It seems like an easy question to solve but could not figure it out:
If it is known that the following system of equations have a solution $x = x_1(t)$, where $x_1(t)$ is a second order polynomial and $y= y_1(t)$ where $y_1(t)$ is a first order polynomial.
$$x' = (t-t^2)x + (t^3-t^2+t+1)y$$ $$y' = (1-t)x + (t^2-t+1)y$$
What is the general solution?
I tried to plug in $x=ax^2+bx+c$ and $y=mx+n$. However could not end up with a solution. I will be glad if you can help.
$$x' = (t-t^2)x + (t^3-t^2+t+1)y$$
$$y' = (1-t)x + (t^2-t+1)y$$
Letting $x = at^2+bt+c$ and $y = mt+n$, we get that
$$2at+b = (t-t^2)(at^2+bt+c) + (t^3-t^2+t+1)(mt+n) =$$ $$(m-a)t^4+(a-b-m+n)t^3+(b-c+m-n)t^2+(c+m+n)t+n$$
and
$$m = (1-t)(at^2+bt+c) + (t^2-t+1)(mt+n)$$
From the first equation, we get that $m-a=0$, $a-b-m+n=0$, $b-c+m-n=0$, $c+m+n=2a$, and $n=b$. From this, we get that $a = m = c$, and $b = n = 0$.
Plugging these back into the second equation, we get
$$a = (1-t)(at^2+a) + (t^2-t+1)at$$
which simplifies to
$$1 = t^2-t^3+1-t + t^3-t^2+t = 1$$
which is a degenerate equation.
So the general solution is $$x = k(t^2+1), y = kt$$ for all $k \ne 0$.