How to find general solution of a DE given its particular solutions? Can you suggest how can I find the general solution of the equation $f(x)y'' + g(x) y'+y = 1$ if $x^2$, $x$, $1$ are solutions of it. My idea is to make somehow a linear combination of the solutions like that $C_1 + C_2x +C_3x^2$ but it does require that this indeed general solution of the equation. Then I thought of finding the roots of the characteristic equation but I have functions, not constant coefficients.

Since $y=x$ is a solution, substituting gives
$$f(x)\times0+g(x)\times1+x=1$$
or $g(x)=1-x$.
Similarly, $$2f(x)+2x(1-x)+x^2=1$$
which gives $$f(x)=\dfrac{1}{2}(1-x)^2$$
so, your ED is 
$$(1-x)^2y''+2(1-x)y'+2y=2$$
To proceed, suppose $u=1-x$.
