If y=y(x) is the solution of the differential | vedclass

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(D) The given differential equation is $x \frac{dy}{dx} + 2y = x^2$.Dividing by $x$,we get $\frac{dy}{dx} + \left(\frac{2}{x}\right)y = x$.This is a l

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$\frac{49}{16}$

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(D) The given differential equation is $x \frac{dy}{dx} + 2y = x^2$.Dividing by $x$,we get $\frac{dy}{dx} + \left(\frac{2}{x}\right)y = x$.This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = \frac{2}{x}$ and $Q = x$.The Integrating Factor ($I$.$F$.) is $e^{\int P dx} = e^{\int \frac{2}{x} dx} = e^{2 \ln|x|} = x^2$.The general solution is $y \cdot (I.F.) = \int Q \cdot (I.F.) dx + C$.$y \cdot x^2 = \int x \cdot x^2 dx + C = \int x^3 dx + C = \frac{x^4}{4} + C$.Given $y(1) = 1$,we substitute $x = 1$ and $y = 1$: $1(1)^2 = \frac{1^4}{4} + C \implies 1 = \frac{1}{4} + C \implies C = \frac{3}{4}$.Thus,the particular solution is $y x^2 = \frac{x^4}{4} + \frac{3}{4}$,or $y = \frac{x^2}{4} + \frac{3}{4x^2}$.Now,we find $y\left(\frac{1}{2}\right) = \frac{(\frac{1}{2})^2}{4} + \frac{3}{4(\frac{1}{2})^2} = \frac{1/4}{4} + \frac{3}{4(1/4)} = \frac{1}{16} + 3 = \frac{1 + 48}{16} = \frac{49}{16}$.

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

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