Find ODE , with constant-coefficient ,homogeneous , with minmum power,linear such that $y_1(x) = xe^x$, $ y_2(x) = \sin^2(x)$ is soultions for it.
how can one construct such ODE ?
my trial :
by finding the roots of the unique polynomal for the Homogeneous ODE :
since $y_1(x) = xe^x$ is solution $\rightarrow$ $(r-1)(r-1)$
since $ y_2(x) = \sin^2(x)$ is solution we rewrite as $ y_2(x) = \frac{1-\cos(2x)}{2}$ so $ 1 , \cos(2x)$ is solutions .
we then write $(r)(r^2 + 4)$ so our polynomal is :
$F(r) = (r-1)(r-1)(r)(r^2+4)$ after simplfying :
$F(r) = r^5 - 2r^4 + 5r^3 - 8r^2 + 4r$
so $ y^{(5)}-2y^{(4)}+5y^{'''}-8y^{''}+4y^{'} = 0 $ is our ODE.
not sure if the solution is right , also not sure if its the minimum powers. much thanks for help
Assuming $\mathcal{D}$ linear we have
$$ \mathcal{D}y_1 = 0\\ \mathcal{D}y_2 = 0 $$
and
$$ \mathcal{D}(y_1+y_2) = \mathcal{D}y_1 +\mathcal{D}y_2 = 0 $$
then $y = y_1+y_2$ satisfies as well the DE. Considering now the more general problem with $\mathcal{D}$ invertible
$$ \mathcal{D}(y) = u\Rightarrow y = \mathcal{D}^{-1}u $$
or using the Laplace transform
$$ Y(s) = \mathcal{L}\left(xe^x+\sin^2 x\right) = \frac{2}{s \left(s^2+4\right)}+\frac{1}{(s-1)^2} = \frac{s^3+2 s^2+2}{(s-1)^2 s \left(s^2+4\right)}=G(s) U(s) $$
making now $u = \delta(t)$ associated to initial conditions we have
$$ G(s) = \mathcal{L}\left(\mathcal{D}^{-1}\right) $$
hence
$$ \mathcal{D} = \partial^{(5)}_t-2\partial^{(4)}_t+5\partial^{(3)}_t-8\partial^{(2)}_t+4\partial_t $$
so your answer is correct.