AP EAMCET · Maths · Differential Equations
The general solution of the differential equation \(\frac{d y}{d x}+\frac{\sec x}{\cos x+\sin x} y=\frac{\cos x}{1+\tan x}\) is
- A \((\cos x+\sin x) y=\sin x+c\)
- B \((\cos x+\sin x) y=\cos x+c\)
- C \((1+\tan x) y=\cos x+c\)
- D \(\sec x(\cos x+\sin x) y=\sin x+c\)
Answer & Solution
Correct Answer
(D) \(\sec x(\cos x+\sin x) y=\sin x+c\)
Step-by-step Solution
Detailed explanation
\( P(x) = \frac{\sec x}{\cos x+\sin x} = \frac{\sec^2 x}{1+\tan x} \) \( \int P(x) dx = \int \frac{\sec^2 x}{1+\tan x} dx = \ln|1+\tan x| \) \( IF = e^{\ln|1+\tan x|} = 1+\tan x \) \( y \cdot IF = \int Q(x) \cdot IF dx + c \)…
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