Question

The state equation of a second order system is $\dot{x}(t)=Ax(t)$, with $x(0)$ as the initial condition. Suppose $\lambda_1$ and $\lambda_2$ are two distinct eigenvalues of $A$ and $v_1$ and $v_2$ are the corresponding eigenvectors. For constants $\alpha_1$ and $\alpha_2$, the solution $x(t)$ of the state equation is _____________

• A. $\displaystyle \sum_{i=1}^{2}\alpha_i e^{\lambda_i t} v_i$
• B. $\displaystyle \sum_{i=1}^{2}\alpha_i e^{2\lambda_i t} v_i$
• C. $\displaystyle \sum_{i=1}^{2}\alpha_i e^{3\lambda_i t} v_i$
• D. $\displaystyle \sum_{i=1}^{2}\alpha_i e^{4\lambda_i t} v_i$

Hint:

Remember $e^{At}$: if $A$ has distinct eigenpairs $(\lambda_i,v_i)$, then $x(t)=\sum \alpha_i e^{\lambda_i t}v_i$; constants $\alpha_i$ are chosen to meet $x(0)$.

Answer 1

The Correct Option is A

Step 1: Form of solutions along eigenvectors.
If $Av_i=\lambda_i v_i$, try $x_i(t)=c_i e^{\lambda_i t} v_i$. Then \[ \dot{x}_i(t)=c_i \lambda_i e^{\lambda_i t} v_i = A(c_i e^{\lambda_i t} v_i)=A x_i(t), \] so each $e^{\lambda_i t}v_i$ satisfies $\dot{x}=Ax$. 
Step 2: Superposition.
Because the system is linear, the general solution (for distinct eigenvalues, hence diagonalizable) is \[ x(t)=\alpha_1 e^{\lambda_1 t} v_1+\alpha_2 e^{\lambda_2 t} v_2 = \sum_{i=1}^{2}\alpha_i e^{\lambda_i t} v_i. \] Step 3: Eliminate other options.
Exponents $2\lambda_i,\ 3\lambda_i,\ 4\lambda_i$ (options B–D) do not satisfy $\dot{x}=Ax$ unless $\lambda_i=0$. Hence they are incorrect. \[ \boxed{\displaystyle x(t)=\sum_{i=1}^{2}\alpha_i e^{\lambda_i t} v_i} \]