The order and degree of ${\left(\frac{{\mathrm{d}}^4\mathrm{y}}{{\mathrm{dx}}^4}+\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}\right)}^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=\mathrm{a} \frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}$  are p,q then p + q = 

 

${\left(\frac{{\mathrm{d}}^4\mathrm{y}}{{\mathrm{dx}}^4}+\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}\right)}^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=\frac{{\mathrm{ad}}^2\mathrm{y}}{{\mathrm{dx}}^2}$

Let order is P and degree is 9.

Order is highest derivation 

$\therefore \mathrm{P}=4$

and the degree is power raised to the highest derivation of differentiation. 

if we open the bracket $\frac{{\mathrm{d}}^4\mathrm{y}}{{\mathrm{dx}}^4}$ get the power $\frac{5}{2}$

Hence the degree is $\frac{5}{2}$

$\therefore \mathrm{q}=\frac{5}{2}$

Now $\mathrm{p}+\mathrm{q}=4+\frac{5}{2}=\frac{13}{2}$