Form the differential equation of the family of circles in the second quadrant and touching the coordinate axes.

OR

Find the particular solution of the differential equation

y = 0 when x = 2

The equation of circle such that coordinate axes touch them and circle lies in the second quadrant can be written with the aid of figure.

Clearly, center of circle is at (-a, a) and radius = a

∴ equation of circle is -

(x + a)² + (y – a)² = a² …(1)

To get the differential equation for a family of similar circles, we need to eliminate the only parameter a from the above equation.

So, Differentiating w.r.t x we get -

2(x + a) + 2 (y – a) = 0

⇒ (x + a) + (y – a) = 0

⇒ (x + a) + y - a = 0

⇒ a (1 – ) = -y - x

⇒ a =

∴ a =

Substituting for a in equation 1 -

– = a²

⇒ +

⇒

⇒

⇒

∴ is the required equation.

OR

Given,

⇒

⇒

As a result can be directly obtained with direct integration. So, Integrating both the sides with the help of partial fractions-

∵

Now applying the integration of both sides-

y = -log x + 0.5 log (x – 1) + 0.5 log (x – 1) + C

y = -log x + log

y = log + C {using log properties}

given that: for y = 0, x = 2

0 = log + C

0 = log + C

∴ C = -log

∴ y = log - log

y = log is the required particular solution.