Question

Differential equation of circle passing through origin and centre on x axis

Answer 1

Since the center of the circle is on the x axis, the center is taken as (a, 0). Let the radius of the circle be r, so we have the equation,

(x - a)² + y² = r²

Given that the origin (0, 0) is a point on the circle. So, taking x = y = 0,

r² = (0 - a)² + 0²

r² = a²

Hence the actual equation of the circle is,

(x - a)² + y² = a² → (1)

Now we have to express the equation as a differential equation in terms of x and y only. For this, first we have to differentiate this equation.

With respect to x,

d ((x - a)² + y²) / dx = d (a²) / dx

[d ((x - a)²) / dx] + [d (y²) / dx] = 0

2(x - a) + [(d (y²) / dy) · (dy / dx)] = 0

2(x - a) + (2y · dy / dx) = 0

x - a + (y · dy / dx) = 0

a = x + (y · dy / dx)

Then (1) becomes,

(x - (x + y · dy / dx))² + y² = (x + y · dy / dx)²

(- y · dy / dx)² + y² = x² + (y² · d²y / dx²) + (2xy · dy / dx)

(y² · d²y / dx²) + y² = x² + (y² · d²y / dx²) + (2xy · dy / dx)

y² = x² + (2xy · dy / dx)

Finally we got the differential equation.

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