The solution of the differential equation e^{ | vedclass

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(B) Given: $e^{-x}(y + 1) dy + (\cos^2 x - \sin 2x)y dx = 0$Divide by $y e^{-x}$: $\frac{y+1}{y} dy + \frac{\cos^2 x - \sin 2x}{e^{-x}} dx = 0$$(1 + \

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$\log y + y + e^x \cos^2 x = 2$

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(B) Given: $e^{-x}(y + 1) dy + (\cos^2 x - \sin 2x)y dx = 0$Divide by $y e^{-x}$: $\frac{y+1}{y} dy + \frac{\cos^2 x - \sin 2x}{e^{-x}} dx = 0$$(1 + \frac{1}{y}) dy + e^x(\cos^2 x - 2 \sin x \cos x) dx = 0$Integrate both sides: $\int (1 + \frac{1}{y}) dy + \int e^x \cos^2 x dx - \int e^x \sin 2x dx = C$Using $\int e^x \cos^2 x dx = \int e^x (\frac{1 + \cos 2x}{2}) dx = \frac{1}{2} e^x + \frac{1}{2} \int e^x \cos 2x dx$Thus, $\int e^x \cos^2 x dx - \int e^x \sin 2x dx = \frac{1}{2} e^x + \frac{1}{2} \int e^x \cos 2x dx - \int e^x \sin 2x dx$. This simplifies to $e^x \cos^2 x$.So, $y + \log y = -e^x \cos^2 x + C$At $x = 0, y = 1$: $1 + \log 1 = -e^0 \cos^2 0 + C \implies 1 = -1 + C \implies C = 2$Final solution: $y + \log y + e^x \cos^2 x = 2$.

The solution of the differential equation $e^{-x}(y + 1) dy + (\cos^2 x - \sin 2x)y dx = 0$, given that $y = 1$ when $x = 0$ is

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