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Exercise 3 · Q6

Q.Form the differential equation of the family of ellipses having their foci on xx-axis and centre at the origin.

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x2a2+y2b2=1\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1 has two constants, so differentiate twice and eliminate a,ba,b to obtain xyy′′+x(y′)2−yy′=0xyy''+x(y')^2-yy'=0.

Ellipse, foci on xx-axis, centre origin: x2a2+y2b2=1\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1 with two arbitrary constants a,ba,b ⇒\Rightarrow order 2.

Steps

  1. Family:

x2a2+y2b2=1.(1)\frac{x^2}{a^2}+\frac{y^2}{b^2}=1.\qquad(1)

  1. Differentiate once:

2xa2+2yb2dydx=0  ⇒  xa2+yb2dydx=0.(2)\frac{2x}{a^2}+\frac{2y}{b^2}\frac{dy}{dx}=0\;\Rightarrow\;\frac{x}{a^2}+\frac{y}{b^2}\frac{dy}{dx}=0.\qquad(2)

From (2): 1a2=−yb2xdydx.(2′)\dfrac{1}{a^2}=-\dfrac{y}{b^2 x}\dfrac{dy}{dx}.\qquad(2')

  1. Differentiate (2) again (product rule on the second term):

1a2+1b2[(dydx)2+yd2ydx2]=0.(3)\frac{1}{a^2}+\frac{1}{b^2}\left[\left(\frac{dy}{dx}\right)^2+y\frac{d^2y}{dx^2}\right]=0.\qquad(3)

  1. Substitute 1a2\dfrac{1}{a^2} from (2') into (3): …

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