Résoudre l'equation différentielle suivante:
\[
y^{\prime} \sin x=y \cos x+\frac{\sin ^{3} x}{1+\sin x+\cos x}
\]
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Answer

5. \(y(x) = v(x) \sin(x) = \sin(x) \left(\int \frac{\sin^3(x)}{(1 + \sin x + \cos x) \sin^2(x)} dx + C\right) \)

Steps

Step 1 ：1. \(v(x) = \frac{y(x)}{\sin(x)} \)

Step 2 ：2. \(v'(x) = \frac{y'(x) \sin(x) - y(x) \cos(x)}{\sin^2(x)} \)

Step 3 ：3. \(v'(x) = \frac{\sin^3(x)}{(1 + \sin x + \cos x) \sin^2(x)} \)

Step 4 ：4. Integrate \(v'(x)\): \(v(x) = \int \frac{\sin^3(x)}{(1 + \sin x + \cos x) \sin^2(x)} dx + C \)

Step 5 ：5. \(y(x) = v(x) \sin(x) = \sin(x) \left(\int \frac{\sin^3(x)}{(1 + \sin x + \cos x) \sin^2(x)} dx + C\right) \)