Question

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

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

Answer 1

The correct option is C $y=\frac{x^2}{4}+\frac{3}{4x^2}$

$x\frac{dy}{dx}+2y=x^{2}andy\left(1\right)=1,\Rightarrow \frac{dy}{dx}+\left(\frac{2}{x}\right)y=x\therefore I.F.=e^{\int \frac{2}{x}dx}=e^{2\log x}=x^2\therefore y\cdot x^2=\int x\cdot x^{2}dx=\frac{x^4}{4}+CBut,y\left(1\right)=1\Rightarrow 1=\frac{1}{4}+C\Rightarrow C=\frac{3}{4}So,y{x}^2=\frac{x^4}{4}+\frac{3}{4}i.e.,y=\frac{x^2}{4}+\frac{3}{4x^2}$