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This is from Introduction to Differential Equations and its applications by Farlow.

Info: The $D$ operator

Let $D$ denote differentiation with respect to $t$, and let $D^2$ denote the second derivative with respect to $t$. That is $ D = \frac{d^2}{dt^2}$. In general, we denote the quantity $D^n =\frac{d^n}{dt^2}$, where $D^0 = I$ is the identity operator. By using the $D$ operator, it is possible to form more general differential operators, such as $D^2+2D-3, (D-1)(D-3),$ and so on.

Here is an example of a linear system

$x_1' = x_1+x_2$

$x_2'=x_1+x_2$

where $x=x_1, y=x_2$ and so forth.

Question: Use the $ D$ operator and the method of elimination to find a general solutions of the given linear system where $x',y'$ and $z$ refer to differentiation with respect to $t$.

12.$x'=5x-6y+1$

$y' = 6x-7y+1$

Attempt:

First I would rewrite this as a linear system

$x_1'=5x_1-6x_2+1$

$x_2'=6x_1-7x_2+1$

$x'_1-5x_1+6x_2-1=0$

$-6x_1+7x_2+x_2'-1=0$

$Dx_1-5x_1+6x_2-1=0$

$-6x_1+7x^2+Dx_2-1=0$

$(D-5)x_1+6x_2-1=0$

$-6x_1+(7+D)x_2-1=0$

I am going to eliminate the $x_2$ and solve $x_1$

$(7+D)(D-5)x_1+6(7+D)x_2-(7+D)=0$

$36x_1-6(7+D)x_2+6=0$

$(7+D)(D-5)+36)x_1 - 7 + D + 6 =0$

$(7D-35+D^2-5D+36)x_1 - D -1 = 0$

$(D^2+2D+1)x_1 = D +1$

$(D^2+2D+1)x_1 = 1$

Now this is where I start to go haywire. The answer to this is $ x(t) = c_1e^{-t}+c_2te^{-t}+1$ and I know that to find the second answer I need to substitute, find the derivatives, and find $y(t)$. Based on what I am seeing on the answer for $x(t)$, I know that I need to use the method of undetermined coefficients. I know that I have my characteristic equation which is $r^2+2r+ 1 = 0$ which factors into $(r+1)(r+1) = 0 $ and I will have a repeated root, so $c_1e^{-t}+c_2te^{-t} = Y_h$.

What I do not understand is the $Y_p$ portion of this problem. At first, I wanted to leave the $D+1$ on the right hand side of the equation intact, but I've read that $D$ is the derivative so it's something like $ D(1) + 1$ and the derivative of 1 is 0 so I am left with 1 and this equation $(D^2+2D+1)x_1 = 1$ to deal with.

I know that the right hand side of this equation is in the form of $x^n$, so my $Y_p$ would be $Y_p = A$... or not... because when I take the first and second derivatives I have 0. If I try $Y_p = Ax+B$, the first derivative will be $A$ but I won't have a second derivative, so I am left with

$0+2(Ax+B) + 1 = 1$

and that's

$2Ax+2B + 1 = 1$.

I got a strange feeling that I may be making a minor mistake somewhere after I find the $Y_p$. I just clarification on this part because I already understand about finding the first answer, using its derivative, and substitution to find $y(t)$. Also, I know that the answer is correct when the left hand side equals the right hand side.

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