Find the particular solution of the differential equation

, given that y = 0, when x = 1

OR

Obtain the differential equation of all circles of radius r.

Given: Differential equation is homogeneous

Proof:

Given differential equation is

…(i)

Putting y = vx

or …(ii)

Differentiating on both sides with respect to x

So, eq. (i) become

Integrating both the sides, we get

…(i)

Solving using integrating by parts

Taking e^-v as first function and sin v as second function

Substituting the value of in eq. (i) we get

⇒ e^-v [cos v + sin v] = 2(log x + C)

⇒ e^-v [cos v + sin v] = log x² + 2C

⇒ e^-v [cos v + sin v] = log x² + C’

As we previously let

For x = 1 and y = 0

⇒ e⁰ = C’ [∵ log (1) = 0]

⇒ C’ = 1 [∵ e⁰ = 1]

Substituting the value of C’, we get

which is the required solution.

OR

Given: Equation of family of circle, (x – a)² + (y – b)² = r² … (i)

Clearly, the given equation have two arbitrary constants, so we differentiate twice with respect to x

Now, again differentiating with respect to x, we get

⇒ 1 – 0 + (y – b)y’’+ y’(y’ – 0) = 0

⇒ 1 + (y – b)y’’ + (y’)² = 0

⇒ (y – b)y’’ = – (y’)² – 1

…(ii)

Substituting the value of (x – a) and (y – b) in eq. (i), we get

[from (ii)]

⇒ [(y’)² + 1]³ = r²(y’’)²

which is the required differential equation