General solution of the differential equation | vedclass

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Question

(B) Given the differential equation $\sin^3 x \frac{dx}{dy} = \sin y$. Separating the variables,we get $\int \sin^3 x \, dx = \int \sin y \, dy$. Usin

Vedclass · Accepted Answer

$\cos y - \frac{3}{4} \cos x + \frac{1}{12} \cos 3x = C$

Vedclass · Answer

(B) Given the differential equation $\sin^3 x \frac{dx}{dy} = \sin y$. Separating the variables,we get $\int \sin^3 x \, dx = \int \sin y \, dy$. Using the trigonometric identity $\sin 3x = 3 \sin x - 4 \sin^3 x$,we have $\sin^3 x = \frac{3 \sin x - \sin 3x}{4}$. Substituting this into the integral,we get $\int \frac{3 \sin x - \sin 3x}{4} \, dx = \int \sin y \, dy$. Integrating both sides,we get $\frac{3}{4} (-\cos x) - \frac{1}{4} (-\frac{\cos 3x}{3}) = -\cos y + C$. This simplifies to $-\frac{3}{4} \cos x + \frac{1}{12} \cos 3x = -\cos y + C$. Rearranging the terms,we get $\cos y - \frac{3}{4} \cos x + \frac{1}{12} \cos 3x = C$.

General solution of the differential equation $\sin^3 x \frac{dx}{dy} = \sin y$ is given by

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