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)

$x\frac{dy}{dx}+2y=x^2:y(1)=1$
$\frac{dy}{dx}+\left(\frac{2}{x}\right)y=x(LDEiny)$
IF = $e^{\int \frac{2}{x}dx}=e^{2\ell nx}=x^2$
$y.(x^2)=\int x.x^{2}dx=\frac{x^4}{4}+C$
y(1) = 1
$1=\frac{1}{4}+C\Rightarrow C=1-\frac{1}{4}=\frac{3}{4}$
$y{x}^2=\frac{x^4}{4}+\frac{3}{4}$
$y=\frac{x^2}{4}+\frac{3}{4x^2}$