Q120 / 160 · TS EAMCET 2023 (13 May Shift 2) · Mathematics
The general solution of the differential equation ( $\sec x+$ $\tan x) \frac{d y}{d x}+\left(\sec ^2 x+\sec x \tan x\right) y=1$ is
- Correct answerA$(1+\sin x) y=n \cos x+c$
- B$(1+\cos x) y=x \sin x+c$
- C$(\sec x+\tan x) y=x \sec x+c$
- D$(\sec x+\tan x) y=x+c$
Solution
$\begin{aligned} & \text {}(\sec x+\tan x) \frac{d y}{d x}+\left(\sec ^2 x+\sec x \tan x\right) y=1 \\ & \Rightarrow \quad(\sec x+\tan x) \frac{d y}{d x}+\sec x(\sec x+\tan x) y=1 \\ & \Rightarrow \quad(\sec x+\tan x)\left(\frac{d y}{d x}+y \sec x\right)=1 \\ & \Rightarrow \quad \frac{d y}{d x}+y \sec x=\frac{1}{\sec x+\tan x} \\ & \Rightarrow \quad \frac{d y}{d x}+y \cdot \sec x=\sec x-\tan x \\ & \Rightarrow \quad \frac{d y}{d x}+P y=Q \\ & \Rightarrow \quad \text { IF }=e^{\int P d x}=e^{\int \sec x d x} \\ & \quad=e^{\log (\sec x+\tan x)}=\sec x+\tan x \\ & \Rightarrow \quad y \cdot \operatorname{IF}=\int Q \cdot \operatorname{IF}+C \\ & \Rightarrow \quad(\sec x+\tan x)=\int(\sec x-\tan x) \cdot(\sec x+\tan x) d x+C \\ & \Rightarrow \quad(\sec x+\tan x) y=\int\left(\sec { }^2 x-\tan { }^2 x\right) d x+C \\ & \Rightarrow(\sec x+\tan x) y=x+C .\end{aligned}$
Hence, the correct answer is option: 'A'.