What are the order and degree respectively of the differential equation $$y = x\frac{{dy}}{{dx}} + \frac{{dx}}{{dy}}\,?$$
A.
$$1,\,1$$
B.
$$1,\,2$$
C.
$$2,\,1$$
D.
$$2,\,2$$
Answer :
$$1,\,2$$
Solution :
The given differential equation is $$y = x\frac{{dy}}{{dx}} + \frac{{dx}}{{dy}}$$
Multiplying both the sides by $$\frac{{dy}}{{dx}}$$ we get
$$\eqalign{
& \left( {\frac{{dy}}{{dx}}} \right)y = x{\left( {\frac{{dy}}{{dx}}} \right)^2} + 1 \cr
& \Rightarrow x{\left( {\frac{{dy}}{{dx}}} \right)^2} - y\left( {\frac{{dy}}{{dx}}} \right) + 1 = 0 \cr} $$
Hence, order and degree of differential equation are $$1$$ and $$2.$$
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-