JEE Mains · Maths · STD 12 - 9. differential equations
If \(y = f(x)\) is the solution of the differential equation \(\frac{{dy}}{{dx}} = \left( {\tan \,x - y} \right){\sec ^2}\,x,\,x \in \left( { - \frac{\pi }{2},\frac{\pi }{2}} \right)\), such that \(y(0) = 0\), then \(y\left( { - \frac{\pi }{4}} \right)\) is equal to
- A \(\frac{1}{2} - e\)
- B \(\frac{1}{e} - 2\)
- C \(e -2\)
- D \(2 + \frac{1}{e}\)
Answer & Solution
Correct Answer
(C) \(e -2\)
Step-by-step Solution
Detailed explanation
\(\frac{d y}{d x}=(\tan x-y) \sec ^{2} x\) \(\frac{d y}{d x}+y \sec ^{2} x=\tan x \sec ^{2} x\) Let \(\tan x=t \Rightarrow \sec ^{2} x=\frac{d t}{d x}\) \(\therefore \frac{d y}{d t}=(t-y)\) \(\frac{d y}{d t}+y=t\) (Linear differential equation) After solving we get…
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