Question

Consider the two statements
Statement-1: $y=\sin$ kt satisfies the differential equation $y^{\prime \prime}+9 y=0$.
Statement-2: $y=e^{k t}$ satisfy the differential equation $y^{\prime \prime}+y^{\prime \prime}-6 y=0$
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 k t, y^{\prime}=k \cos k t ; y^{\prime \prime}=-k^2 \sin k t $
$\therefore -k^2 \sin k t+9 \sin k t=0 $
$ \sin k t\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 }+ ke ^{ kt }-6 e ^{ kt }=0 $
$ e ^{ kt }\left[ k ^2+ k -6\right]=0$
$ ( k +3)( k -2)=0$
$ k =-3 \text { or } 2$
common value is $k =-3$