The general solution of the differential equa | vedclass

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Question

(D) Given the differential equation: $(y \sin x + y) \frac{dy}{dx} - \cos^2 x = 0$ Separate the variables: $y(1 + \sin x) \frac{dy}{dx} = \cos^2 x$ $y

Vedclass · Accepted Answer

$y^2 = 2x + 2 \cos x + c$

Vedclass · Answer

(D) Given the differential equation: $(y \sin x + y) \frac{dy}{dx} - \cos^2 x = 0$ Separate the variables: $y(1 + \sin x) \frac{dy}{dx} = \cos^2 x$ $y \, dy = \frac{\cos^2 x}{1 + \sin x} dx$ Using the identity $\cos^2 x = 1 - \sin^2 x = (1 - \sin x)(1 + \sin x)$: $y \, dy = \frac{(1 - \sin x)(1 + \sin x)}{1 + \sin x} dx$ $y \, dy = (1 - \sin x) dx$ Integrating both sides: $\int y \, dy = \int (1 - \sin x) dx$ $\frac{y^2}{2} = x - (-\cos x) + c$ $\frac{y^2}{2} = x + \cos x + c$ $y^2 = 2x + 2 \cos x + c$

The general solution of the differential equation $(y \sin x + y) \frac{dy}{dx} - \cos^2 x = 0$ is:

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