Solution of $7$-th order ODE Let , $y=x\sin x+x^2$ be a solution of the $7$-th order ODE $$a_7y^{(7)}+a_6y^{(6)}+\cdots +a_0y=0.$$Then which of the following are correct ?
(A) $a_7+a_3=a_5$.
(B) $\displaystyle \sum_{i=0}^7 a_i=4$.
(C) $\displaystyle \sum_{i=0}^7 a_i=3$.
First I differentiate $y$ ,$7$-times and then put the values in the given equation. But from there I got no benefit.
Any hint please ..
Edit :
From StephenG's answer I get , $$a_0-a_2+a_4-a_6=0$$ $$a_1-a_3+a_5-a_7=0$$ $$a_1-3a_3-5a_5-7a_7=0$$ $$2a_2-4a_4+6a_6=0$$ $$a_0x^2+2x+2a_2=0$$Am I right now? From here how I can proceed ?
 A: Assuming the $a_k$ are all real we know that $\bar z$ is a solution to the characteristic equation given that $z$ is. 
By the nature of solutions to homogeneous linear differential equations the term $x\sin x$ implies that $\pm i$ are solutions to the characteristic equation. Moreover, the $x$ in front of $\sin x$ implies that the multiplicity is at least $2$. Thus, the characteristic polynomial has factors $(r-i)^2(r+i)^2=(r^2+1)^2$.
On the other hand, the $x^2$ part of the solution implies that $0$ is also a solution to the characteristic polynomial, and the degree $2$ in $x^2$ implies that the multiplicity is at least $3$. Thus, $r^3$ is also a factor of the characteristic polynomial.
We have found factors whose total degree add up to 7, and thus, since the order of the differential equation is 7, we have found the characteristic polynomial $p$ up to a multiplicative constant $a(=a_7)$,
$$
p(r)=a(r^2+1)^2r^3.
$$
Can you take it from here?
A: If you substitute in the expression for $y\left(x\right)$, you end up with a set of terms with different functions, $\sin \left(x\right), x\cos \left(x\right)$ and so on.  The coefficients of each term must be zero.
This gives you a set of simultaneous equations for the $a_k$ from which you can work out the solution you require ( and it's an easy set of equations to deal with ).
