JEE Mains · Maths · STD 12 - 9. differential equations
Let \(y = y\left( x \right)\) be the solutions of the differential equation, \(\left( {{x^2} + 1} \right)^2\,\frac{{dy}}{{dx}} + 2x\left( {{x^2} + 1} \right)\,y = 1\) such that \(y\left( 0 \right) = 0\). If \(\sqrt a y\left( 1 \right) = \frac{\pi }{{32}}\), then the value of \(‘a’\) is
- A \(\frac{1}{2}\)
- B \(1\)
- C \(\frac{1}{16}\)
- D \(\frac{1}{4}\)
Answer & Solution
Correct Answer
(C) \(\frac{1}{16}\)
Step-by-step Solution
Detailed explanation
\(\frac{d y}{d x}+\left(\frac{2 x}{x^{2}+1}\right) y=\frac{1}{\left(x^{2}+1\right)^{2}}\) (Linear differential equation) \(\therefore \quad I.{\rm{F}} = {e^{\ell n\left( {{x^2} + 1} \right)}} = \left( {{x^2} + 1} \right)\) So, general solution is…
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