The curve y=(cos x+y)^{1 / 2} satisfies the d | vedclass

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(A) Given curve is $y=(\cos x+y)^{1/2}$.Squaring both sides, we get $y^2 = \cos x + y$.Differentiating both sides with respect to $x$:$2y \frac{dy}{dx

Vedclass · Accepted Answer

$(2 y-1) \frac{d^{2} y}{d x^{2}}+2\left(\frac{d y}{d x}\right)^{2}+\cos x=0$

Vedclass · Answer

(A) Given curve is $y=(\cos x+y)^{1/2}$.Squaring both sides, we get $y^2 = \cos x + y$.Differentiating both sides with respect to $x$:$2y \frac{dy}{dx} = -\sin x + \frac{dy}{dx}$.Rearranging the terms:$(2y - 1) \frac{dy}{dx} = -\sin x$.Differentiating again with respect to $x$ using the product rule:$(2y - 1) \frac{d^2y}{dx^2} + \frac{dy}{dx} \cdot \frac{d}{dx}(2y - 1) = -\cos x$.$(2y - 1) \frac{d^2y}{dx^2} + \frac{dy}{dx} \cdot (2 \frac{dy}{dx}) = -\cos x$.$(2y - 1) \frac{d^2y}{dx^2} + 2 \left(\frac{dy}{dx}\right)^2 + \cos x = 0$.

The curve $y=(\cos x+y)^{1 / 2}$ satisfies the differential equation

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