The general solution of the differential equa | vedclass

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(A) Given differential equation is $\frac{d y}{d x}+\frac{\sin (2 x+y)}{\cos x}+2=0$ ....$(i)$Let $2 x+y=t$. Differentiating with respect to $x$, we g

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$(\sec x+	an x)[\operatorname{cosec}(2 x+y)-\cot (2 x+y)]=c$

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(A) Given differential equation is $\frac{d y}{d x}+\frac{\sin (2 x+y)}{\cos x}+2=0$ ....$(i)$Let $2 x+y=t$. Differentiating with respect to $x$, we get $2+\frac{d y}{d x}=\frac{d t}{d x}$, so $\frac{d y}{d x}=\frac{d t}{d x}-2$.Substituting this into equation $(i)$:$(\frac{d t}{d x}-2) + \frac{\sin t}{\cos x} + 2 = 0$$\frac{d t}{d x} + \frac{\sin t}{\cos x} = 0$$\frac{d t}{\sin t} = -\frac{d x}{\cos x}$$\operatorname{cosec} t \, dt = -\sec x \, dx$Integrating both sides:$\int \operatorname{cosec} t \, dt = -\int \sec x \, dx$$\ln|\operatorname{cosec} t - \cot t| = -\ln|\sec x + 	an x| + C_1$$\ln|\operatorname{cosec} t - \cot t| + \ln|\sec x + 	an x| = C_1$$\ln|(\sec x + 	an x)(\operatorname{cosec} t - \cot t)| = C_1$$(\sec x + 	an x)(\operatorname{cosec} t - \cot t) = e^{C_1} = C$Substituting $t = 2x+y$ back, we get:$(\sec x + 	an x)[\operatorname{cosec}(2x+y) - \cot(2x+y)] = C$.

The general solution of the differential equation $\frac{d y}{d x}+\frac{\sin (2 x+y)}{\cos x}+2=0$ is:

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