Question

Let be an matrix anda scalar. Consider the following two systems:

$$\begin{gathered} \frac{\mathbf{d}\overrightarrow{\mathbf{x}}}{\mathbf{dt}}\mathbf{=}\mathbf{A}\overrightarrow{\mathbf{x}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\left(l\right) \\ \frac{\mathbf{d}\overrightarrow{\mathbf{c}}}{\mathbf{dt}}\mathbf{=}\left(A+{\mathrm{kl}}_n\right)\overrightarrow{\mathbf{c}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\left(\mathrm{ll}\right) \end{gathered}$$

Show that if $\overrightarrow{\mathbf{x}}\left(t\right)$ is a solution of the system (l)then role="math" localid="1659701582223" $\overrightarrow{\mathbf{c}}\left(t\right)\mathbf{=}{\mathit{e}}^{\mathbf{k}\mathbf{t}}\overrightarrow{\mathbf{x}}\left(t\right)$is a solution of the system (ll).

Short answer

Yes,$\overrightarrow{c}\left(t\right)=e^{kt}\overrightarrow{x}\left(t\right)$ is a solution of the system $\frac{\mathrm{d}\overrightarrow{\mathrm{c}}}{\mathrm{dt}}=(\mathrm{A}+{\mathrm{kl}}_{\mathrm{n}})\overrightarrow{\mathrm{c}}$.

Step-by-step solution

Determine the equation of the solution.

Consider$\overrightarrow{x}\left(t\right)$and$\overrightarrow{c}\left(t\right)$is a solution of the system$\frac{d\overrightarrow{x}}{dt}=A\overrightarrow{x}$and$\frac{\mathrm{d}\overrightarrow{\mathrm{c}}}{\mathrm{dt}}=\left(\mathrm{A}+{\mathrm{kl}}_{\mathrm{n}}\right)\overrightarrow{\mathrm{c}}$respectively.

Assume${\lambda }_1,{\lambda }_2,{\lambda }_3,\dots ,{\lambda }_n$be the Eigen values of the matrix A then there exist Eigen vectors${\overrightarrow{v}}_1,{\overrightarrow{v}}_2.{\overrightarrow{v}}_3,\dots ,{\overrightarrow{v}}_n$such that ${\overrightarrow{x}}_1\left(t\right)=c_1{e}^{{\lambda }_{1}t}{\overrightarrow{v}}_1+c_2{e}^{{\lambda }_{2}t}{\overrightarrow{v}}_2+...+c_n{e}^{{\lambda }_{n}t}{\overrightarrow{v}}_n$and$\overrightarrow{c}\left(t\right)=c_1{e}^{\left({\lambda }_1+k\right)}{\overrightarrow{v}}_1+c_2{e}^{\left({\lambda }_2+k\right)}{\overrightarrow{v}}_2+...+c_n{e}^{\left({\lambda }_n+k\right)}{\overrightarrow{v}}_n$where$c_1,c_2,\dots ,c_n$is constant.

If x(t) is the solution of the linear system localid="1659702528208" $\frac{d\overrightarrow{x}}{dt}=A\overrightarrow{x}$then${\overrightarrow{x}}_1\left(t\right)=c_1{e}^{{\lambda }_{1}t}{\overrightarrow{v}}_1+c_2{e}^{{\lambda }_{2}t}{\overrightarrow{v}}_2+...+c_n{e}^{{\lambda }_{n}t}{\overrightarrow{v}}_n$where${\lambda }_1,{\lambda }_2,{\lambda }_3,\dots ,{\lambda }_n$be the Eigen values and of$n\times n$matrix A.

Show that c→(t)=ektx→(t) is a solution of the system.

Simplify the equation${\overrightarrow{x}}_1\left(t\right)=c_1{e}^{{\lambda }_{1}t}{\overrightarrow{v}}_1+c_2{e}^{{\lambda }_{2}t}{\overrightarrow{v}}_2+...+c_n{e}^{{\lambda }_{n}t}{\overrightarrow{v}}_n$as follows.

$$\begin{gathered} \overrightarrow{c}\left(t\right)=c_1{e}^{\left({\lambda }_1+k\right)}{\overrightarrow{v}}_1+c_2{e}^{\left({\lambda }_2+k\right)}{\overrightarrow{v}}_2+...+c_n{e}^{\left({\lambda }_n+k\right)}{\overrightarrow{v}}_n \\ \overrightarrow{c}\left(t\right)=c_1{e}^{{\lambda }_{1}t}{e}^{kt}{\overrightarrow{v}}_1+c_2{e}^{{\lambda }_{2}t}{e}^{kt}{\overrightarrow{v}}_2+...+c_n{e}^{{\lambda }_{n}t}{e}^{kt}{\overrightarrow{v}}_n \\ \overrightarrow{c}\left(t\right)=e^{kt}\left\{c_1{e}^{{\lambda }_{1}t}{\overrightarrow{v}}_1+c_2{e}^{{\lambda }_{2}t}{\overrightarrow{v}}_2+...+c_n{e}^{{\lambda }_{n}t}{\overrightarrow{v}}_n\right\} \\ \overrightarrow{c}\left(t\right)=e^{kt}\overrightarrow{x}\left(t\right) \end{gathered}$$

By the definition of solution of the linear system,$\overrightarrow{c}\left(t\right)=e^{kt}\overrightarrow{x}\left(t\right)$is a solution.

Hence, if $\overrightarrow{x}\left(t\right)$and$\overrightarrow{c}\left(t\right)$is a solution of the system$\frac{d\overrightarrow{x}}{dt}=A\overrightarrow{x}$and$\frac{\mathrm{d}\overrightarrow{\mathrm{c}}}{\mathrm{dt}}=\left(\mathrm{A}+{\mathrm{kl}}_{\mathrm{n}}\right)\overrightarrow{\mathrm{c}}$respectively then is a solution.