FInd the general solution of $(1+\tan y)(dx-dy)+2xdy=0$

$\begin{array}{1 1} (A)\;x(\sin y+\cos y)=\sin y+ce^{-y} \\ (B)\;x(\sin y-\cos y)=\sin y-ce^{-y}\\(C)\;x(\sin y-\cos y)=\sin y+ce^{-y} \\ (D)\;x(\sin y+\cos y)=\sin y-c\end{array}$

Toolbox:
• A linear differential equation of the form $\large\frac{dy}{dx}$$=f(x) can be solved by seperating the variables and then integrating it. • A linear differential equation of the form \large\frac{dx}{dy}$$+Px=Q$ . has the general solution $xe^{\int pdy}=\int Q e^{\int pdy}.dy+c$
• $\int e^{ax}.\sin bx dx=\large\frac{e^{\Large ax}}{a^2+b^2}$$[a \sin bx-b \cos bx]+c Step 1: Given (1+\tan y)(dx-dy)+2xdy=0 This can be written as dx+\tan y dx-dy-\tan y dy+2x dy=0 (ie) -dy(1+\tan y -2x)+dx(1+\tan y)=0 dy(1+\tan y-2x)=(1+\tan y)dx \large\frac{dy}{dx}=\frac{1+\tan y}{1+\tan y -2x} (or) \large\frac{dx}{dy}=\frac{1+\tan y-2x}{1+\tan y } \large\frac{dx}{dy}=\frac{1+\tan y}{1+\tan y }-\frac{2x}{1+\tan y} Therefore I.F=\sin y+\cos y.e^y Hence the required solution is x \times I.F=\int Q \times I.F \;dy+c x. \sin y+\cos y.e^y=\int 1 \times \sin y+\cos y.e^y dy+c x .e^y( \sin y+\cos y)=\int e^y( \sin y+\cos y) dy+c x .e^y( \sin y+\cos y)=\int e^y \sin y dy +\int e^y \cos y dy+c Step 2: Consider \int e^y \sin y dy This is of the form \int e^{ax} \sin bx dx =\large\frac{e^{\Large ax}}{a^2+b^2}$$(a \sin bx-b \cos bx)$
Here $a=1\; and\; b=1$
Hence $\int e^y \sin y dy=\large\frac{\Large e^y}{2}$$(\sin x-\cos x) Similarly \int e^{ax} \cos bx dx=\large\frac{e^{ax}}{a^2+b^2}$$(a \cos bx +b \sin bx)$
Here $a=1\; and \; b=1$
Hence $\int e^y \cos y dy=\large\frac{\Large e^y}{2}$$(\cos y+\sin y) Now combining and substituting in the required solution we get, xe^y(\sin y+\cos y)=\large\frac{\Large e^y}{2}$$ (\sin y-\cos y)$$+\large\frac{\Large e^y}{2}$$(\cos y+\sin y)+c$