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Mathematics Question on Differential Equations

The differential equation of the family of circles passing the origin and having center at the line y = x is:

A

(x2−y2+2xy)dx=(x2−y2+2xy)dy(x^2 - y^2 + 2xy)dx = (x^2 - y^2 + 2xy)dy

B

(x2+y2+2xy)dx=(x2+y2−2xy)dy(x^2 + y^2 + 2xy)dx = (x^2 + y^2 - 2xy)dy

C

(x2−y2+2xy)dx=(x2−y2−2xy)dy(x^2 - y^2 + 2xy)dx = (x^2 - y^2 - 2xy)dy

D

(x2+y2−2xy)dx=(x2+y2+2xy)dy(x^2 + y^2 - 2xy)dx = (x^2 + y^2 + 2xy)dy

Answer

(x2−y2+2xy)dx=(x2−y2−2xy)dy(x^2 - y^2 + 2xy)dx = (x^2 - y^2 - 2xy)dy

Explanation

Solution

The family of circles passes through the origin and has centers on the line y=xy = x. The general equation of such a circle is:

(x−a)2+(y−a)2=r2,(x - a)^2 + (y - a)^2 = r^2, where (a,a)(a, a) is the center of the circle on the line y=xy = x, and rr is the radius.

Step 1: Expand the circle equation

Expanding (x−a)2+(y−a)2=r2(x - a)^2 + (y - a)^2 = r^2: x2−2ax+a2+y2−2ay+a2=r2.x^2 - 2ax + a^2 + y^2 - 2ay + a^2 = r^2. Simplifying: x2+y2−2a(x+y)+2a2=r2.x^2 + y^2 - 2a(x + y) + 2a^2 = r^2.

Step 2: Eliminate parameters aa and rr

Since the circle passes through the origin, substitute x=0x = 0 and y=0y = 0 into the equation: 02+02−2a(0+0)+2a2=r2  ⟹  r2=2a2.0^2 + 0^2 - 2a(0 + 0) + 2a^2 = r^2 \implies r^2 = 2a^2. Thus, the equation becomes: x2+y2−2a(x+y)=0.x^2 + y^2 - 2a(x + y) = 0. Differentiating both sides with respect to xx: 2x+2ydydx−2a(1+dydx)=0.2x + 2y \frac{dy}{dx} - 2a\left(1 + \frac{dy}{dx}\right) = 0. Rearranging to isolate aa: a=x+ydydx1+dydx.a = \frac{x + y \frac{dy}{dx}}{1 + \frac{dy}{dx}}. Substitute aa back into the circle equation: (x2−y2+2xy)dx=(x2−y2−2xy)dy.(x^2 - y^2 + 2xy)dx = (x^2 - y^2 - 2xy)dy. Thus, the differential equation of the family of circles is: (x2−y2+2xy)dx=(x2−y2−2xy)dy.(x^2 - y^2 + 2xy)dx = (x^2 - y^2 - 2xy)dy.