Question

The solution of the differential equation $x \frac{dy}{dx} + 2y = x^2 \, (x \ne 0)$ with $y(1) = 1$, is

• A. $y = \frac{x^3}{5} + \frac{1}{5x^2}$
• B. $y = \frac{4}{5}x^3 + \frac{1}{5x^2}$
• C. $y = \frac{3}{5}x^2 + \frac{1}{4x^2}$
• D. $y = \frac{x^2}{4} + \frac{3}{4x^2}$

Answer 1

Answer: (D)

$x \frac{dy}{dx} + 2y = x^2 : y(1) = 1$
$\frac{dy}{dx} + \bigg(\frac{2}{x}\bigg)y \, = \, x \, \, (LDE \, in \, y) $
IF = $e^{\int \frac{2}{x} dx} \, = \, e^{2\ell n x} = x^2$
$y.(x^2) = \int x.x^2 dx = \frac{x^4}{4} + C$
y(1) = 1
$1 = \frac{1}{4} + C \Rightarrow C =1 - \frac{1}{4} = \frac{3}{4}$
$yx^2 = \frac{x^4}{4} + \frac{3}{4}$
$y = \frac{x^2}{4} + \frac{3}{4x^2}$