Question

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

• A. $x(\sin y+\cos y)=\sin y+c{e}^{-y}$
• B. $x(\sin y-\cos y)=\sin y+c{e}^{-y}$
• C. $x(\sin y+\cos y)=\cos y+c{e}^{-y}$
• D. None of the above

Answer 1

The correct option is A $x(\sin y+\cos y)=\sin y+c{e}^{-y}$

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

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

Here $P=\frac{2}{1+\tan y}\Rightarrow \int PdP=\int \frac{2}{1+\tan y}dy=\int \frac{2\cos y}{\sin y+\cos y}dy$

$=\int (1+\frac{\cos y-\sin y}{\cos y+\sin y})dy=y+\log (\cos y+\sin y)$

$\therefore I.F.=e^{y+\log (\cos y+\sin y)}=e^y.e^{\log (\cos y+\sin y)}=(\cos y+\sin y)e^y$

Multiplying (1) by $I.F.$ we get

$(\cos y+\sin y)e^y\frac{dx}{dy}+(\cos y+\sin y)e^y.\frac{2}{1+\tan y}x=(\cos y+\sin y)e^y$

Integrating both sides, we get

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

$\Rightarrow x(-\sin y-\cos y)\sin y+{ce}^{-y}$