The degree of the differential equation ${\left[ {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} \right]^{3/2}} = \frac{{{d^2}y}}{{d{x^2}}}$ is

Given that ${\left[ {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} \right]^{3/2}} = \frac{{{d^2}y}}{{d{x^2}}}$

On squaring both sides,

we get
${\left[ {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} \right]^3} = {\left( {\frac{{{d^2}y}}{{d{x^2}}}} \right)^2}$

So, the degree of differential equation is 2 .