Question

if the below diff eq for f not equal to 0 is a family of circles then

Answer 1

\[d ^{2}y/dx ^{2} =g/f+c\]

Answer 2

options ::::

a) g,f hv sme sign

b>g,f hv opposite sign

c>mod g < mod f

d> mod g =mod f

Answer 3

I don't know what the question is

Answer 4

for the given ques nd a diff eq...select nd calculate the correct option

Answer 5

Well, the way to evaluate this equation in general is to integrate
y'' = g/f + c
y' = (g/f)x + cx + c1
y = (g/f + c)x^2 + c1x + c2

This then is the equation of a parabola.

So I don't see how to recover a circle from the ODE. In fact, let me say outright this is not the ODE that results from a circle.

Answer 6

hmm what if g/f + c = 1, then it must result into a circle right?

Answer 7

To see that, let's start with a circle
(x-f)^2 + (y-g)^2 = c^2

Then differentiating once
2(x-f) + 2(y-g)y' = 0 ----- (*)

and now again
2x + 2y'^2 + 2(y-g)y'' = 0

You can manipulate this now to obtain an equation only in y'' by substituting (*) into the second equation. Do that, and you won't recover the original ODE.

Answer 8

if g/f + c = 1 then y'' = 1
=> y' = x + c1
=> y = x^2/2 + c1x + c2

which is not a circle for any c1 and c2, but a parabola.

Answer 9

if g/f + c = 2 then it must circle for sure

y" = 2
y' = 2x + c1
y = x^2 + c1x + c2 :-D

I don't know, must be mathematics

Answer 10

No, that is NOT the equation of a circle. It is the equation of a parabola.

Answer 11

oh sorry please ignore me

Answer 12

In other words, I think there's something wrong with the question. Follow the procedure I outlined above to find the expression for y'' from a circle, and perhaps you can figure out what the question should have been.