The differential equation frac{dy}{dx} + frac | vedclass

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(B) Given the differential equation: $\frac{dy}{dx} + \frac{1}{x} \sin 2y = x^3 \cos^2 y$. Using $\sin 2y = 2 \sin y \cos y$,we have $\frac{dy}{dx} +

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$6x^2 	an y = x^6 + C$

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(B) Given the differential equation: $\frac{dy}{dx} + \frac{1}{x} \sin 2y = x^3 \cos^2 y$. Using $\sin 2y = 2 \sin y \cos y$,we have $\frac{dy}{dx} + \frac{2}{x} \sin y \cos y = x^3 \cos^2 y$. Dividing both sides by $\cos^2 y$,we get: $\sec^2 y \frac{dy}{dx} + \frac{2}{x} 	an y = x^3$. Let $z = 	an y$,then $\frac{dz}{dx} = \sec^2 y \frac{dy}{dx}$. The equation becomes: $\frac{dz}{dx} + \frac{2}{x} z = x^3$. This is a linear differential equation of the form $\frac{dz}{dx} + P(x)z = Q(x)$,where $P(x) = \frac{2}{x}$ and $Q(x) = x^3$. The integrating factor ($I$.$F$.) is $e^{\int P(x) dx} = e^{\int \frac{2}{x} dx} = e^{2 \ln x} = x^2$. The solution is $z \cdot (I.F.) = \int Q(x) \cdot (I.F.) dx + C$. $z \cdot x^2 = \int x^3 \cdot x^2 dx + C = \int x^5 dx + C$. $z x^2 = \frac{x^6}{6} + C$. Substituting $z = 	an y$: $(	an y) x^2 = \frac{x^6}{6} + C$. Multiplying by $6$: $6x^2 	an y = x^6 + 6C$. Since $6C$ is an arbitrary constant,we can write it as $C$. Thus,$6x^2 	an y = x^6 + C$.

The differential equation $\frac{dy}{dx} + \frac{1}{x} \sin 2y = x^3 \cos^2 y$ represents a family of curves given by:

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