The differential equation representing the family of curves y2=a(ax+b) where a and b are arbitrary constants, is of
A
order 1,degree 1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
order 1,degree 3
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
order 2,degree 3
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
order 1,degree 4
No worries! We‘ve got your back. Try BYJU‘S free classes today!
E
order 2,degree 1
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution
The correct option is D order 2,degree 1 we have y2=a(ax+b) ⇒y2=a2x+ab On differentiating w.r.t x we get 2ydydx=a2 ⇒ydydx=a22 ⇒yd2ydx2+dydx.dydx=0 y.y2+(y1)2=0 Hence order and degree of the given equation is 2 and 1, respectively