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Check the continuity of the function at.

The first derivative of the function is,

\(y' = - 2x\)

Substitute \(y\) and \(y'\) in the differential equation.

\(\begin{aligned} x( - 2x) - 2( - {x^2}) &= 0\\0 &= 0\end{aligned}\)

Hence, the left-hand limit is equal to the right-hand limit, so the function is continuous at and the solution is verified.

Check the continuity of the function at\(x \ge 0\).

The first derivative of the function is,

\(y' = 2x\)

Substitute \(y\) and \(y'\) in the differential equation.

\(\begin{aligned} x(2x) - 2({x^2}) &= 0\\0 &= 0\end{aligned}\)

Hence, the left-hand limit is equal to the right-hand limit, so the function is continuous at \(x \ge 0\) and the solution is verified.

Hence, the piecewise-defined function is a solution of the differential equation on the interval \(( - \infty ,\infty )\).