The differential equation representing the family of curves $${y^2} = 2c\left( {x + \sqrt c } \right),$$ where $$c > 0,$$ is a parameter, is of order and degree as follows :
A.
order 1, degree 2
B.
order 1, degree 1
C.
order 1, degree 3
D.
order 2, degree 2
Answer :
order 1, degree 3
Solution :
$$\eqalign{
& {y^2} = 2c\left( {x + \sqrt c } \right)\,.....({\text{i}}) \cr
& 2yy' = 2c.1\,\,{\text{or}}\,\,yy' = c\,.....({\text{ii}}) \cr
& \Rightarrow {y^2} = 2yy'\left( {x + \sqrt {yy'} } \right) \cr} $$
[On putting value of $$c$$ from (ii) in (i)]
On simplifying, we get
$${\left( {y - 2xy'} \right)^2} = 4yy{'^3}\,.....({\text{iii}})$$
Hence equation (iii) is of order 1 and degree 3
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-