Number of values of m in N for which y = e^{m | vedclass

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(C) The given differential equation is $D^3y - 3D^2y - 4Dy + 12y = 0$. Substituting $y = e^{mx}$,we get $Dy = me^{mx}$,$D^2y = m^2e^{mx}$,and $D^3y =

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$2$

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(C) The given differential equation is $D^3y - 3D^2y - 4Dy + 12y = 0$. Substituting $y = e^{mx}$,we get $Dy = me^{mx}$,$D^2y = m^2e^{mx}$,and $D^3y = m^3e^{mx}$. Substituting these into the equation: $m^3e^{mx} - 3m^2e^{mx} - 4me^{mx} + 12e^{mx} = 0$. Since $e^{mx} 
eq 0$,we have the characteristic equation: $m^3 - 3m^2 - 4m + 12 = 0$. Factoring the equation: $m^2(m - 3) - 4(m - 3) = 0 \Rightarrow (m^2 - 4)(m - 3) = 0$. This gives $(m - 2)(m + 2)(m - 3) = 0$. The roots are $m = 2, m = -2, m = 3$. Since $m \in N$ (natural numbers),we only consider $m = 2$ and $m = 3$. Thus,there are $2$ such values.

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

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