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Example Solve the following initial value problem．

$\displaystyle y^{\prime\prime\prime} - 3y^{\prime} - 2y = 0, y(0) = y'(0) = y''(0) = 1$

Answer Let a solution be $y = e^{mx}$. Then the characteristic equation is $m^3 - 3m - 2 = 0 $.

$\displaystyle m^3 - 3m - 2 = (m + 1)(m^2 - m - 2) = ( m + 1)(m + 1)(m - 2)$

implies $m = -1, -1, 2$, where $m = -1,-1$ are repeated roots. Then the fundamental solutions are given by

$\displaystyle \{ e^{-x}, xe^{-x}, e^{2x} \} $

Thus, the general solution is

$\displaystyle y = c_{1}e^{-x} + c_{2}xe^{-x} + c_{3}e^{2x}$

We apply the initial conditions to get
$\displaystyle 1$ $\displaystyle =$ $\displaystyle c_{1} + c_{3}$  
$\displaystyle 1$ $\displaystyle =$ $\displaystyle -c_{1} + c_{2} + 2c_{3}$  
$\displaystyle 1$ $\displaystyle =$ $\displaystyle c_{1} - 2c_{2} + 4c_{3}$  

Using the Cramer's rule to solve this system of linear equations, we obtain，

$\displaystyle c_{1} = \frac{\begin{vmatrix}1&0&1\\ 1&1&2\\ 1&-2&4\end{vmatrix}}{\begin{vmatrix}1&0&1\\ 1&1&2\\ 1&-2&4\end{vmatrix}} = \frac{5}{9}$

Similarly, $c_{2} = \frac{2}{3}, c_{3} = \frac{4}{9}$.　Thus,

$\displaystyle y = \frac{5}{9}e^{-x} + \frac{2}{3}xe^{-x} + \frac{4}{9}e^{2x}$