JEE PYQ: Differential Equation Question 34
Question 34 - 2019 (08 Apr Shift 1)
Let $y = y(x)$ be the solution of the differential equation $(x^2+1)^2\frac{dy}{dx} + 2x(x^2+1)y = 1$ such that $y(0) = 0$. If $\sqrt{a},y(1) = \frac{\pi}{32}$, then the value of ‘$a$’ is:
(1) $\frac{1}{4}$
(2) $\frac{1}{2}$
(3) 1
(4) $\frac{1}{16}$
Show Answer
Answer: (4)
Solution
$\frac{dy}{dx} + \frac{2x}{1+x^2}y = \frac{1}{(1+x^2)^2}$. I.F. $= 1+x^2$. Solution: $y(1+x^2) = \tan^{-1}x + c$. $y(0) = 0$: $c = 0$. $y(1) = \frac{\pi/4}{2} = \frac{\pi}{8}$. $\sqrt{a}\cdot\frac{\pi}{8} = \frac{\pi}{32}$, so $\sqrt{a} = \frac{1}{4}$, $a = \frac{1}{16}$.
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