Correct Answer - Option 4 :

\(\frac{{{d^2}y}}{{d{x^2}}} + {a^2}y = 0\)
__Calculation:__

Given:

y = p cos (ax) + q sin (ax) ….(1)

Now differentiating both sides, we get

\(\Rightarrow \frac{{{\rm{dy}}}}{{{\rm{dx}}}} = {\rm{}} - {\rm{pa\;}}\sin \left( {{\rm{ax}}} \right) + {\rm{qa\;}}\cos \left( {{\rm{ax}}} \right)\)

Again differentiating both sides, we get

\(\Rightarrow \frac{{{{\rm{d}}^2}{\rm{y}}}}{{{\rm{d}}{{\rm{x}}^2}}} = {\rm{}} - {\rm{p}}{{\rm{a}}^2}\cos \left( {{\rm{ax}}} \right) - {\rm{q}}{{\rm{a}}^2}{\rm{\;}}\sin \left( {{\rm{ax}}} \right) = - {{\rm{a}}^2}\left( {{\rm{p\;cos\;}}\left( {{\rm{ax}}} \right){\rm{}} + {\rm{\;q\;sin\;}}\left( {{\rm{ax}}} \right)} \right)\)

From equation 1^{st},

\(\Rightarrow \frac{{{{\rm{d}}^2}{\rm{y}}}}{{{\rm{d}}{{\rm{x}}^2}}} = {\rm{}} - {{\rm{a}}^2}{\rm{y\;}}\)

\(\therefore \frac{{{{\rm{d}}^2}{\rm{y}}}}{{{\rm{d}}{{\rm{x}}^2}}} + {\rm{\;}}{{\rm{a}}^2}{\rm{y}}\)