The general solution of the differential equa | vedclass

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Question

(D) Given differential equation: $(\sin y \cos^2 y - x \sec^2 y) dy = 	an y dx$ Rearranging the terms: $	an y \frac{dx}{dy} + x \sec^2 y = \sin y \c

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$3x 	an y + \cos^3 y = c$

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(D) Given differential equation: $(\sin y \cos^2 y - x \sec^2 y) dy = 	an y dx$ Rearranging the terms: $	an y \frac{dx}{dy} + x \sec^2 y = \sin y \cos^2 y$ Dividing by $	an y$: $\frac{dx}{dy} + x \frac{\sec^2 y}{	an y} = \frac{\sin y \cos^2 y}{	an y} = \cos^3 y$ This is a linear differential equation of the form $\frac{dx}{dy} + P(y)x = Q(y)$,where $P(y) = \frac{\sec^2 y}{	an y}$ and $Q(y) = \cos^3 y$. Integrating factor ($I$.$F$.) $= e^{\int P(y) dy} = e^{\int \frac{\sec^2 y}{	an y} dy} = e^{\ln(	an y)} = 	an y$. The solution is $x \cdot (I.F.) = \int Q(y) \cdot (I.F.) dy + C$. $x 	an y = \int \cos^3 y \cdot 	an y dy = \int \cos^3 y \cdot \frac{\sin y}{\cos y} dy = \int \cos^2 y \sin y dy$. Let $u = \cos y$,then $du = -\sin y dy$. $x 	an y = -\int u^2 du = -\frac{u^3}{3} + C = -\frac{\cos^3 y}{3} + C$. Multiplying by $3$: $3x 	an y = -\cos^3 y + 3C$. Thus,$3x 	an y + \cos^3 y = C$ (where $C$ is a constant).

The general solution of the differential equation $(\sin y \cos^2 y - x \sec^2 y) dy = (\tan y) dx$ is

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