The solution of the differential equation e^{ | vedclass

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(D) Given differential equation: $e^{-x}(y+1) dy + (\cos^2 x - \sin 2x) y dx = 0$ Dividing by $y e^{-x}$,we get: $\frac{y+1}{y} dy + e^x(\cos^2 x - \s

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

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(D) Given differential equation: $e^{-x}(y+1) dy + (\cos^2 x - \sin 2x) y dx = 0$ Dividing by $y e^{-x}$,we get: $\frac{y+1}{y} dy + e^x(\cos^2 x - \sin 2x) dx = 0$ Integrating both sides: $\int (1 + \frac{1}{y}) dy + \int e^x(\cos^2 x - 2 \sin x \cos x) dx = C$ Using the integral formula $\int e^x(f(x) + f'(x)) dx = e^x f(x) + C$,where $f(x) = \cos^2 x$ and $f'(x) = -2 \cos x \sin x = -\sin 2x$: $y + \log |y| + e^x \cos^2 x = C$ Given $x=0, y=1$: $1 + \log(1) + e^0 \cos^2(0) = C$ $1 + 0 + 1(1) = C \Rightarrow C = 2$ Thus,the solution is $y + \log |y| + e^x \cos^2 x = 2$.

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

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