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The solutions of the differential equation $\mathit{f}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{+}\mathit{a}\mathit{f}\mathbf{\text{'}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{+}\mathit{b}\mathit{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathbf{0}$, where a and b are constants form a two dimensional subspace of the space${\mathit{C}}^{\mathbf{鈭�}}$of smooth functions.

Generally the solutions of the differential equations are

${\mathit{f}}^{\mathbf{\left(}\mathbf{n}\mathbf{\right)}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{+}{\mathit{a}}_{\mathbf{n}\mathbf{-}\mathbf{1}}{\mathit{f}}^{\mathbf{(}\mathbf{n}\mathbf{-}\mathbf{1}\mathbf{)}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{+}\mathbf{.}\mathbf{....}\mathbf{+}{\mathit{a}}_{\mathbf{1}}{\mathit{f}}^{\mathbf{1}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{+}{\mathit{a}}_{\mathbf{0}}\mathit{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathbf{0}$

Consider the differential equation as

$f\text{'}\text{'}\left(x\right)+8f\text{'}\left(x\right)鈭�20f\left(x\right)=0$

Take the exponential form of the equation be$f\left(x\right)=e^{kx}$

$\begin{array}{c}f\text{'}\text{'}\left(x\right)+8f\text{'}\left(x\right)鈭�20f\left(x\right)=0\\ k^2{e}^{kx}+8k{e}^{kx}鈭�20e^{kx}=0\\ (k^2+8k鈭�20)e^{kx}=0\\ k^2+10k鈭�2k鈭�20=0\end{array}$

Further simplification can be as follows

$\begin{array}{l}k(k+10)鈭�2(k+10)=0\\ (k+10)(k鈭�2)=0\\ k=2,鈭�10\end{array}$

The solutions are$k=2$and$k=鈭�10$be of the exponential form as$e^{2x}\text{鈥夆赌夆赌�}and\text{鈥夆赌�}{e}^{鈭�10x}$.

By a theorem we know that the solution space is two dimensional.

Thus, the linear independent functions$e^{2x}\text{鈥夆赌夆赌�}and\text{鈥夆赌�}{e}^{鈭�10x}$form a basis of V and all solutions of the form is$f\left(x\right)=C_1{e}^{2x}+C_2{e}^{鈭�10x}$.

Hence the solution.