Finding the general solution of a system of Differential Equations I need helping find the general solution to the following systen
$$ x'=2t^2+2-4x+6y $$
$$ y'=-2t^2-t+6-3x+5y $$
I know i need to turn the 2 equations into a matrix, but I can't figure out how to do it.
So far I have the matrix for the $x$ and $y$ values and for $t$, but I don't know what to do with the $+2$ and $+6$ in each equation.
 A: I presume you are using $x'$ to denote $\frac{\mathrm{d}x}{\mathrm{d}t}$. Now in general, you try to represent a coupled differential equation of $x$ and $y$ in the variable $t$ in the format
$$\begin{bmatrix} x'(t) \\ y'(t)\end{bmatrix} = A(t)\begin{bmatrix} x(t) \\ y(t)\end{bmatrix}+ \begin{bmatrix}f_1(t) \\ f_2(t)\end{bmatrix},$$
where $A(t)\in \mathbb{R}^{2\times2}$.
For instance, in your case, you would write the differential equation as
$$ \begin{bmatrix}x'(t) \\ y'(t)\end{bmatrix} = \begin{bmatrix}-4 & 6\\ -3 & 5\end{bmatrix}\begin{bmatrix}x(t) \\ y(t)\end{bmatrix} + \begin{bmatrix}2t^2+2 \\ -2t^2 -t +6\end{bmatrix}.$$
Since your $A$ matrix is time-invariant the solution of the above equation can be found easily. There are a couple of ways you can use, one way would be to transform $A$ using a similarity transformation to the Jordan cannonical form $A_J = PAP^{-1}$ and writing the above equation in terms
$$ \begin{bmatrix} z_1 \\ z_2 \end{bmatrix} = P \begin{bmatrix} x \\ y\end{bmatrix}, \qquad 
 \begin{bmatrix} f_1(t) \\ f_2(t) \end{bmatrix} = P\begin{bmatrix}2t^2+2 \\ -2t^2 -t +6\end{bmatrix},$$
as
$$ \begin{bmatrix} z_1'(t) \\ z_2'(t)\end{bmatrix} = A_J\begin{bmatrix} z_1(t) \\ z_2(t)\end{bmatrix}+ \begin{bmatrix}f_1(t) \\ f_2(t)\end{bmatrix}.$$
The solution to the above equation is straight forward. You can obtain back $x$ and $y$ back using
$$\begin{bmatrix} x \\ y \end{bmatrix} = P^{-1} \begin{bmatrix} z_1 \\ z_2\end{bmatrix}.$$
A: $$ x'=2t^2+2-4x+6y $$
$$ y'=-2t^2-t+6-3x+5y $$
You can also solve it this way. Substract both DE:
$$ x'-y'=4t^2+t-4-x+y $$
$$u'+u=4(t^2-1)+t$$
Where $u=x-y$. This is a first order linear DE that's easy to solve.
