The order and degree of the differential equation [,( 1 + 3dy dx )^2/3 = 4d^3y dx^3] are

Question

The order and degree of the differential equation

$$\,{\left( {1 + 3{{dy} \over {dx}}} \right)^{2/3}} = 4{{{d^3}y} \over {d{x^3}}}$$ are

Accepted Answer

$${\left( {1 + 3{{d\,y} \over {d\,x}}} \right)^2} =$$ (4)3$${\left( {{{{d^3}y} \over {d\,{x^3}}}} \right)^3}$$

$$ \Rightarrow {\left( {1 + 3{{d\,y} \over {d\,x}}} \right)^2} = 16{\left( {{{{d^3}y} \over {d\,{x^3}}}} \right)^3}$$

$$ \therefore $$ Order = 3 and Degree = 3

Solutions · 1