If $y=y(x)$ is the solution of the differential equation $x \frac{\mathrm{dy}}{\mathrm{d} x}+2 y=x^2$ satisfying $y(1)=1$, then the value of $y\left(\frac{1}{2}\right)$ is

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Aicharya · Accepted Answer

$\begin{aligned} & x \frac{\mathrm{~d} y}{\mathrm{~d} x}+2 y=x^2 \ & \frac{\mathrm{~d} y}{\mathrm{~d} x}+\left(\frac{2}{x}ight) y=x \ & 	ext { I.F. }=\mathrm{e}^{\int \frac{2}{x} \mathrm{~d} x}=\mathrm{e}^{2 \log x}=x^2\end{aligned}$ $	herefore \quad$ Solution of differential equation is $\begin{aligned} & \quad y \cdot x^2=\int x \cdot x^2 \mathrm{~d} x+\mathrm{c} \ & \quad y x^2=\frac{x^4}{4}+\mathrm{c} ...(i)\ & 	herefore \quad \ & y=\frac{x^2}{4}+\frac{\mathrm{c}}{x^2} ...(ii)\end{aligned}$ $\begin{aligned} & 	ext { Given, } y(1)=1 \ & \Rightarrow 1=\frac{1}{4}+\mathrm{c} \ & \Rightarrow \mathrm{c}=\frac{3}{4}\end{aligned}$ $	herefore \quad$ equation (i) becomes, $\begin{aligned} & y=\frac{x^2}{4}+\frac{3}{4 x^2} \ 	herefore \quad & y\left(\frac{1}{2}ight)=\frac{\left(\frac{1}{2}ight)^2}{4}+\frac{3}{4 	imes\left(\frac{1}{2}ight)^2}=\frac{1}{16}+3=\frac{49}{16} \end{aligned}$

If $y=y(x)$ is the solution of the differential equation $x \frac{\mathrm{dy}}{\mathrm{d} x}+2 y=x^2$…

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