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Thread: 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 $\displaystyle L[y] =ty'' + (2-t)y' -y = e^t$ by factorizing L into two first order linear operators.

    My thoughts on what to do, is it divide through by t, then make $\displaystyle z_{1}$= y, $\displaystyle z_{2}$ = y' and $\displaystyle z_{3}$ = y''.
    That makes $\displaystyle z_{3}= -\frac{2-t}{t}z_{2} -\frac{1}{t} z_{1} +\frac{e^t}{t}$. 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 $\displaystyle L[y] =ty'' + (2-t)y' -y = e^t$ by factorizing L into two first order linear operators.

    My thoughts on what to do, is it divide through by t, then make $\displaystyle z_{1}$= y, $\displaystyle z_{2}$ = y' and $\displaystyle z_{3}$ = y''.
    That makes $\displaystyle z_{3}= -\frac{2-t}{t}z_{2} -\frac{1}{t} z_{1} +\frac{e^t}{t}$. 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. $\displaystyle z_1, z_2$ or $\displaystyle z_3$). I think the idea is to seek operators of the form

    $\displaystyle
    L_1 = t \frac{d}{dt} + a,\;\;\; L_2 = \frac{d}{dt} + b
    $

    and find $\displaystyle a $ and $\displaystyle b$ such that

    $\displaystyle
    L_1(L_2(y(t))) = e^t.
    $

    If you're close try

    $\displaystyle
    \frac{L_1(L_2(ty(t))) }{t} = e^t.
    $
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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 $\displaystyle L[y] =ty'' + (2-t)y' -y = e^t$ by factorizing L into two first order linear operators.
    Also if I may add what I would do: if we have $\displaystyle y''+\frac{2-t}{t}y'-\frac{y}{t}=\frac{e^t}{t}$, then we seek the operator equation:

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

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

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

    then

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

    I don't know. Looks mighty messy to me. I might be wrong on this.
    Last edited by shawsend; Mar 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 $\displaystyle y''+\frac{2-t}{t}y'-\frac{y}{t}=\frac{e^t}{t}$, then we seek the operator equation:

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

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

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

    then

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

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

    $\displaystyle
    \begin{aligned}
    \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) \\
    &=y''+gy'+yg'+fy'+fgy \\
    &=y''+(f+g)y'+(fg+g')y
    \end{aligned}
    $

    Therefore:

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

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

    $\displaystyle
    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}
    $
    and therefore we can solve:

    $\displaystyle \left(D+\frac{1}{t}\right)v=\frac{e^t}{t}$
    $\displaystyle \left(D+\frac{1}{t}-1\right)y=v$
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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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