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Math Help - Second order LDE operators

  1. #1
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    Second order LDE operators

    Hi there,
    I have a question that states:
    Obtain the general solution of  L[y] =ty'' + (2-t)y' -y = e^t by factorizing L into two first order linear operators.

    My thoughts on what to do, is it divide through by t, then make z_{1}= y, z_{2} = y' and z_{3} = y''.
    That makes z_{3}= -\frac{2-t}{t}z_{2} -\frac{1}{t} z_{1} +\frac{e^t}{t}. Then solve for this first order DE. Once finding a solution to that DE, replace y'' with the answer, than solve the first order linear equation then.

    I was wondering whether my thought process was right. Is my method correct?
    Thanks for your time.
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  2. #2
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    Quote Originally Posted by Silverflow View Post
    Hi there,
    I have a question that states:
    Obtain the general solution of  L[y] =ty'' + (2-t)y' -y = e^t by factorizing L into two first order linear operators.

    My thoughts on what to do, is it divide through by t, then make z_{1}= y, z_{2} = y' and z_{3} = y''.
    That makes z_{3}= -\frac{2-t}{t}z_{2} -\frac{1}{t} z_{1} +\frac{e^t}{t}. Then solve for this first order DE. Once finding a solution to that DE, replace y'' with the answer, than solve the first order linear equation then.

    I was wondering whether my thought process was right. Is my method correct?
    Thanks for your time.
    But you don't have a ODE in a single variable (i.e. z_1, z_2 or z_3). I think the idea is to seek operators of the form

     <br />
L_1 = t \frac{d}{dt} + a,\;\;\; L_2 = \frac{d}{dt} + b <br />

    and find a and b such that

     <br />
L_1(L_2(y(t))) = e^t.<br />

    If you're close try

     <br />
\frac{L_1(L_2(ty(t))) }{t} = e^t.<br />
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  3. #3
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    Quote Originally Posted by Silverflow View Post
    Hi there,
    I have a question that states:
    Obtain the general solution of  L[y] =ty'' + (2-t)y' -y = e^t by factorizing L into two first order linear operators.
    Also if I may add what I would do: if we have y''+\frac{2-t}{t}y'-\frac{y}{t}=\frac{e^t}{t}, then we seek the operator equation:

    \left(D-f(t)\right)\left(D-g(t)\right)y=\frac{e^t}{t}

    But y''+\frac{2-t}{t}y'-\frac{y}{t}=\frac{e^t}{t} can be written as \left(D^2+\frac{2-t}{t}D-\frac{1}{t}\right)y=\frac{e^t}{t} and therefore:

    \left\{f(t),g(t)\right\}=\frac{\frac{t-2}{t}\pm \sqrt{\frac{(2-t)^2}{t^2}+\frac{4}{t}}}{2}

    then

    \left(D-f(t)\right)v=\frac{e^t}{t}
    \left(D-g(t)\right)y=v

    I don't know. Looks mighty messy to me. I might be wrong on this.
    Last edited by shawsend; March 20th 2010 at 07:12 AM.
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  4. #4
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    Quote Originally Posted by shawsend View Post
    Also if I may add what I would do: if we have y''+\frac{2-t}{t}y'-\frac{y}{t}=\frac{e^t}{t}, then we seek the operator equation:

    \left(D-f(t)\right)\left(D-g(t)\right)y=\frac{e^t}{t}

    But y''+\frac{2-t}{t}y'-\frac{y}{t}=\frac{e^t}{t} can be written as \left(D^2+\frac{2-t}{t}D-\frac{1}{t}\right)y=\frac{e^t}{t} and therefore:

    \left\{f(t),g(t)\right\}=\frac{\frac{t-2}{t}\pm \sqrt{\frac{(2-t)^2}{t^2}+\frac{4}{t}}}{2}

    then

    \left(D-f(t)\right)v=\frac{e^t}{t}
    \left(D-g(t)\right)y=v

    I don't know. Looks mighty messy to me. I might be wrong on this.
    I messed up interpreting the differential operators and mistakenly treated them as algebraic. So I'd like to start over please.

    We have ty''+(2-t)y'-y=e^t or y''+\frac{2-t}{t}y'-\frac{1}{t}y=\frac{e^t}{t} and I want to find a product of two linear operators of the form:

    <br />
\begin{aligned}<br />
\left(\frac{d}{dt}+f\right)\left(\frac{d}{dt}+g\ri  ght)y&=\left(\frac{d}{dt}+f\right)\left(y'+gy\righ  t) \\<br />
&=y''+gy'+yg'+fy'+fgy \\<br />
&=y''+(f+g)y'+(fg+g')y<br />
\end{aligned}<br />

    Therefore:

    f+g=\frac{2}{t}-1
    fg+g'=-\frac{1}{t}

    and by inspection and a little messin' around, I get: f=\frac{1}{t} and g=\frac{1}{t}-1 so I can now write the DE as:

    <br />
y''+\frac{2-t}{t}y'-\frac{1}{t}y=\left(D+\frac{1}{t}\right)\left(D+\fr  ac{1}{t}-1\right)y=\frac{e^t}{t}<br />
    and therefore we can solve:

    \left(D+\frac{1}{t}\right)v=\frac{e^t}{t}
    \left(D+\frac{1}{t}-1\right)y=v
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  5. #5
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    Thank you both, very much. This has given me a lot of help.
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