Question

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

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

Answer 1

Answer: (B)

We have, $(1+\tan y)(d x-d y)+2 x d y=0$
$\Rightarrow (1+\tan y) d x=(1+\tan y-2 x) d y$
$\Rightarrow \frac{d x}{d y}+\frac{2}{1+\tan y} x=1,$ which is linear differential equation.
I.F. $=e^{2 \int \frac{d y}{1+\tan y}}=e^{\int \frac{2 \cos y}{\sin y+\cos y} d y} $
$=e^{\int\left(1+\frac{\cos y-\sin y}{\sin y+\cos y}\right) d y}=e^{y+\log (\cos y+\sin y)} $
$=(\cos y+\sin y) e^{y}$
So, the solution is
$x e^{y}(\sin y+\cos y)=\int e^{y}(\sin y+\cos y) d y+c$
i.e. $x e^{y}(\sin y+\cos y)=e^{y} \sin y+c .$
i.e. $x(\sin y+\cos y)=\sin y+c e^{-y}$