First order differential equation with non constant coefficients I have the following system :
$$\begin{cases}(t^2+1)x'(t)=tx+y+2t^2+1\\(t^2+1)y'(t)=-x+ty+3t\end{cases}$$
How can it be solved ?

What I have tried so far :


*

*polynomials of the first, second degree as solutions - didn't work

*One can notice that if we use $X=\begin{bmatrix}x\\y\end{bmatrix},A=\begin{bmatrix}t&1\\-1&t\end{bmatrix},B=\begin{bmatrix}2t^2-1\\3t\end{bmatrix}$ then the system becomes $(t^2+1)X'=AX+B$, and $t^2+1=\det A$. I'm pretty sure that this last result is supposed to help, but I haven't been able to find a way to use it.


We are given:
$$\begin{cases}  (t^2+1)x'(t)=tx+y+2t^2+1\\ (t^2+1)y'(t)=-x+ty+3t\end{cases}$$
From the first equation $(t^2+1)x'(t)=tx+y+2t^2+1$, we have:
$$\tag 1 y = t^2 x' + x' - t x - 2 t^2 -1$$
Taking the derivative of $(1)$, yields:
$$\tag 2 y' = t^2 x'' + x'' + t x' -x - 4 t$$
Substituting $(1)$ and $(2)$ into the original second equation yields:
$$\tag 3 (t^2+1)((t^2+1)x'' + t x' - x - 4 t) = -x +t((t^2+1) x' -  t x - 2 t^2 -1) + 3 t$$
Simplifying $(3)$, yields:
$$\tag 4 x'' = 2 \dfrac{t^3}{(t^2+1)^2} + 6 \dfrac{t}{(t^2+1)^2}$$
Integrating $(4)$ twice yields:
$$x(t) = c_1 + c_2 t + t \ln(t^2 + 1)$$
We can now use this result and $(1)$ to yield:
$$y(t) = -c_1 t + c_2 -1 + \ln(t^2+1)$$
