JEE Mains · Maths · STD 12 - 9. differential equations
Let \(y = y(x)\) be the solution of the differential equation \(\frac{{dy}}{{dx}} + y\,\tan \,x = 2x\, + \,{x^2}\,\tan \,x\,,\,x\, \in \,\left( { - \frac{\pi }{2},\frac{\pi }{2}} \right),\) such that \(y(0) = 1.\) Then
- A \(y'\,\left( {\frac{\pi }{4}} \right)\, + \,y'\left( {\frac{{ - \pi }}{4}} \right) = - \sqrt 2 \)
- B \(y'\,\left( {\frac{\pi }{4}} \right)\, - \,y'\left( {\frac{{ - \pi }}{4}} \right) = \pi \, - \,2\)
- C \(y\,\left( {\frac{\pi }{4}} \right)\, - \,y\left( { - \frac{\pi }{4}} \right) = \sqrt 2 \)
- D \(y\,\left( {\frac{\pi }{4}} \right)\, + \,y\left( { - \frac{\pi }{4}} \right) = \frac{{{\pi ^2}}}{2}\, + \,2\)
Answer & Solution
Correct Answer
(B) \(y'\,\left( {\frac{\pi }{4}} \right)\, - \,y'\left( {\frac{{ - \pi }}{4}} \right) = \pi \, - \,2\)
Step-by-step Solution
Detailed explanation
\(\frac{d y}{d x}+y(\tan x)=2 x+x^{2} \tan x\) I.F. \( = {e^{ \pm \int {\tan } xdx}} = {e^{\ln x{\mathop{\rm sen}\nolimits} x}} = \sec x\) \(\therefore y . \sec x=\int\left(2 x+x^{2} \tan x\right) \sec x \cdot d x\) \(=\int 2 x \sec x d x+\int x^{2}(\sec x \cdot \tan x) d x\) y…
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