Differential EquationHard
Question
The differential equation representing the family of curves y2 = 2c (x + √c), where c > 0, is a parameter, is of order and degree as follows:
Options
A.order 1, degree 2
B.order 1, degree 1
C.order 1, degree 3
D.order 2, degree 2
Solution
y2 = 2c(x + √c) ....(i)
2yy′ = 2c.1 or yy′ = c ....(ii)
y2 = 2yy′
[on putting value of c from (ii) in (i)]
On simplifying, we get
(y - 2xy′)2 = 4yy′3 ....(iii)
Hence equation (iii) is of order 1 and degree 3.
2yy′ = 2c.1 or yy′ = c ....(ii)
y2 = 2yy′
[on putting value of c from (ii) in (i)]On simplifying, we get
(y - 2xy′)2 = 4yy′3 ....(iii)
Hence equation (iii) is of order 1 and degree 3.
Create a free account to view solution
View Solution FreeMore Differential Equation Questions
The differential equation for the line y = mx + c is ( where c is arbitrary constant)-...The differential equation of the family of curves y2 = 4a (x + a) , where a is an arbitrary constant, is-...The differential equation of all ′Simple Harmonic Motions′ of given period is-...The solution of the differential equation (1 + x2) = x is-...The solution of = 0 is...