The general solution of the differential equa | vedclass

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(C) Given differential equation is $(1+	an y)(dx-dy)+2x dy=0$. Rearranging the terms,we get $(1+	an y)dx = (1+	an y - 2x)dy$. Dividing by $(1+	an

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$e^y(x \cos y+x \sin y-\sin y)=c$

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(C) Given differential equation is $(1+	an y)(dx-dy)+2x dy=0$. Rearranging the terms,we get $(1+	an y)dx = (1+	an y - 2x)dy$. Dividing by $(1+	an y)dy$,we obtain $\frac{dx}{dy} = 1 - \frac{2x}{1+	an y}$,which simplifies to $\frac{dx}{dy} + \left(\frac{2}{1+	an y}ight)x = 1$. This is a linear differential equation of the form $\frac{dx}{dy} + P(y)x = Q(y)$,where $P(y) = \frac{2}{1+	an y} = \frac{2\cos y}{\sin y + \cos y}$. The Integrating Factor ($I$.$F$.) is $e^{\int P(y)dy} = e^{\int \frac{2\cos y}{\sin y + \cos y} dy}$. Since $\int \frac{2\cos y}{\sin y + \cos y} dy = \int \frac{(\cos y - \sin y) + (\cos y + \sin y)}{\sin y + \cos y} dy = \int \left(\frac{\cos y - \sin y}{\sin y + \cos y} + 1ight) dy = \ln|\sin y + \cos y| + y$. Thus,$I.F. = e^{\ln(\sin y + \cos y) + y} = e^y(\sin y + \cos y)$. The solution is $x \cdot (I.F.) = \int Q(y) \cdot (I.F.) dy + c$. $x e^y(\sin y + \cos y) = \int e^y(\sin y + \cos y) dy + c$. Using the formula $\int e^y(f(y) + f'(y)) dy = e^y f(y) + c$,where $f(y) = \sin y$ and $f'(y) = \cos y$,we get: $x e^y(\sin y + \cos y) = e^y \sin y + c$. Rearranging gives $e^y(x \sin y + x \cos y - \sin y) = c$.

The general solution of the differential equation $(1+\tan y)(dx-dy)+2x dy=0$ is

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