Solution: The given differential equation can be written as \begin{equation}\label{eq:27Jan2023-1} y^{-\frac{1}{2}}y^\prime -\frac{3y^{\frac{1}{2}}}{t} = t\quad \left( \text{divide the given ODE by } ty^{\frac{1}{2}} \right) \end{equation} Take $y^{\frac{1}{2}} = u$. Then differentiating both sides with respect to $t$, we have \begin{align*} \frac{1}{2}y^{-\frac{1}{2}} \frac{\mathrm{d} y}{\mathrm{d} t} = \frac{\mathrm{d} u}{\mathrm{d} t} \implies \frac{\mathrm{d}y}{\mathrm{d}t } = 2u \frac{\mathrm{d} u}{\mathrm{d} t}. \end{align*}

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Substituting the above in \eqref{eq:27Jan2023-1}, we obtain \begin{align*} 2\frac{\mathrm{d} u}{\mathrm{d} t} - \frac{3u}{t} = t & \implies \frac{\mathrm{d} u}{\mathrm{d} t} - \frac{3}{2t}u = \frac{1}{2}t. \end{align*} Thus, we obtained a linear differential equation in the form \[ \frac{\mathrm{d} y}{\mathrm{d} t} + P(t)y = Q(t). \]

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In order to solve this differential equation, we multiply both sides by the integrating factor $I$ which is defined as \[ I = e^{\int P(t) \mathrm{d} t}. \] Here in the given differential equation, we have \[ P(t) = -\frac{3}{2t},~\text{and }~ Q(t) = \frac{t}{2}. \] Thus, the integrating factor will be \begin{align*} I & = e^{\int P(t)~\mathrm{d} t} \\ & = e^{\int -\frac{3t}{2}~\mathrm{d} t} \\ & = e^{-\frac{3}{2} \ln t} \\ & = t^{\frac{-3}{2}}. \end{align*}

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Multiply both sides of the equation by the integrating factor to \begin{align*} & t^{-\frac{3}{2}} \frac{\mathrm{d} u}{\mathrm{d} t} - t^{-\frac{3}{2}} \frac{3}{2t}u = t^{-\frac{3}{2}} \frac{1}{2}t \\ \implies & \frac{\mathrm{d} }{\mathrm{d} t}\left( t^{-\frac{3}{2}} u(t) \right) = \frac{1}{2\sqrt{t} } \\ \implies & \int \frac{\mathrm{d} }{\mathrm{d} t}\left( t^{-\frac{3}{2}} u(t) \right)~\mathrm{d} t = \int \frac{1}{2\sqrt{t} }\mathrm{d} t \\ \implies & t^{-\frac{3}{2}} u(t) = \sqrt{t} + C . \end{align*}

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Note that \[ y(1) = 1 \implies u(1) = 1. \] So, \[ t^{-\frac{3}{2}} u(t) = \sqrt{t} + C \implies C = 0. \] Thus, \[ u(t) = t^2 \implies y(t) = t^4. \] Hence, \[ \textcolor{blue}{\boxed{y(2) = 16}} \]