The differential equation of the family of ci | vedclass

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(D) The general equation of a family of circles with center on the $x$-axis and passing through the origin is given by $(x-a)^2 + y^2 = a^2$,which sim

Vedclass · Accepted Answer

$y \frac{d^{2} y}{d x^{2}}+\left(\frac{d y}{d x}\right)^{2}+1=0$

Vedclass · Answer

(D) The general equation of a family of circles with center on the $x$-axis and passing through the origin is given by $(x-a)^2 + y^2 = a^2$,which simplifies to $x^2 - 2ax + a^2 + y^2 = a^2$,or $x^2 + y^2 - 2ax = 0$.Differentiating with respect to $x$,we get $2x + 2y \frac{dy}{dx} - 2a = 0$,which implies $a = x + y \frac{dy}{dx}$.Substituting the value of $a$ back into the original equation: $x^2 + y^2 - 2x(x + y \frac{dy}{dx}) = 0$.$x^2 + y^2 - 2x^2 - 2xy \frac{dy}{dx} = 0 \Rightarrow y^2 - x^2 - 2xy \frac{dy}{dx} = 0$.However,for a general family of circles with center $(a, 0)$ and radius $r$,the equation is $(x-a)^2 + y^2 = r^2$. Differentiating twice eliminates two constants $a$ and $r$. The standard form for circles with center on the $x$-axis is $y \frac{d^2y}{dx^2} + (\frac{dy}{dx})^2 + 1 = 0$.

The differential equation of the family of circles whose center lies on the $x$-axis is:

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