Another DE problem.

So, this is our DE -

$\displaystyle y''' + 3y'' - 4y' - 12y = 0$

Doing Laplace transform both sides

$\displaystyle L(y''') + 3L(y'') - 4L(y') - 12(L(y) = 0$

We can find Laplace transform for ${f}^{(n)}(x)$ through integrating by parts.

A general formula is

$\displaystyle L({f}^{(n)}(x)) = {s}^{n}L(f(x)) - {s}^{n-1}f(0) - {s}^{n-2}f'(0) -.... - {f}^{(n-1)}(0)$

Hence, our DE becomes

$\displaystyle {s}^{3}L(y) - {s}^{2}y(0) - sy'(0) - y''(0) + 3{s}^{2}L(y) - 3sy(0) - 3sy'(0) -4sL(y) + 4y(0) - 12L(y) = 0$

$\displaystyle L(y) = \frac{y''(0) + y'(0)[4s] + y(0)[{s}^{2}+3s-4]}{{s}^{3} + 3{s}^{2}-4s-12}$

$\displaystyle y = {L}^{-1}\left(\frac{y''(0) + y'(0)[4s] + y(0)[{s}^{2}+3s-4]}{{s}^{3} + 3{s}^{2}-4s-12}\right)$

$\displaystyle y = 2{e}^{-3t} + 4sinh(2t) + 3cosh(2t)$

And you are done!

Characteristic polynomial = $x^3+3x^2-4x-12$

Solutions of characteristic polynomial = -3, -2, 2

$\therefore y= A e^{-3t} + Be^{-2t} +C e^{2t}$

Substituting the given initial values after taking the appropriate derivatives, we can form the following system of linear equations:

$\begin{bmatrix} 1 & 1 & 1 \\ -3 & -2 & 2 \\ 9 & 4 & 4 \end{bmatrix}\begin{bmatrix} A \\ B \\ C \end{bmatrix}=\begin{bmatrix} 5 \\ 2 \\ 30 \end{bmatrix}$

The solution of this system is given by:

$\begin{bmatrix} A \\ B \\ C \end{bmatrix}=\begin{bmatrix} 2 \\ -\frac { 1 }{ 2 } \\ \frac { 7 }{ 2 } \end{bmatrix}$

So this means that the solution of the differential equation is:

$y = 2 e^{-3t} -\frac 1 2 e^{-2t} +\frac 7 2 e^{2t}$

From the identity $e^{at} = cosh(at)+sinh(at)$ we get,

$y = 2e^{-3t}+3cosh(2t)+4sinh(2t)$

$\therefore$

$a=2;\quad c=4;\quad e=3;\\ b=-3;\quad d=2;\quad f=2.$

and the value of the determinant= $\boxed5.$