The equation of the curve passing through (3, | vedclass

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(D) Given the differential equation: $y \left( \frac{dy}{dx} \right)^2 + (x - y) \frac{dy}{dx} - x = 0$. This is a quadratic equation in $\frac{dy}{dx

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$A$ and $B$ both

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(D) Given the differential equation: $y \left( \frac{dy}{dx} \right)^2 + (x - y) \frac{dy}{dx} - x = 0$. This is a quadratic equation in $\frac{dy}{dx}$. Using the quadratic formula $\frac{dy}{dx} = \frac{-(x-y) \pm \sqrt{(x-y)^2 - 4(y)(-x)}}{2y} = \frac{y-x \pm \sqrt{x^2 - 2xy + y^2 + 4xy}}{2y} = \frac{y-x \pm \sqrt{(x+y)^2}}{2y} = \frac{y-x \pm (x+y)}{2y}$. Case $1$: $\frac{dy}{dx} = \frac{y-x+x+y}{2y} = \frac{2y}{2y} = 1$. Integrating,$y = x + c$. Since it passes through $(3, 4)$,$4 = 3 + c \Rightarrow c = 1$. So,$y = x + 1$ or $x - y + 1 = 0$. Case $2$: $\frac{dy}{dx} = \frac{y-x-x-y}{2y} = \frac{-2x}{2y} = -\frac{x}{y}$. Integrating,$y dy = -x dx \Rightarrow \frac{y^2}{2} = -\frac{x^2}{2} + C' \Rightarrow x^2 + y^2 = C$. Since it passes through $(3, 4)$,$3^2 + 4^2 = C \Rightarrow 9 + 16 = 25$. So,$x^2 + y^2 = 25$. Thus,both $A$ and $B$ are correct.

The equation of the curve passing through $(3, 4)$ and satisfying the differential equation $y \left( \frac{dy}{dx} \right)^2 + (x - y) \frac{dy}{dx} - x = 0$ can be:

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