The degree of the differential equation {left | vedclass

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Question

(B) Given differential equation is ${\left( {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} \right)^{3/4}} = {\left( {\frac{{{d^2}y}}{{d{x^2}}}} \right

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(B) Given differential equation is ${\left( {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} \right)^{3/4}} = {\left( {\frac{{{d^2}y}}{{d{x^2}}}} \right)^{1/3}}$.To find the degree,we must eliminate the fractional exponents by raising both sides to the power of the least common multiple of the denominators,which is $12$.Raising both sides to the power of $12$:${\left[ {{\left( {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} \right)^{3/4}}} \right]^{12}} = {\left[ {{\left( {\frac{{{d^2}y}}{{d{x^2}}}} \right)^{1/3}}} \right]^{12}}$${\left( {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} \right)^9} = {\left( {\frac{{{d^2}y}}{{d{x^2}}}} \right)^4}$.The highest order derivative present is $\frac{{{d^2}y}}{{d{x^2}}}$,which is of order $2$.The degree of a differential equation is the power of the highest order derivative when the equation is expressed as a polynomial in derivatives.Here,the power of the highest order derivative $\frac{{{d^2}y}}{{d{x^2}}}$ is $4$.Therefore,the degree is $4$.

The degree of the differential equation ${\left( {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} \right)^{3/4}} = {\left( {\frac{{{d^2}y}}{{d{x^2}}}} \right)^{1/3}}$ is

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