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aditya.sinha

  • 4 years ago

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

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  1. aditya.sinha
    • 4 years ago
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    \[d ^{2}y/dx ^{2} =g/f+c\]

  2. aditya.sinha
    • 4 years ago
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    options :::: a) g,f hv sme sign b>g,f hv opposite sign c>mod g < mod f d> mod g =mod f

  3. alexray19
    • 4 years ago
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    I don't know what the question is

  4. aditya.sinha
    • 4 years ago
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    for the given ques nd a diff eq...select nd calculate the correct option

  5. JamesJ
    • 4 years ago
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    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.

  6. Ishaan94
    • 4 years ago
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    hmm what if g/f + c = 1, then it must result into a circle right?

  7. JamesJ
    • 4 years ago
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    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.

  8. JamesJ
    • 4 years ago
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    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.

  9. Ishaan94
    • 4 years ago
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    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

  10. JamesJ
    • 4 years ago
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    No, that is NOT the equation of a circle. It is the equation of a parabola.

  11. Ishaan94
    • 4 years ago
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    oh sorry please ignore me

  12. JamesJ
    • 4 years ago
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    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.

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