What is the degree of the differential equation $$y = x\frac{{dy}}{{dx}} + {\left( {\frac{{dy}}{{dx}}} \right)^{ - 1}}\,?$$
A.
$$1$$
B.
$$2$$
C.
$$ - 1$$
D.
Degree does not exist.
Answer :
$$2$$
Solution :
Given differential equation is $$y = x\frac{{dy}}{{dx}} + {\left( {\frac{{dy}}{{dx}}} \right)^{ - 1}}$$
Multiply by $$\frac{{dy}}{{dx}}\,;\,\,\,y\frac{{dy}}{{dx}} = x{\left( {\frac{{dy}}{{dx}}} \right)^2} + 1$$
Since power of highest order derivative is $$2$$
$$\therefore $$ 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-