Question

find the D square y by D x square + mu square y is equal to zero​

Answer 1

$: \implies \tt \:Evaluate : \: \dfrac{ {d}^{2}y }{ {dx}^{2} } + {\mu}^{2} \: y = 0$

$\large\underline\purple{\bold{Solution :- }}$

Its symbolic form is

$: \implies \tt \: ({D}^{2} + {\mu}^{2} )y = 0$

Its Auxiliary equation is

$: \implies \tt \: {D}^{2} + {\mu}^{2} = 0$

$: \implies \tt \: {D}^{2} = - {\mu}^{2}$

$: \implies \tt \: D \: = \: \pm \: i \: \mu$

So, solution is given by

$: \implies \bf \: y \: = \: a \: cos\mu \: x \: + \: b \: sin\mu \: x$