Question

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

Answer 1

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

$\Rightarrow dx+\tan ydx-dy-\tan ydy+2xdy=0$

$\Rightarrow dy(1+\tan y-2x)=(1+\tan y)dx$

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

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

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

I.F$=\sin y+\cos y{e}^y$

$x\times$ I.F$=\int Q\times$ I.F$dy+c$

$x\sin y+\cos y{e}^y=\int 1\times \sin y+\cos y{e}^{y}dy+c$

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

Consider $\int e^y\sin ydy$ is of the form

$\int e^{ax}\sin bxdy=\frac{e^{ax}}{a^2+b^2}(a\sin bx-b\cos bx)$

Here $a=1$ and $b=1$

Hence $\int e^y\cos ydy=\frac{e^y}{2}(\cos y+\sin y)$

Now combining and substituting in the required solution we get

$x{e}^y(\sin y+\cos y)=\frac{e^y}{2}(\sin y-\cos y)+\frac{e^y}{2}(\sin y+\cos y)+c$

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

Dividing throughout by $e^y$ we get

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