If y=y(x) and left(frac{2+sin x}{y+1}right) f | vedclass

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(A) Given the differential equation: $\left(\frac{2+\sin x}{y+1}\right) \frac{dy}{dx} = -\cos x$. Separating the variables,we get: $\frac{dy}{y+1} = -

Vedclass · Accepted Answer

$\frac{1}{3}$

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(A) Given the differential equation: $\left(\frac{2+\sin x}{y+1}\right) \frac{dy}{dx} = -\cos x$. Separating the variables,we get: $\frac{dy}{y+1} = -\frac{\cos x}{2+\sin x} dx$. Integrating both sides: $\int \frac{dy}{y+1} = -\int \frac{\cos x}{2+\sin x} dx$. Let $u = 2+\sin x$,then $du = \cos x dx$. So,$\ln|y+1| = -\ln|2+\sin x| + C$. This simplifies to $\ln|y+1| + \ln|2+\sin x| = C$,or $\ln|(y+1)(2+\sin x)| = C$. Thus,$(y+1)(2+\sin x) = K$ (where $K = e^C$). Using the initial condition $y(0)=1$: $(1+1)(2+\sin 0) = K \implies 2(2+0) = K \implies K = 4$. So,$(y+1)(2+\sin x) = 4$. Now,find $y\left(\frac{\pi}{2}\right)$: $(y(\frac{\pi}{2})+1)(2+\sin(\frac{\pi}{2})) = 4$. $(y(\frac{\pi}{2})+1)(2+1) = 4$. $3(y(\frac{\pi}{2})+1) = 4$. $y(\frac{\pi}{2})+1 = \frac{4}{3}$. $y(\frac{\pi}{2}) = \frac{4}{3} - 1 = \frac{1}{3}$.

If $y=y(x)$ and $\left(\frac{2+\sin x}{y+1}\right) \frac{dy}{dx} = -\cos x$,$y(0)=1$,then $y\left(\frac{\pi}{2}\right) = $

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