JEE PYQ: Differential Equation Question 41
Question 41 - 2019 (09 Jan Shift 1)
If $y = y(x)$ is the solution of the differential equation $x\frac{dy}{dx} + 2y = x^2$ satisfying $y(1) = 1$, then $y\left(\frac{1}{2}\right)$ is equal to:
(1) $\frac{7}{64}$
(2) $\frac{1}{4}$
(3) $\frac{49}{16}$
(4) $\frac{13}{16}$
Show Answer
Answer: (3)
Solution
I.F. $= x^2$. Solution: $yx^2 = \frac{x^4}{4} + C$. $y(1) = 1$: $C = \frac{3}{4}$. $y = \frac{x^2}{4} + \frac{3}{4x^2}$. $y\left(\frac{1}{2}\right) = \frac{1}{16} + 3 = \frac{49}{16}$.
BETA
