If the differential equation frac{dy}{dx} + f | vedclass

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(A) Given the differential equation: $\frac{dy}{dx} + \frac{x}{y} = \frac{a}{y}$.Multiply both sides by $y$: $y \frac{dy}{dx} + x = a$.This is a varia

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$\sqrt{a^2 + 2c}$,where $c$ is the constant of integration

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(A) Given the differential equation: $\frac{dy}{dx} + \frac{x}{y} = \frac{a}{y}$.Multiply both sides by $y$: $y \frac{dy}{dx} + x = a$.This is a variable separable form: $y dy = (a - x) dx$.Integrate both sides: $\int y dy = \int (a - x) dx$.$\frac{y^2}{2} = ax - \frac{x^2}{2} + C$,where $C$ is the constant of integration.Multiply by $2$: $y^2 = 2ax - x^2 + 2C$.Rearrange the terms: $x^2 - 2ax + y^2 = 2C$.Complete the square for $x$: $(x^2 - 2ax + a^2) + y^2 = 2C + a^2$.$(x - a)^2 + y^2 = a^2 + 2C$.This represents a circle with center $(a, 0)$ and radius squared $r^2 = a^2 + 2C$.Therefore,the radius $r = \sqrt{a^2 + 2C}$.

If the differential equation $\frac{dy}{dx} + \frac{x}{y} = \frac{a}{y}$,where $a$ is a constant,represents a family of circles,then the radius of the circle is ......

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