In Problems 19 and 20 verify that the indicated expression is an implicit solution of the given first-order differential equation. Find at least one explicit solution y (x) in each case. Use a graphing utility to obtain the graph of an explicit solution. Give an interval I of definition of each solution ./frac (dX)(dt)= (X -1)(1- 2X); ln /frac (2X- 1)( X -1)

The given differential equation is dx/dt = (x - 1)(1 - 2x) and the solution provided is ln|(2x - 1)/(x - 1)|.

Step 1: Verify the solution To verify the solution, we need to differentiate the given solution and check if it matches the given differential equation.

The derivative of ln|(2x - 1)/(x - 1)| using the chain rule is 1/(2x - 1) * 2 - 1/(x - 1) = (2/(2x - 1) - 1/(x - 1)).

Simplifying this gives us (2x - 1 - 2)/(x - 1)(2x - 1) = (x - 1)(1 - 2x), which is the given differential equation. So, the solution is verified.

Step 2: Find an explicit solution The given solution is already in explicit form y(x) = ln|(2x - 1)/(x - 1)|.

Step 3: Graph the solution To graph the solution, you would need to use a graphing utility like Desmos or a graphing calculator. Input the function y(x) = ln|(2x - 1)/(x - 1)| and observe the graph.

Step 4: Interval of definition The interval of definition is the set of all x-values for which the function is defined. In this case, the function is undefined when the denominator of the fraction inside the logarithm is zero, i.e., x = 1 or x = 1/2. Therefore, the interval of definition is (-∞, 1/2) U (1/2, 1) U (1, ∞).