The differential equation of all circles passing through the origin and having their centres on the x-axis is

  • Option 1)

    y^{2}=x^{2}+2xy\frac{dy}{dx}

  • Option 2)

    y^{2}=x^{2}-2xy\frac{dy}{dx}

  • Option 3)

    x^{2}=y^{2}+xy\frac{dy}{dx}

  • Option 4)

    x^{2}=y^{2}+3xy\frac{dy}{dx}

 

Answers (1)
S Sabhrant Ambastha

As we learnt in

Formation of Differential Equations -

A differential equation can be derived from its equation by the process of differentiation and other algebraical process of elimination

-

 

 Let the equation of circle is 

(x-a)^{2}+y^{2}=a^{2} where (a,0) is center and a is radius.

2(x-a) +2y\frac{dy}{dx}=0

\Rightarrow (x-a) +y\frac{dy}{dx}=0

\Rightarrow x+yy'=a

\Rightarrow (yy')^{2}+ y^{2}= (x+yy')^{2}= x^{2}+ (yy')^{2}+2xyy'

\Rightarrow \frac{y^{2}-x^{2}}{2xy}= \frac{dy}{dx}

 

 


Option 1)

y^{2}=x^{2}+2xy\frac{dy}{dx}

Correct option

Option 2)

y^{2}=x^{2}-2xy\frac{dy}{dx}

Incorrect option

Option 3)

x^{2}=y^{2}+xy\frac{dy}{dx}

Incorrect option

Option 4)

x^{2}=y^{2}+3xy\frac{dy}{dx}

Incorrect option

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