91Ó°ÊÓ

To check whether the function \(y \equiv y_{0}\) is a solution to the differential equation \(y'=f(x,y)\), we need to calculate its derivative and check if it's equal to \(f(x, y_0)\) for all \(a < x < b\), given that \(f(x,y_0)=0\) for all \(x\) in \((a, b)\). Since \(y \equiv y_{0}\) is a constant function, its derivative is: $$ y^\prime = \frac{d}{dx}(y_0) = 0 $$ It satisfies the given differential equation because \(y^\prime = 0 = f(x, y_0)\) for all \(a < x < b\). So, \(y \equiv y_{0}\) solves the differential equation in the interval \((a, b)\).

Assume that there exists another solution, \(y(x)\), of \(y'=f(x,y)\) that is equal to \(y_0\) at some point \(c\) in \((a, b)\). Then, $$ y(c) = y_0 $$ Now, consider the function \(g(x) = y(x) - y_0\). From the above assumption, we have \(g(c) = 0\). Taking the derivative of \(g(x)\), we get: $$ g'(x) = \frac{d}{dx} (y(x) - y_0) = \frac{dy}{dx} = f(x, y) $$ We know that \(f(x, y_0) = 0\) by the given condition. By the Mean Value Theorem (since \(f\) and \(f_y\) are continuous on an open rectangle containing \((x_{0}+y_{0})\)), there exists a \(y^* = y(c) + k (y(x) - y(c))\), where \(0 < k < 1\), such that: $$ f_x(x, y^*) = \frac{f(x, y(x)) - f(x,y_0)}{y(x) - y_0} = \frac{f(x, y) - 0}{g(x)} = \frac{g'(x)}{g(x)} $$ Since \(f_y\) is continuous, \(f_x(x, y^*)\) is continuous in \((a, b)\). Furthermore, \(g'(x) = 0 = g(c)\). So on applying the uniqueness theorem, we conclude that the only possible solution \(y(x)\) is \(y \equiv y_0\), as there can be no other solution in the interval \((a, b)\) that equals \(y_0\) at some point. Hence, the proof is complete.