Examine the stability of the trivial solution of the…
Examine the stability of the trivial solution of the equation $$y^{(4)}+y'''+3y''+2y'+y=0.$$
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
$$P(\lambda)=\lambda^4+\lambda^3+3\lambda^2+2\lambda+1, \qquad a_0=1,a_1=1,a_2=3,a_3=2,a_4=1.$$
Step 2 · Hurwitz determinants
$$\Delta_1=1,\qquad \Delta_2=\begin{vmatrix}1&2\\1&3\end{vmatrix}=1,\qquad \Delta_3=\begin{vmatrix}1&2&0\\1&3&1\\0&1&2\end{vmatrix}=1,\qquad \Delta_4=a_4\Delta_3=1.$$ All four Hurwitz determinants are strictly positive.
Step 3 · Routh array (verification)
$$\begin{array}{c|ccc} \lambda^4 & 1 & 3 & 1\\ \lambda^3 & 1 & 2 & 0\\ \lambda^2 & 1 & 1 & 0\\ \lambda^1 & 1 & 0 & 0\\ \lambda^0 & 1 & 0 & 0 \end{array}$$ First column $1,1,1,1,1$: all positive, no sign changes.
Step 4 · Conclusion
$$\boxed{ \text{The trivial solution is \textbf{asymptotically stable}.} }$$ All Hurwitz determinants $\Delta_1=\Delta_2=\Delta_3=\Delta_4=1$ are positive, confirming all four roots of $P(\lambda)$ lie in the open left half-plane.
Summary Table
| Quantity | Value |
|---|---|
| Eigenvalues | — |
| Classification | asymptotically stable |
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: