Let
$$
y=x^2 \exp (3x) + \sin x
$$
be solution of initial value problem with constant coefficients then what is the least possible order of differential equation if y solves the homogeneous linear differential equation with constant coefficients?
Is it 2 or 3 or 4 or 5?
I know that the first term of the solution corresponds to the factor $$(D-3) ^3$$ & the second term corresponds to the factor $$((D^2 )+1)$$ of D.E $$f(D)y =0$$ where
$f(D)$ is linear differential order.
So the D.E is $$[ ((D-3) ^3) ((D^2) +1) ]y = 0$$
Further solving the above D.E gives order =5.
So the answer comes to be 5.
It is asked to find least possible order and order is nothing but the highest order of differential coefficient. Then why the word 'least possible' used here?
What role it does in this question?
Best Answer
You have an argument that $$ f(D)=(D-3)^3(D^2+1)P(D) $$ for some polynomial $P$. Thus, the degree of $f$ is at least $5$. But, we do not know if there are other solutions in $f(D)y=0$, origin from $P$. Thus, we know that the least degree of $f$ is $5$, but we have no upper bound.
For example, the function you mention also solves $$ (D-3)^3(D^2+1)(D+2)(D-7)y=0. $$ and $$ (D-3)^3(D^2+1)y=0. $$ The first one of these has degree 7, the second degree 5.
Does this answers your question completely?