The question is the following : $$x''-3x'+2x=t^2$$ However, my book does not provide any answer to this equation
I tried to solve it but I feel lost after writing down the following system $$x_1=x$$$$x_2=x'=x_1'$$ $$x_2'=t^2-2x_1+3x_2$$
The book that expects me to write the system of equations so it does not give the answer to the differential equation. I was wondering how can I solve it by using operators?
On
A simple googling of "solving second order differential equations" reveals a multitude of links. Reading the first two pages of, for example,
will answer your question. In short, you always try to find a solution $x = C_1 e^{r_1 t} + C_2 e^{r_2 t}$, where $C_1$ and $C_2$ are constants, but $r_1$ and $r_2$ are solutions to the specific 2nd order polynomial. After having found the solution to the homogeneous equation this way, you will have to guess the solution to the inhomogeneous equation. I think you will succeed by trying out an arbitrary 2nd order polynomial as an inhomogeneous solution
$x_{inhomog}(t) = at^2 +bt+c$
The constants can be figured out by substituting it into the original equation. The final solution will be the sum of homogeneous and inhomogeneous one.
I don't feel like solving the equation explicitly here. There is really a lot of material on this type of problems online
Substitute $D=\frac{d}{dx}$ such that you have $$(D^2-3D+2)x = t^2$$ But we know that $D^2-3D+2 = (D-2)(D-1)$, if we set $x_1 = (D-2)x$ then we get: \begin{align} (D-1)x_1 &= t^2\\ (D-2)x &= x_1 \end{align} The first ode back to the original form is $x_1' -x_1 = t^2$ we can solve this ode by first solve the homogeneous and then use $x_1 = a+bt+ct^2$ with $a,b,c$ constants to solve the particular solution and add these two. Use this solution and substitute it in the second ode and solve this ode as well.