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Question

(B) Given the differential equation: $\frac{dy}{dx} + x \sin^2 y = \sin y \cos y$Divide both sides by $\sin^2 y$: $\csc^2 y \frac{dy}{dx} + x = \cot y

Vedclass · Accepted Answer

$\cot\,y = (x - 1) + Ce^{-x}$

Vedclass · Answer

(B) Given the differential equation: $\frac{dy}{dx} + x \sin^2 y = \sin y \cos y$Divide both sides by $\sin^2 y$: $\csc^2 y \frac{dy}{dx} + x = \cot y$Rearrange the terms: $\csc^2 y \frac{dy}{dx} - \cot y = -x$Let $v = \cot y$. Then $\frac{dv}{dx} = -\csc^2 y \frac{dy}{dx}$,which implies $-\frac{dv}{dx} = \csc^2 y \frac{dy}{dx}$.Substituting this into the equation: $-\frac{dv}{dx} - v = -x$,or $\frac{dv}{dx} + v = x$.This is a linear differential equation of the form $\frac{dv}{dx} + Pv = Q$,where $P = 1$ and $Q = x$.The integrating factor is $IF = e^{\int 1 dx} = e^x$.The general solution is $v \cdot e^x = \int x e^x dx + C$.Using integration by parts: $\int x e^x dx = x e^x - \int e^x dx = x e^x - e^x = (x - 1)e^x$.So,$v e^x = (x - 1)e^x + C$.Dividing by $e^x$: $v = (x - 1) + Ce^{-x}$.Substituting $v = \cot y$ back: $\cot y = (x - 1) + Ce^{-x}$.

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

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