91Ó°ÊÓ

The derivatives of a function \(y = f(x)\) of a quantity x is a measurement of the rate at which the function's value y changes when the variable x varies. The derivatives of

f with respect to x is what it's called.

Let the given function be\(xy'' + 2y' = 0\).

Then, the first derivative of the function is \(y' = m{x^{m - 1}}\).

The second derivative of the function is \(y'' = m(m - 1){x^{m - 2}}\).

Substitute\(y\),\(y'\)and \(y''\) into the differential equation.

\(\begin{aligned}{c}x(m(m - 1){x^{m - 2}}) + 2(m{x^{m - 1}}) = 0\\m(m - 1){x^{m - 1}} + 2m{x^{m - 1}} = 0\\(m(m - 1) + 2m){x^{m - 1}} = 0\\({m^2} + m){x^{m - 1}} = 0\end{aligned}\)

This implies that \({m^2} + m = 0\).

\(\begin{aligned}{c}{m^2} + m = 0\\m(m + 1) = 0\\m = 0, - 1\end{aligned}\)

Thus, the value of \(m\) is \(0\) or \( - 1\).