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

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

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$\frac{49}{16}$

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(C) The given differential equation is $x \frac{dy}{dx} + 2y = x^2$.Dividing by $x$,we get $\frac{dy}{dx} + \frac{2}{x}y = x$.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 $(IF)$ is $e^{\int P dx} = e^{\int \frac{2}{x} dx} = e^{2 \ln|x|} = x^2$.The solution is given by $y \cdot (IF) = \int Q \cdot (IF) dx + C$.$y \cdot x^2 = \int x \cdot x^2 dx + C = \int x^3 dx + C$.$y x^2 = \frac{x^4}{4} + C$.Given $y(1) = 1$,we substitute $x=1$ and $y=1$ to find $C$: $1(1)^2 = \frac{1^4}{4} + C \implies 1 = \frac{1}{4} + C \implies C = \frac{3}{4}$.Thus,the solution is $y x^2 = \frac{x^4}{4} + \frac{3}{4} = \frac{x^4 + 3}{4}$.To find $y\left(\frac{1}{2}ight)$,substitute $x = \frac{1}{2}$:$y \left(\frac{1}{2}ight)^2 = \frac{(\frac{1}{2})^4 + 3}{4} = \frac{\frac{1}{16} + 3}{4} = \frac{\frac{49}{16}}{4} = \frac{49}{64}$.Wait,re-evaluating: $y \cdot \frac{1}{4} = \frac{\frac{1}{16} + 3}{4} = \frac{49}{64} \implies y = \frac{49}{64} 	imes 4 = \frac{49}{16}$.

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

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