How to get parameter value given the form of the homogeneous differential equation solution

Question

The question is: if I know the differential equation has a solution, which has the form of the quadratic polynomial, how do I get to solve the unknown of the equation? For example, for the differential equation $$(x^2-1)y''+\alpha y = 0$$ has a solution of the form : cubic polynomial. How to find all values of alpha for the equation.

Answer 1

Since a cubic $$ y=ax^3+bx^2+cx+d$$ satisfies $$(x^2-1)y''+\alpha y = 0$$

We substitute $y$ and $$y'' = 6ax +2b $$ in our equation to get $$(x^2-1) (6ax +2b)+\alpha (ax^3+bx^2+cx+d) = 0$$

Notice that coefficient of $x^3$ is $ 6a+\alpha a $ which should be zero, so we get $$ \alpha = -6$$