Question

The curve for which the subnormal is equal to the square of radius vector (i.e. distance from origin).

• A. $y^2 = x^2 - x - \frac{1}{2}ce^{-2x}$
• B. $y^2 = -(x^2 + x + \frac{1}{2})+ ce^{-2x}$
• C. $y^2 = -(x^2 - x + \frac{1}{2})+ ce^{2x}$
• D. $y^2 = -(x^2 - x - \frac{1}{2})+ ce^{2x}$

Answer 1

The curve is $y^2 = -(x^2 + x + \frac{1}{2}) + C e^{2x}$. None of the provided options exactly matches this result.

Answer 2

The problem translates to the differential equation $y \frac{dy}{dx} = x^2 + y^2$. By substituting $z = y^2$, it becomes a linear first-order differential equation $\frac{dz}{dx} - 2z = 2x^2$. Solving this using an integrating factor $e^{-2x}$ yields $z = -(x^2 + x + \frac{1}{2}) + C e^{2x}$, so $y^2 = -(x^2 + x + \frac{1}{2}) + C e^{2x}$.