The order and degree of the differential equation$\frac{{{d^2}y}}{{d{x^2}}} + {\left( {\frac{{dy}}{{dx}}} \right)^{1/4}} + {x^{1/5}} = 0$
respectively, are

Given that, $\frac{{{d^2}y}}{{d{x^2}}} + {\left( {\frac{{dy}}{{dx}}} \right)^{1/4}} = - {x^{1/5}}$

$\Rightarrow$ $\frac{{{d^2}y}}{{d{x^2}}} + {\left( {\frac{{dy}}{{dx}}} \right)^{1/4}} = - {x^{1/5}}$

$\Rightarrow$ ${\left( {\frac{{dy}}{{dx}}} \right)^{1/4}} = - \left( {{x^{1/5}} + \frac{{{d^2}y}}{{d{x^2}}}} \right)$

On squaring both sides,

we get
${\left( {\frac{{dy}}{{dx}}} \right)^{1/2}} = {\left( {{x^{1/5}} + \frac{{{d^2}y}}{{d{x^2}}}} \right)^2}$

Again, on squaring both sides,

we have
$\frac{{dy}}{{dx}} = {\left( {{x^{1/5}} + \frac{{{d^2}y}}{{d{x^2}}}} \right)^4}$

order $= 2$, degree $= 4$