JEE PYQ: Differential Equation Question 35
Question 35 - 2019 (09 Apr Shift 1)
The solution of the differential equation $x\frac{dy}{dx} + 2y = x^2$ ($x \neq 0$) with $y(1) = 1$, is:
(1) $y = \frac{4}{5}x^3 + \frac{1}{5x^2}$
(2) $y = \frac{x^3}{5} + \frac{1}{5x^2}$
(3) $y = \frac{x^2}{4} + \frac{3}{4x^2}$
(4) $y = \frac{3}{4}x^2 + \frac{3}{4x^2}$
Show Answer
Answer: (3)
Solution
$\frac{dy}{dx} + \frac{2}{x}y = x$. I.F. $= x^2$. Solution: $yx^2 = \int x^3,dx = \frac{x^4}{4} + C$. $y(1) = 1$: $1 = \frac{1}{4} + C$, $C = \frac{3}{4}$. So $y = \frac{x^2}{4} + \frac{3}{4x^2}$.
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