The differential equation whose solution is `A x^2+B y^2=1` , where `A` and `B` are arbitrary constants, is of (a) second order and second degree (b) first order and second degree (c) first order and first degree (d) second order and first degree

Options

A. second order and second degree
B. first order and second degree
C. first order and first degree
D. second order and first degree

Answer

D

Explanation

We have,
`Ax^(2)+By^(2)=1" …(i)"`
Differentiating w.r. to x, we get
`2Ax+2By(dy)/(dx)=0 rArr Ax +By(dy)/(dx)=0" …(ii)"`
Differentiating w.r. to x, we get
`A+B((dy)/(dx))^(2)+By(d^(2)y)/(dx^(2))=0" …(iii)"`
Multiplying (iii) by x and subtracting (ii) from it, we get
`x((dy)/(dx))^(2)+xy(d^(2)y)/(dx^(2))-y(dy)/(dx)=0`
Clearly, it is a second order and first degree differential equation.