{-2}
    \end{bmatrix}$

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

    $\displaystyle w = det \begin{bmatrix}
    cos(t) & sin(t) \\
    -sin(t) & cos(t)
    \end{bmatrix}$


    which drops to be $\displaystyle 1$

    this leaves

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

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

    It does not seem to possible to integrate $\displaystyle t^{-2}sin(t)$.... have I made an error in my working or gone off track somewhere?
    Last edited by padawan; Aug 13th 2010 at 12:54 PM.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor
    Jester's Avatar
    Joined
    Dec 2008
    From
    Conway AR
    Posts
    2,470
    Thanks
    83
    I might add that

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

    has the particular solution $\displaystyle z_p = \ln t$.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Junior Member
    Joined
    Aug 2010
    Posts
    53
    Cheers danny.... I see that works via subn in and evaluating the DE.... was that by observation or just through some other way?
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor
    Jester's Avatar
    Joined
    Dec 2008
    From
    Conway AR
    Posts
    2,470
    Thanks
    83
    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

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

    I saw

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

    and $\displaystyle y = x$ identically satisfies your DE.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. first order homogenous equations
    Posted in the Differential Equations Forum
    Replies: 5
    Last Post: Jan 3rd 2012, 07:56 AM
  2. Linear first order homogenous DE
    Posted in the Differential Equations Forum
    Replies: 5
    Last Post: Feb 23rd 2010, 03:36 PM
  3. second order non-homogenous IVP
    Posted in the Differential Equations Forum
    Replies: 3
    Last Post: Jul 23rd 2009, 05:44 AM
  4. Third Order Non-Homogenous ODE
    Posted in the Differential Equations Forum
    Replies: 1
    Last Post: Jun 25th 2009, 04:08 AM
  5. Replies: 1
    Last Post: May 11th 2007, 03:01 AM

Search Tags


/mathhelpforum @mathhelpforum