Question

The solution of $\cos y+(x\sin y-1)\frac{dy}{dx}=0$ is:

• A. $x\sec y=\tan y+e$
• B. $\tan y-\sec y=ex$
• C. $\tan y+\sec y=ex$
• D. $x\sec y+\tan y=e$

Answer 1

Answer: (A)

$x\sec y=\tan y+e$

$\frac{dy}{dx}=\frac{cosy}{1-xsiny}$

$\Rightarrow \frac{dx}{dy}+xtany=secy$

$I.F.=e^{\int pdy}=e^{\int tanydy}=e^{\ln secy}=secy$

$\therefore xsecy=\int secy.secydy=tany+e$

$\therefore xsecy=tany+e$