Question

The general solution of $(1+\tan y)(dx-dy)+2xdy=0$, is

• A. $x=\frac{-1}{1+\tan y}+\frac{C.e^{-y}}{1+\tan y}$
• B. $x=\frac{1}{1+\tan y}+\frac{C.e^y}{1+\tan y}$
• C. $x=\frac{1}{1+\tan y}+\frac{C.e^{-y}}{1+\tan y}\sec y$
• D. $noneofthese$

Answer 1

Answer: (A)

$x=\frac{-1}{1+\tan y}+\frac{C.e^{-y}}{1+\tan y}$

Given equtaion is,

$(1+tany)(dx-dy)+2xdy=0$

Dividing above equation by (1+tany)dy

$\therefore \frac{dx}{dy}+\frac{2x}{1+tany}=1$

P=$\frac{2}{1+tany}$,Q=1

I.F.=$e^{\int P.dy}$

=$e^{\int \frac{2}{1+tany}dy}$

=$e^y(siny+cosy)$

Complete solution is given by,

$x\times e^y(siny+cosy)=\int e^{(}siny+cosy)dy+c$

$x.e^y(siny+cosy)=\int e^{y}sinydy+\int e^{y}cosydy+c$

$x.e^y(siny+cosy)=\frac{e^y}{2}(siny-cosy)+\frac{e^y}{2}(siny-cosy)+c$

$x.e^y(siny+cosy)=\frac{e^y}{2}(-2cosy)+c$

$x.e^y(siny+cosy)=-e^{y}cosy+c$

$x=\frac{-1}{1+tany}+\frac{c{e}^{-y}}{1+tany}$