Number of values of $m$ $\in N$ for which $y = e^{m x}$ is a solution of the differential equation $D^{3} y - 3 D^{2} y - 4 D y + 12 y = 0$ , is

0

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1

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2

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more than

2

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Solution

The correct option is C $2$
Let,

$y = e^{m x}$

$t h e n D (y) = m e^{m x}, D^{2} (y) = m^{2} e^{m x}, D^{3} (y) = m^{3} e^{m x}$

then given
$D^{3} (y) - 3 D^{2} (y) - 4 D (y) + 12 y = 0$

$\Rightarrow m^{3} e^{m x} - 3 m^{3} e^{m x} - 4 m e^{m x} + 12 e^{m x} = 0$

$∴ m^{3} - 3 m^{2} - 4 m + 12 = 0 (∵ e^{m x} \neq 0)$

$\Rightarrow (m - 2) (m - 3) (m + 2) = 0$

$\Rightarrow m = 2, 3, - 2$

Hence, the number of values of m $\in$ N will be $2$ ......... $(∵ - 2 d o e s n o t b e l o n g t o t h e s e t o f n a t u r a l n u m b e r s)$

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