The given function is
[tex]y=sin(kt)[/tex]
Differentiating
[tex]y'=kcos(kt)[/tex]
Again differentiating
[tex]y''=-k^2 sin(kt)[/tex]
Substituting the values of y '' and y in
[tex]y''+19y=0[/tex]
We will get
[tex]-k^2 sin(kt)+19sin(kt)=0 \\ sin(kt) (19-k^2)=0 sin(kt) =0, 19-k^2=0 \\ kt = \pi n , k =+- \sqrt{19} \\ k = \frac{ \pi n}{t} , - \sqrt{17}, \sqrt{17}[/tex]
Applying the derivatives, it is found that the differential equation is satisfied for:
[tex]k = n\frac{\pi}{2t}, n = \pm 1, \pm 3, \pm 5, \cdots[/tex]
[tex]k = \pm \sqrt{19}[/tex]
It is given by:
[tex]y^{\prime\prime} + 19y = 0[/tex]
The format of the solution is:
[tex]y = \sin{(kt)}[/tex]
The derivatives are given by:
[tex]y^{\prime} = k\cos{(kt)}[/tex]
[tex]y^{\prime\prime} = -k^2\sin{(kt)}[/tex]
Hence, applying the derivatives into the differential equation:
[tex]y^{\prime\prime} + 19y = 0[/tex]
[tex]-k^2\sin{(kt)} + 19\sin{(kt)} = 0[/tex]
[tex]\sin{(kt)}(-k^2 + 19) = 0[/tex]
Then:
[tex]\sin{(kt)} = 0[/tex]
[tex]kt = n\frac{\pi}{2}, n = \pm 1, \pm 3, \pm 5, \cdots[/tex]
[tex]k = n\frac{\pi}{2t}, n = \pm 1, \pm 3, \pm 5, \cdots[/tex]
Or also:
[tex]-k^2 + 19 = 0[/tex]
[tex]k^2 = 19[/tex]
[tex]k = \pm \sqrt{19}[/tex]
More can be learned about differential equations at https://brainly.com/question/14423176
