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\({x^2}y'' - 7xy' + 15y = 0\).

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.

\(\begin{aligned}{c}{x^2}(m(m - 1){x^{m - 2}}) - 7x(m{x^{m - 1}}) + 15({x^m}) = 0\\m(m - 1){x^m} - 7m{x^m} + 15{x^m} = 0\\(m(m - 1) - 7m + 15){x^m} = 0\\({m^2} - 8m + 15){x^m} = 0\end{aligned}\)

This implies that\({m^2} - 8m + 15\).

\(\begin{aligned}{c}{m^2} - 8m + 15 = 0\\(m - 3)(m - 5) = 0\\m = 3,5\end{aligned}\)

Thus, the value of \(m\) is \(3\) or \(5\).