Question

Consider a linear system $\frac{\mathbf{d}\overrightarrow{\mathbf{x}}}{\mathbf{dt}}=\mathbf{A}\overrightarrow{\mathbf{x}}$ of arbitrary size. Suppose${\overrightarrow{\mathbf{x}}}_{\mathbf{1}}\left(t\right)$ is a solution of the system and k is an arbitrary constant. Is $\overrightarrow{\mathbf{x}}\left(\mathbf{t}\right)=\mathbf{k}{\overrightarrow{\mathbf{x}}}_1\left(\mathbf{t}\right)$a solution as well? How do you know?

Short answer

Yes, $\overrightarrow{x}\left(t\right)=k{\overrightarrow{x}}_1\left(t\right)$would be the solution as well since$\frac{d\left(k{\overrightarrow{x}}_1\right)}{dt}=k\frac{d{\overrightarrow{x}}_1}{dt}$ .

Step-by-step solution

The matrix form of the derivatives

The rates of change of the components of the state vector $\overrightarrow{x}\left(t\right)$ , or its derivatives $\frac{d{x}_1}{dt},\frac{d{x}_2}{dt},\dots ,\frac{d{x}_n}{dt}$.

When these rates depend linearly on $x_1,x_2,\dots ,x_n$, write the matrix equation as$\frac{d\overrightarrow{x}}{dt}=A\overrightarrow{x}$

.

Check whether x→t=kx→1tis a solution as well

When ${\overrightarrow{x}}_1\left(t\right)$would be a solution of the linear system $\frac{d\overrightarrow{y}}{dt}=A\overrightarrow{y}$would satisfy this matrix equation.

Consider k as an arbitrary constant. Utilize the linearity of the differentiation as follow:

$\begin{array}{rcl}\frac{d\overrightarrow{x}}{dt}& =& \frac{d\left(k{\overrightarrow{x}}_1\right)}{dt}\\ & =& k\frac{d{\overrightarrow{x}}_1}{dt}\\ & =& kA{\overrightarrow{x}}_1\\ & =& Ak{\overrightarrow{x}}_1\\ & =& A\overrightarrow{x}\end{array}$

Therefore,$\overrightarrow{x}\left(t\right)=k{\overrightarrow{x}}_1\left(t\right)$ would be a solution of the given system.

Thus, yes,$\overrightarrow{x}\left(t\right)=k{\overrightarrow{x}}_1\left(t\right)$ would be the solution as well since$\frac{d\left(k{\overrightarrow{x}}_1\right)}{dt}=k\frac{d{\overrightarrow{x}}_1}{dt}$ .