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(A) The given differential equation is $\frac{dy}{dx} = \frac{a + bx}{c + dy}$.Rearranging the terms, we get $(c + dy) dy = (a + bx) dx$.Integrating b

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$a = c = 0, b + d = 0$

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(A) The given differential equation is $\frac{dy}{dx} = \frac{a + bx}{c + dy}$.Rearranging the terms, we get $(c + dy) dy = (a + bx) dx$.Integrating both sides, we have $\int (c + dy) dy = \int (a + bx) dx$.This gives $cy + \frac{d}{2} y^2 = ax + \frac{b}{2} x^2 + C$.For the equation to represent a family of circles centered at the origin, the equation must be of the form $x^2 + y^2 = R^2$.Comparing the terms, we must have $a = 0$ and $c = 0$.Substituting these, we get $\frac{d}{2} y^2 = \frac{b}{2} x^2 + C$, which simplifies to $d y^2 - b x^2 = 2C$.For this to be a circle, the coefficients of $x^2$ and $y^2$ must be equal and have the same sign. Thus, $d = -b$, which implies $b + d = 0$.

The solution of the differential equation $\frac{dy}{dx} = \frac{a + bx}{c + dy}$ represents a family of circles centered at the origin if...

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