The solution of the differential equation e^{ | vedclass

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

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

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(B) Given equation: $e^{-x}(y + 1)dy + (\cos^2 x - \sin 2x)y dx = 0$Divide by $y e^{-x}$: $\frac{y+1}{y} dy + e^x (\cos^2 x - \sin 2x) dx = 0$$(1 + \frac{1}{y}) dy + e^x \cos^2 x dx - e^x \sin 2x dx = 0$Integrate both sides: $\int (1 + \frac{1}{y}) dy + \int e^x \cos^2 x dx - \int e^x \sin 2x dx = C$Note that $\int e^x \cos^2 x dx = \int e^x (\frac{1 + \cos 2x}{2}) dx = \frac{1}{2} e^x + \frac{1}{2} \int e^x \cos 2x dx$Using $\int e^x \cos 2x dx = \frac{e^x}{1^2 + 2^2} (\cos 2x + 2 \sin 2x) = \frac{e^x}{5} (\cos 2x + 2 \sin 2x)$So, $\int e^x \cos^2 x dx = \frac{e^x}{2} + \frac{e^x}{10} (\cos 2x + 2 \sin 2x) = \frac{e^x}{10} (5 + \cos 2x + 2 \sin 2x)$Also $\int e^x \sin 2x dx = \frac{e^x}{5} (\sin 2x - 2 \cos 2x)$Substituting back: $y + \log y + \frac{e^x}{10} (5 + \cos 2x + 2 \sin 2x - 2 \sin 2x + 4 \cos 2x) = C$$y + \log y + \frac{e^x}{10} (5 + 5 \cos 2x) = C \implies y + \log y + e^x \frac{1 + \cos 2x}{2} = C \implies y + \log y + e^x \cos^2 x = C$At $x = 0, y = 1$: $1 + \log 1 + e^0 \cos^2 0 = C \implies 1 + 0 + 1(1) = C \implies C = 2$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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