# Thread: Second order LDE operators

1. ## 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?

2. Originally Posted by Silverflow
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?
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

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

and find $a$ and $b$ such that

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

If you're close try

$
\frac{L_1(L_2(ty(t))) }{t} = e^t.
$

3. Originally Posted by Silverflow
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.

4. Originally Posted by shawsend
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:


\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:

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

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:

$
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:

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

5. Thank you both, very much. This has given me a lot of help.