Exmple Solve the following initial value problem．

$$y^{\prime\prime} - 8y^{\prime} - 9y = 0, y(0) = y'(0) = 1$$

Answer Put $y = e^{mx}$. Then we obtain the characteristic equation $m^2 - 8m - 9 = 0$ by letting $y^{(n)} \Leftrightarrow m^n$。Thus, characteristic values are $m = -1, 9$ distinct real values. From this, the fundamental solutions are give by

$$\{e^{-x}, e^{9x} \}$$

This implies the general solution can be given by

$$y = c_{1}e^{-x} + c_{2}e^{9x}$$

Now applying initial conditions $y(0) = y'(0) = 1$, we obtain

$$1 = c_{1} + c_{2}$$

$$1 = -c_{1} + 9c_{2}$$

Solving this system, we have $c_{2} = \frac{1}{5}, c_{1} = \frac{4}{5}$T herefore,

$$y = \frac{4}{5}e^{-x} + \frac{1}{5}e^{9x}$$