The general solution of the differential equa | vedclass

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Question

(C) The given differential equation is $(1+\sin^2 x) \frac{dy}{dx} + y \sin 2x = \cos x(1+\sin^2 x)$.Divide by $(1+\sin^2 x)$ to get the linear form $

Vedclass · Accepted Answer

$(1+\sin^2 x) y = \sin x + \frac{\sin^3 x}{3} + c$

Vedclass · Answer

(C) The given differential equation is $(1+\sin^2 x) \frac{dy}{dx} + y \sin 2x = \cos x(1+\sin^2 x)$.Divide by $(1+\sin^2 x)$ to get the linear form $\frac{dy}{dx} + y \frac{\sin 2x}{1+\sin^2 x} = \cos x$.This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = \frac{\sin 2x}{1+\sin^2 x}$ and $Q = \cos x$.The integrating factor $IF = e^{\int P dx} = e^{\int \frac{\sin 2x}{1+\sin^2 x} dx}$.Let $u = 1+\sin^2 x$,then $du = 2 \sin x \cos x dx = \sin 2x dx$.So,$IF = e^{\int \frac{du}{u}} = e^{\ln u} = u = 1+\sin^2 x$.The solution is $y(IF) = \int Q(IF) dx + c$.$y(1+\sin^2 x) = \int \cos x (1+\sin^2 x) dx + c$.Let $t = \sin x$,then $dt = \cos x dx$.$y(1+\sin^2 x) = \int (1+t^2) dt + c = t + \frac{t^3}{3} + c$.Substituting $t = \sin x$,we get $(1+\sin^2 x) y = \sin x + \frac{\sin^3 x}{3} + c$.

The general solution of the differential equation $(1+\sin^2 x) \frac{dy}{dx} + y \sin 2x = \cos x + \sin^2 x \cos x$ is

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