The general solution of the differential equa | vedclass

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(B) Given the differential equation: $\frac{dy}{dx} = \sin(x-y) + \cos(x-y)$.Let $x-y = v$. Then $1 - \frac{dy}{dx} = \frac{dv}{dx}$,which implies $\f

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$\log \left|\frac{	an \frac{(x-y)}{2}-1}{	an \frac{(x-y)}{2}}ight|=x+c$

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(B) Given the differential equation: $\frac{dy}{dx} = \sin(x-y) + \cos(x-y)$.Let $x-y = v$. Then $1 - \frac{dy}{dx} = \frac{dv}{dx}$,which implies $\frac{dy}{dx} = 1 - \frac{dv}{dx}$.Substituting this into the equation: $1 - \frac{dv}{dx} = \sin v + \cos v$.Rearranging gives: $\frac{dv}{dx} = 1 - (\sin v + \cos v)$,so $\frac{dv}{1 - (\sin v + \cos v)} = dx$.Using half-angle formulas $\sin v = \frac{2 	an(v/2)}{1 + 	an^2(v/2)}$ and $\cos v = \frac{1 - 	an^2(v/2)}{1 + 	an^2(v/2)}$:$\frac{dv}{1 - \left(\frac{2 	an(v/2) + 1 - 	an^2(v/2)}{1 + 	an^2(v/2)}ight)} = dx$.Simplifying the denominator: $\frac{(1 + 	an^2(v/2)) dv}{1 + 	an^2(v/2) - 2 	an(v/2) - 1 + 	an^2(v/2)} = dx$.This simplifies to $\frac{\sec^2(v/2) dv}{2 	an^2(v/2) - 2 	an(v/2)} = dx$.Let $u = 	an(v/2)$,then $du = \frac{1}{2} \sec^2(v/2) dv$,so $\sec^2(v/2) dv = 2 du$.Substituting: $\frac{2 du}{2(u^2 - u)} = dx \Rightarrow \frac{du}{u(u-1)} = dx$.Using partial fractions: $\int (\frac{1}{u-1} - \frac{1}{u}) du = \int dx$.Integrating gives $\log|u-1| - \log|u| = x + C$.Substituting $u = 	an(\frac{x-y}{2})$: $\log \left| \frac{	an(\frac{x-y}{2}) - 1}{	an(\frac{x-y}{2})} ight| = x + C$.

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

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