anonymous
  • anonymous
if the below diff eq for f not equal to 0 is a family of circles then
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
\[d ^{2}y/dx ^{2} =g/f+c\]
anonymous
  • anonymous
options :::: a) g,f hv sme sign b>g,f hv opposite sign c>mod g < mod f d> mod g =mod f
anonymous
  • anonymous
I don't know what the question is

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
for the given ques nd a diff eq...select nd calculate the correct option
JamesJ
  • JamesJ
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.
anonymous
  • anonymous
hmm what if g/f + c = 1, then it must result into a circle right?
JamesJ
  • JamesJ
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.
JamesJ
  • JamesJ
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.
anonymous
  • anonymous
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
JamesJ
  • JamesJ
No, that is NOT the equation of a circle. It is the equation of a parabola.
anonymous
  • anonymous
oh sorry please ignore me
JamesJ
  • JamesJ
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.

Looking for something else?

Not the answer you are looking for? Search for more explanations.