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JEE MainMathematicsModerateDifferential equation

If y=y(x) is the solution of the differential equation,
xdydx+2y=x2x\frac{dy}{dx} + 2y = x^{2} satisfying y

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xdydx+2y=x2x\frac{dy}{dx} + 2y = x^{2}
dydx+2xy=x\frac{dy}{dx} + \frac{2}{x}y = x
This is linear differential equation
 I. F.=e∫2xdx=e2lnx=x2\ I.\ F_{.} = e^{\int\frac{2}{x}dx} = e^{2lnx} = x^{2}
 Solution of differential equation is \ Solution\ of\ differential\ equation\ is\
y⋅x2=∫x⋅x2dx+Cy \cdot x^{2} = \int x \cdot x^{2}dx + C
yx2=x44+Cyx^{2} = \frac{x^{4}}{4} + C
y=x44+Cx2y = \frac{x^{4}}{4} + \frac{C}{x^{2}}
given y (1) = 1
1=14+C⇒C=341 = \frac{1}{4} + C \Rightarrow C = \frac{3}{4}
Hence  y(x)=x24+34x2\ y(x) = \frac{x^{2}}{4} + \frac{3}{4x^{2}}
y(12)=116+34×14y(12)=4916\begin{matrix} y\left( \frac{1}{2} \right) = \frac{1}{16} + \frac{3}{4 \times \frac{1}{4}} \\ y\left( \frac{1}{2} \right) = \frac{49}{16} \end{matrix}

Source: Grass-G / M-28 Mathematics Paper FST 16 Jan.docx

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