Examine the stability of the trivial solution of the…
Examine the stability of the trivial solution of the equation $$y^{(5)}+2y^{(4)}+3y'''+2y''+y'+y=0.$$
Explanation of Method
Rather than solving the characteristic polynomial explicitly (which may have no rational or easily closed-form roots), the Routh–Hurwitz criterion is applied directly to its coefficients. The signs of a sequence of determinants built from those coefficients (equivalently, the signs in the first column of the Routh array) determine whether every root lies in the open left half-plane, without ever computing the roots themselves.
Step 1 · Characteristic polynomial
$$P(\lambda)=\lambda^5+2\lambda^4+3\lambda^3+2\lambda^2+\lambda+1.$$
Step 2 · Factorisation
Testing $\lambda=-1$ gives $P(-1)=0$, so $$P(\lambda)=(\lambda+1)(\lambda^4+\lambda^3+2\lambda^2+1)=(\lambda+1)Q(\lambda).$$
Step 3 · Routh--Hurwitz analysis of $Q(\lambda)=\lambda^4+\lambda^3+2\lambda^2+1$
The coefficient of $\lambda^1$ in $Q$ is $0$, violating the necessary all-positive-coefficients condition. The Routh array for $Q$ is $$\begin{array}{c|ccc} \lambda^4 & 1 & 2 & 1\\ \lambda^3 & 1 & 0 & 0\\ \lambda^2 & 2 & 1 & 0\\ \lambda^1 & -\tfrac12 & 0 & 0\\ \lambda^0 & 1 & 0 & 0 \end{array}$$ with first column $1,1,2,-\tfrac12,1$: two sign changes, so $Q$ has exactly 2 roots with positive real part. Independently, the Hurwitz determinant $\Delta_3^Q=-1<0$ confirms this.
Step 4 · Routh array for the full polynomial $P(\lambda)$
$$\begin{array}{c|ccc} \lambda^5 & 1 & 3 & 1\\ \lambda^4 & 2 & 2 & 1\\ \lambda^3 & 2 & \tfrac12 & 0\\ \lambda^2 & \tfrac32 & 1 & 0\\ \lambda^1 & -\tfrac56 & 0 & 0\\ \lambda^0 & 1 & 0 & 0 \end{array}$$ first column $1,2,2,\tfrac32,-\tfrac56,1$: two sign changes, confirming $P(\lambda)$ has exactly 2 roots with positive real part.
Step 5 · Conclusion
$$\boxed{ \text{The trivial solution is \textbf{unstable}.} }$$ The characteristic polynomial factors as $(\lambda+1)Q(\lambda)$ with $\lambda=-1$ stable but $Q(\lambda)$ contributing two roots of strictly positive real part (Hurwitz determinant $\Delta_3^Q=-1<0$, two sign changes in the Routh array), so the trivial solution is unstable.
Summary Table
| Quantity | Value |
|---|---|
| Eigenvalues | — |
| Classification | unstable |
Additional Notes
- The Routh–Hurwitz determinants (or, equivalently, the first column of the Routh array) provide a purely algebraic stability test that requires only the coefficients of the characteristic polynomial, with no need to locate its roots.
Result
The final classification is: