$\begin{array}{1 1} (A)\;y'' - y' - 6y = 0. \\(B)\;y'' - y' - y = 0. \\ (C)\;y'' - y - 6y' = 0. \\ (D)\;y''' - y' - 6y = 0. \end{array} $

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- After differentiating, to eliminate the arbitrary constants we can convert the equations, into a determinant form and expand along their rows or columns to eliminate the arbitrary constants.

Step 1:

Given : $y = ae^3x + be^{-2x}$ -----------(1)

Differentiate on both sides

$y' 3ae^{3x} - 2be^{-2x}$-------------(2)

Again differentate equ (2) on both sides

$y'' = 9ae^{3x} + 4be^{-2x}$-----------(3)

Step 2:

As the hint in the tool box says, we can write the three equations in the form a determinant to eliminate the arbitrary constants

$\begin{vmatrix}y&y'&y''\\1&3&9\\1&-2&4\end{vmatrix}=0$

Step 3:

On expanding the determinant along the rows we get

$y(12 + 18) - y'(4 - 9) + y''(4-9) = 0$

$30y +5y' -5y'' = 0$

Dividing throughout by -5 we get,

$y'' - y' - 6y = 0.$

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