JEE Mains · Maths · STD 12 - 9. differential equations
Let \(y=y(x)\) be the solution of the differential equation, \(x y^{\prime}-y=x^{2}(x \cos x+\sin x), x>0\) If \(y (\pi)=\pi,\) then \(y ^{\prime \prime}\left(\frac{\pi}{2}\right)+ y \left(\frac{\pi}{2}\right)\) is equal to
- A \(2+\frac{\pi}{2}\)
- B \(1+\frac{\pi}{2}\)
- C \(1+\frac{\pi}{2}+\frac{\pi^{2}}{4}\)
- D \(2+\frac{\pi}{2}+\frac{\pi^{2}}{4}\)
Answer & Solution
Correct Answer
(A) \(2+\frac{\pi}{2}\)
Step-by-step Solution
Detailed explanation
\(x \frac{d y}{d x}-y=x^{2}(x \cos x+\sin x), x>0\) \(\frac{d y}{d x}-\frac{y}{x}=x(x \cos x+\sin x) \Rightarrow \frac{d y}{d x}+P y=Q\) so, \(I \cdot F \cdot= e ^{\int-\frac{1}{ x } d x }=\frac{1}{| x |}=\frac{1}{ x }( x >0)\) Thus,…
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