JEE MAIN - Mathematics (2002 - No. 25)
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
$$\,{\left( {1 + 3{{dy} \over {dx}}} \right)^{2/3}} = 4{{{d^3}y} \over {d{x^3}}}$$ are
$$\left( {1,{2 \over 3}} \right)$$
$$(3, 1)$$
$$(3,3)$$
$$(1,2)$$
Explanation
$${\left( {1 + 3{{d\,y} \over {d\,x}}} \right)^2} =$$ (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
$$ \Rightarrow {\left( {1 + 3{{d\,y} \over {d\,x}}} \right)^2} = 16{\left( {{{{d^3}y} \over {d\,{x^3}}}} \right)^3}$$
$$ \therefore $$ Order = 3 and Degree = 3
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