Exercise · ZW-LINEAR-SYSTEM-AND-EIGENVALUES-007 · ★★☆ Intermediate
For \(A=\begin{pmatrix}0&1\\-2&-3\end{pmatrix}\), the solution of \(\mathbf x'=A\mathbf x\) with \(\mathbf x(0)=(1,0)\) has first component:
Choose one option.
A \(e^{-t}-2e^{-2t}\) B \(2e^{t}-e^{2t}\) C \(2e^{-t}-e^{-2t}\) D \(e^{-t}+e^{-2t}\) Show answer ← Back to the question
Explanation The eigenvalues are \(-1\) and \(-2\), with vectors \((1,-1)\) and \((1,-2)\). From \(c_1+c_2=1\) and \(-c_1-2c_2=0\) we get \(c_1=2,\ c_2=-1\).
Homogeneous system with a constant \(n\times n\) matrix \(A\), whose elementary solutions are built from an eigenvalue \(\lambda\) and an eigenvector \(\mathbf v\) of \(A\).
\[\mathbf x'=A\mathbf x,\qquad \mathbf x(t)=e^{\lambda t}\mathbf v,\quad A\mathbf v=\lambda\mathbf v\]
More practice problems: linear system and eigenvalues