Question

Find the general solution of the differential equation.
$(1+\tan y)(dx-dy)+2xdy=0$.

Answer 1

$(1+\tan y)(dx-dy)+2xdy=0$

$\frac{(1+\tan y)}{2x}=\frac{-dy}{dx-dy}$

$\frac{dx-dy}{-dy}=\frac{2x}{1+\tan y}$

$\frac{-dx}{dy}+1=\frac{2x}{1+\tan y}$

$\frac{dx}{dy}+\frac{2x}{1+\tan y}=1$

$I.F=\int e^{\frac{2}{1+\tan y}.dy}$

$=\int e^{\frac{2\cos y}{1+\sin y+\cos y}.dy}$

$=\int e^{1+\frac{\cos y-\sin y}{\sin y+\cos y}.dx}$

$=e^{y+\left(n\right(\sin y+\cos y\left)\right)}$

$=e^y.(\sin y+\cos y)$

$x.e^y.(\sin y.\cos y)=\int e^y(\sin y+\cos y).dy$

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

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