If f is an linear operator $f:ℝ[x]\toℝ^2$ which is defined with: $f:u\to(u(2), u'(2)) $ find all the solutions to $f(u)=(3,-2)$.
So I started like this:
$u(x)=a_0+a_1x+\dots+a_nx^n$
$u'(x)=a_1+2a_2x+3a_3x^2+\dots+na_nx^{n-1}$
$u(2)=a_0+2a_1+\dots+2^na_n=3$
$u'(2)=a_1+2^2a_2+3\times2^2a_3+\dots+na_n2^{n-1}=-2$
So the solution to the equation are all polynomials whose coefficients satisfy the above equations. We can see that the zero polynomial doesn't satisfy the equations so the solutions are not a subspace.
$\sum_{k=0}^{n} 2^{k}a_{k} = 3$
$\sum_{k=1}^{n} 2^{k-1}ka_{k} = -2$
From the second sum, $a_1=-2-\sum_{k=2}^{n} 2^{k-1}ka_{k}$. Now we use that in the first sum:
$a_0+2a_1+\sum_{k=2}^{n} 2^{k}a_{k}=3$
$a_{0}=3-2\times (-2-\sum_{k=2}^{n} 2^{k-1} k a_{k})=7+\sum_{k=2}^{n} 2^ka_k(k-1)$
So all the solutions are in
$\begin{bmatrix} 7 \\ -2 \\ 0 \\ 0 \\ \vdots \\ 0 \end{bmatrix} +L(\{ \begin{bmatrix} 8 \\ -4 \\ 1 \\ 0 \\ \vdots \\ 0 \end{bmatrix}, \begin{bmatrix} 24 \\ -12 \\ 0 \\ 1 \\ \vdots \\ 0 \end{bmatrix}, \dots , \begin{bmatrix} 2^nn \\ -2^{n-1}n \\ 0 \\ 0 \\ \vdots \\ 1 \end{bmatrix}\})$ in respect to basis $\{1,x,x^2,\dots,x^n\}$.
Is this correct and is there any better way of solving this?
Your solution is correct. A smoother way would be the following:
As you have observed this is a linear inhomogeneous problem: Find all $u(x)\in{\mathbb R}[x]$ satisfying $u(2)=3$, $u'(2)=-2$. The philosophy for such problems is the following: The general solution is the sum of the general solution of the associated homogeneous problem plus a particular solution of the given inhomogeneous problem.
The associated homogeneous problem is the following: Find all polynomials $u(x)$ satisfying $$u(2)=0,\quad u'(2)=0\ .$$ These are obviouslay the polynomials having $(x-2)^2$ as a factor: $$u_{\rm hom}(x)=(x-2)^2 p(x),\qquad p(x)\in{\mathbb R}[x]\ .$$ A particular solution can be guessed as $$u_{\rm part}(x)=3-2(x-2)=7-2x\ .$$ It follows that the general solution of your problem is given by $$u(x)=u_{\rm hom}(x)+u_{\rm part}(x)=(x-2)^2 p(x)+7-2x,\qquad p(x)\in{\mathbb R}[x]\ .$$