Question

Form the differential equation of the family of circles touching the y-axis at origin.

Answer 1

Answer:

The required differential equation is

$\bf\,(yy')^2+y^2=(x+yy')^2$

Step-by-step explanation:

Differential equation is formed by eliminating arbitrary constant occur in the given equation

$\text{The equation of family of circles touching y-axis at origin is }$

$(x-\lambda)^2+y^2={\lambda}^2$..........(1)

$\text{differentiate with respect to x}$

$2(x-\lambda)+2y\,y'=0$

$\implies\,(x-\lambda)+y\,y'=0$

$\implies\,x-\lambda=-y\,y'$....(2)

$\text{using (2), equation (1) can be written as }$

$(-yy')^2+y^2=(x+yy')^2$

$\implies\bf\,(yy')^2+y^2=(x+yy')^2$

Answer 2

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

TO DETERMINE

Form the differential equation of the family of circles touching the y-axis at origin

CALCULATION

Let radius of the circle = a unit

Since the circles are touching the y-axis at origin

So the centre of the circle is ( a , 0)

So the equation of the circle is

$\sf{ \: {(x - a)}^{2} + {y}^{2} = {a}^{2} \: }$

$\implies\sf{{x}^{2} - 2ax + {a}^{2} + {y}^{2} = {a}^{2} \: }$

$\implies\sf{{x}^{2} - 2ax + {y}^{2} = 0 \: \: \: \: ......(1) \: }$

Differentiating both sides with respect to x we get

$\displaystyle \: \sf{2x - 2a + 2y \frac{dy}{dx} = 0 \: }$

$\implies \displaystyle \: \sf{2a \: = \: 2x + 2y \frac{dy}{dx} \: }$

From Equation (1) we get

$\displaystyle \: \sf{ {x}^{2} - x( \: 2x + 2y \frac{dy}{dx} ) + {y}^{2} = 0\: }$

$\implies \displaystyle \: \sf{ {x}^{2} -2 {x}^{2} - 2xy \frac{dy}{dx} + {y}^{2} = 0\: }$

$\implies \displaystyle \: \sf{ 2xy \frac{dy}{dx} = {y}^{2} - {x}^{2} \: }$

RESULT

The required differential equation is

$\boxed{ \displaystyle \: \sf{ \: \: 2xy \frac{dy}{dx} = {y}^{2} - {x}^{2} \: \: \: }}$

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