Let y=y(x) be the solution of the differential equation, (x2+1)2dydx+2 – InterviewSolution

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1.	Let y=y(x) be the solution of the differential equation, (x2+1)2dydx+2x(x2+1)y=1 such that y(0)=0. If √a y(1)=π32, then the value of a is:
Answer» Let $y = y (x)$ be the solution of the differential equation, $(x^{2} + 1)^{2} \frac{d y}{d x} + 2 x (x^{2} + 1) y = 1$ such that $y (0) = 0$ . If $\sqrt{a} y (1) = \frac{π}{32}$ , then the value of $a$ is:

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