91Ó°ÊÓ

To plot a direction field, we need to create a mesh of points and draw arrows based on the value of the differential equation, which is given by \(y'(t)=t(y-1)\). To plot this using a graphing utility, we need to first set the range for t and y. In this exercise, we are given the ranges as \(0 \leq t \leq 2\) and \(0 \leq y \leq 2\). Using these ranges, plot the direction field for the equation \(y'(t)=t(y-1)\) using a graphing utility such as Desmos or Wolfram Alpha.

To find the constant solutions, we should look for the situations when the function \(y(t)\) does not change. Mathematically, it means that we need to find the values of \(y\) for which \(y'(t) = 0\). From the given equation, the derivative is given as: $$y'(t) = t(y - 1)$$ Setting \(y'(t)\) to zero, we get: $$0 = t(y - 1)$$ As we can see, if \(y= 1\), the function \(y'(t)\) will be zero, regardless of the value of \(t\). Hence, the constant solution is \(y(t) = 1\).

Now that we have the direction field plot and the constant solutions, our task is to find the initial conditions \(y(0) = A\) that lead to increasing solutions. To do that, we will study the given differential equation: $$y'(t) = t(y - 1)$$ In order for a solution to be increasing, its derivative, \(y'(t)\) must be positive. Therefore, we want to find the conditions for which \(y'(t) > 0\). We know that: $$y'(t) = t(y - 1)$$ Given the range \(0 \leq t \leq 2\), we know that \(t\) will always be non-negative. Thus, the sign of \(y'(t)\) will be determined by the factor \((y - 1)\). If \((y-1) > 0\), then \(y'(t) > 0\) and the solution will be increasing. When \(t=0\), \(y(0)=A\). So, to find the initial conditions that lead to increasing solutions, we must find when \((y(0)-1) > 0\) or \((A-1) > 0\). From this inequality, we can say that: $$A > 1$$ Thus, any initial condition \(y(0)=A\) where \(A>1\) will lead to solutions that are increasing in time.