Question

Solution of the differential equations $\cos xdy=y(\sin x-y)dx,0<x<\frac{\pi }{2}$ is

• A. $\sec x=(\tan x+C)y$
• B. $y\sec x=\tan x+C$
• C. $y\tan x=\sec x+C$
• D. $\tan x=(\sec x+C)y$

Answer 1

The correct option is A

$\sec x=(\tan x+C)y$

Explanation for the correct option:

Simplifying and Integrating the given expression:

$$\begin{gathered} \cos xdy=y(\sin x-y)dx \\ \frac{dy}{dx}=y\frac{(\sin x-y)}{\cos x}\sec x \\ \frac{dy}{dx}=y\tan x-y^2\sec x\left[\because \frac{\sin \theta }{\cos \theta }=\tan \theta \text{and}\frac{1}{\cos \theta }=\sec \theta \right] \\ \frac{dy}{dx}=y^2\left(\frac{1}{y}\tan x-\sec x\right) \\ \frac{1}{y^2}\frac{dy}{dx}-\frac{1}{y}\tan x=-\sec x \end{gathered}$$

Substituting

$$\begin{gathered} \frac{1}{y}=t \\ \therefore \frac{1}{y^2}\frac{dy}{dx}=\frac{dt}{dx} \end{gathered}$$

$$\begin{gathered} -\frac{dt}{dx}-t\left(\tan x\right)=-\sec x \\ \frac{dt}{dx}+t\left(\tan x\right)=\sec x \end{gathered}$$

Now, Integrating the above expression we get

$IF=e^{\tan xdx}=\sec \left(x\right)\left[\text{where}IF=\text{integrating factor}\right]$

Solution is

$$\begin{gathered} t\left(IF\right) \\ =\int \left(IF\right)\sec \left(x\right)dx \\ =\frac{1}{y}\sec \left(x\right) \\ =\tan x+c \end{gathered}$$

Therefore, option (A) is the correct answer.