Let y = y(x) be the solution of the differential equation, d y d x +y x=2 x+x^ 2 x, x (…
Mathematics · JEE Main · NTA Exams — Differential Equations
Let y = y(x) be the solution of the differential equation, \(\frac{d y}{d x}+y \tan x=2 x+x^{2} \tan x,\) \(x \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\), such that \(y(0)=1\). Then:
- \(v\left(\frac{\pi}{4}\right)-v\left(-\frac{\pi}{4}\right)=\sqrt{2}\)
- \(y^{\prime}\left(\frac{\pi}{4}\right)-y^{\prime}\left(-\frac{\pi}{4}\right)=\pi-\sqrt{2}\)
- \(y\left(\frac{\pi}{4}\right)+y\left(-\frac{\pi}{4}\right)=\frac{\pi^{2}}{2}+z\)
- \(y^{\prime}\left(\frac{\pi}{4}\right)+y^{\prime}\left(-\frac{\pi}{4}\right)=-\sqrt{2}\)
Answer
(B) y^ ( 4 )-y^ (- 4 )= - 2
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