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Exercise 3 · Q3

Q.Form the differential equation of the family of circles having centre on yy-axis and passing through origin.

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A circle with centre on the yy-axis through the origin is x2+y2−2ay=0x^2+y^2-2ay=0; eliminating the single constant aa yields (x2−y2)dydx=2xy(x^2-y^2)\dfrac{dy}{dx}=2xy.

Centre on yy-axis =(0,a)=(0,a), and passing through origin forces radius =a=a. Equation: x2+(y−a)2=a2⇒x2+y2−2ay=0x^2+(y-a)^2=a^2\Rightarrow x^2+y^2-2ay=0, with one arbitrary constant aa (order 1).

Steps

  1. Centre (0,a)(0,a), radius aa (so the circle passes through the origin):

x2+(y−a)2=a2  ⇒  x2+y2−2ay=0.(1)x^2+(y-a)^2=a^2\;\Rightarrow\; x^2+y^2-2ay=0.\qquad(1)

  1. From (1) solve for the constant:

a=x2+y22y.(2)a=\frac{x^2+y^2}{2y}.\qquad(2)

  1. Differentiate (1) with respect to xx:

2x+2ydydx−2adydx=0  ⇒  x+ydydx=adydx.2x+2y\frac{dy}{dx}-2a\frac{dy}{dx}=0\;\Rightarrow\; x+y\frac{dy}{dx}=a\frac{dy}{dx}.

⇒  a=x+y dydxdydx.(3)\Rightarrow\; a=\frac{x+y\,\frac{dy}{dx}}{\frac{dy}{dx}}.\qquad(3)

  1. Equate (2) and (3): …

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