Question

Consider the two statements
Statement-1: $y=\sin$ kt satisfies the differential equation $y^{\prime \prime }+9y=0$.
Statement-2: $y=e^{kt}$ satisfy the differential equation $y^{\prime \prime }+y^{\prime \prime }-6y=0$
The value of $k$ for which both the statements are correct is

• A. -3
• B. 0
• C. 2
• D. 3

Answer 1

Answer: (A)

$\text{ S-1: }y=\sin kt,y^{\prime }=k\cos kt;y^{\prime \prime }=-k^2\sin kt$
$\therefore -k^2\sin kt+9\sin kt=0$
$\sin kt\left[9-k^2\right]=0\Rightarrow k=0,k=3,k=-3$
S-2: $y=e^{kt},y^{\prime }=k{e}^{kt};y^{\prime \prime }=k^2{e}^{kt}$
$\therefore k^2{e}^{kt}+k{e}^{kt}-6e^{kt}=0$
$e^{kt}\left[k^2+k-6\right]=0$
$(k+3)(k-2)=0$
$k=-3\text{ or }2$
common value is $k=-3$