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Consider the differential equation $\mathbf{f}\mathbf{\text{'}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{-}\mathbf{af}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{g}\mathbf{\left(}\mathbf{t}\mathbf{\right)}$where$\mathit{g}\mathbf{\left(}\mathit{t}\mathbf{\right)}$is a smooth function and 'a'is a constant. Then the general solution will berole="math" localid="1660806568260" $\mathbf{f}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}{\mathbf{e}}^{\mathbf{at}}\mathbf{∫}{\mathbf{e}}^{\mathbf{-}\mathbf{at}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{dt}$.

Consider the linear differential operator

$\mathrm{T}\left(\mathrm{f}\right)={\mathrm{f}}^{\left(\mathrm{n}\right)}+{\mathrm{a}}_{\mathrm{n}-1}{\mathrm{f}}^{\left(\mathrm{n}-1\right)}+...+{\mathrm{a}}_1\mathrm{f}\text{'}+{\mathrm{a}}_0\mathrm{f}$.

The characteristic polynomial of T is defined as

${\mathrm{p}}_{\mathrm{τ}}\left(\mathrm{λ}\right)={\mathrm{λ}}^{\mathrm{n}}+{\mathrm{a}}_{\mathrm{n}-1}\mathrm{λ}+...+{\mathrm{a}}_1\mathrm{λ}+{\mathrm{a}}_0$.

Consider the IVP $\frac{d\stackrel{→}{x}}{dt}=A\stackrel{→}{x}$with $\stackrel{→}{x}\left(0\right)={\stackrel{→}{x}}_0$where A is an upper triangular$n×n$ matrix with m distinct diagonal entries. ${\mathrm{λ}}_1,….,{\mathrm{λ}}_m$

Hence by the above definitions we get the solution as follows

$x_i\left(t\right)=P_1\left(t\right)e^{{\mathrm{λ}}_{1}t}+...+P_m\left(t\right)e^{{\mathrm{λ}}_{m}t}$

Hence the solution.

For a system$\frac{d\stackrel{→}{x}}{dt}=A\stackrel{→}{x}$, here A is the matrix form.

The zero state is an asymptotically stable equilibrium solution if and only if the real parts of all eigen values of A are negative.

Here A has no real eigen value, which means the real part of all the eigen values are negative.

Thus, the system$\frac{d\stackrel{→}{x}}{dt}=A\stackrel{→}{x}$is stable.