Question

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

• A. $xsecy=\tan y+C$
• B. $\tan y–secy=Cx$
• C. $\tan y+secy=Cx$
• D. $xsecy+t\mathrm{an}y=C$

Answer 1

The correct option is A

$xsecy=\tan y+C$

Find the solution of the given expression

Given: $\cos y+(x\sin y-1)\frac{dy}{dx}=0$

$$\begin{gathered} \Rightarrow \frac{\cos y}{\cos y}+\frac{(x\sin y-1)}{\cos y}\frac{dy}{dx}=\frac{0}{\cos y} \\ \Rightarrow 1+\frac{(x\sin y-1)}{\cos y}\frac{dy}{dx}=0 \\ \Rightarrow \left(\frac{x\sin y}{\cos y}-\frac{1}{\cos y}\right)\frac{dy}{dx}=-1 \\ \Rightarrow \left(x\frac{\sin y}{\cos y}-\frac{1}{\cos y}\right)=-\frac{dx}{dy} \end{gathered}$$

$\Rightarrow \frac{dx}{dy}+x\frac{\sin y}{\cos y}-\frac{1}{\cos y}=0$

$\Rightarrow \frac{dx}{dy}+\tan yx=secy$ , linear in $x$

So, $I.F.=e^{\int \tan ydy}$

$$\begin{gathered} =e^{\int \frac{\sin y}{\cos y}dy} \\ =e^{\ln secy} \\ =secy \end{gathered}$$

Therefore, solution is $xsecy=\int se{c}^{2}ydy+C$

$xsecy=\tan y+C$

Hence, option (A) is the correct answer.