Question

If $y=y\left(x\right)$ is the solution of the differential equation, $x\frac{dy}{dx}+2y=x^2$ satisfying $y\left(1\right)=1$, then $y\left(\frac{1}{2}\right)$ is equal to:

• A. $\frac{1}{4}$
• B. $\frac{7}{64}$
• C. $\frac{13}{16}$
• D. $\frac{49}{16}$

Answer 1

The correct option is D $\frac{49}{16}$

The given differential equation can be rewritten as $\frac{dy}{dx}+\left(\frac{2}{x}\right)y=x$
which is a linear differential equation.
Then the integrating factor,
$I.F.=e^{\int \frac{2}{x}dx}=e^{\left(2logx\right)}=e^{\left(log{x}^2\right)}=x^2$

$\therefore$ The solution of the given differential equation is
$y\cdot x^2=\int x\cdot x^2+C$
where $C$ is the constant of integration
$\Rightarrow y\cdot x^2=\frac{x^4}{4}+C$
Given $y\left(1\right)=1\Rightarrow C=\frac{3}{4}$
$\therefore y=\frac{x^2}{4}+\frac{3}{4x^2}$
Hence,
$y\left(\frac{1}{2}\right)=\frac{1}{4\times 4}+\frac{3\times 4}{4}=\frac{49}{16}$