What is the degree of the differential equation $${\left( {\frac{{{d^3}y}}{{d{x^3}}}} \right)^{\frac{2}{3}}} + 4 - 3\left( {\frac{{{d^2}y}}{{d{x^2}}}} \right) + 5\left( {\frac{{dy}}{{dx}}} \right) = 0\,?$$
A.
$$3$$
B.
$$2$$
C.
$$\frac{2}{3}$$
D.
Not defined
Answer :
$$2$$
Solution :
Degree of a differential equation is the power to which the highest derivative is raised when it is expressed as polynomial of derivatives.
Given equation is
$$\eqalign{
& {\left( {\frac{{{d^3}y}}{{d{x^3}}}} \right)^{\frac{2}{3}}} - 3\left( {\frac{{{d^2}y}}{{d{x^2}}}} \right) + 5\left( {\frac{{dy}}{{dx}}} \right) + 4 = 0 \cr
& \Rightarrow {\left( {\frac{{{d^3}y}}{{d{x^3}}}} \right)^{\frac{2}{3}}} = 3\frac{{{d^2}y}}{{d{x^2}}} - 5\frac{{dy}}{{dx}} - 4 \cr} $$
Cube on both side, $${\left( {\frac{{{d^3}y}}{{d{x^3}}}} \right)^2} = {\left[ {3\frac{{{d^2}y}}{{d{x^2}}} - 5\frac{{dy}}{{dx}} - 4} \right]^3}$$
Hence, degree $$ = 2.$$
Releted MCQ Question on Calculus >> Differential Equations
Releted Question 1
A solution of the differential equation $${\left( {\frac{{dy}}{{dx}}} \right)^2} - x\frac{{dy}}{{dx}} + y = 0$$ is-
If $$y\left( t \right)$$ is a solution $$\left( {1 + t} \right)\frac{{dy}}{{dt}} - ty = 1$$ and $$y\left( 0 \right) = - 1,$$ then $$y\left( 1 \right)$$ is equal to-