Find the degree of the differential equation | vedclass

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(B) The given differential equation is $y_3^{2/3} + 2 + 3y_2 + y_1 = 0$. To find the degree,we must first eliminate the fractional exponent of the hig

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(B) The given differential equation is $y_3^{2/3} + 2 + 3y_2 + y_1 = 0$. To find the degree,we must first eliminate the fractional exponent of the highest order derivative. Rearranging the equation,we get $y_3^{2/3} = -(2 + 3y_2 + y_1)$. Cubing both sides to remove the fractional power,we obtain $(y_3^{2/3})^3 = (-(2 + 3y_2 + y_1))^3$,which simplifies to $y_3^2 = -(2 + 3y_2 + y_1)^3$. The highest order derivative present is $y_3$ (the third derivative),and its exponent after rationalizing the equation is $2$. Therefore,the degree of the differential equation is $2$. Hence,option $B$ is correct.

Find the degree of the differential equation $y_3^{2/3} + 2 + 3y_2 + y_1 = 0$.

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