1. The problem is to find the direction field for the first order differential equation $$\frac{dx}{dt} = (t^2 - 1)x.$$\n\n2. The direction field shows the slope of the solution curves at various points $(t,x)$ in the plane. The slope at each point is given by the right side of the differential equation: $$\text{slope} = (t^2 - 1)x.$$\n\n3. Important rules:\n- When $x=0$, the slope is zero regardless of $t$, so the direction field has horizontal segments along the $t$-axis.\n- When $t^2 - 1 < 0$, i.e., for $-1 < t < 1$, the slope changes sign depending on $x$. For $x>0$, slope is negative (downward), for $x<0$, slope is positive (upward).\n- When $t^2 - 1 > 0$, i.e., for $t < -1$ or $t > 1$, the slope has the same sign as $x$, so for $x>0$ slope is positive (upward), for $x<0$ slope is negative (downward).\n\n4. To draw the direction field, plot short line segments at grid points $(t,x)$ with $t$ and $x$ from -2 to 2 at intervals of 1. The slope at each point is calculated as $$m = (t^2 - 1)x.$$\n\n5. For example, at $(t,x) = (0,1)$, slope is $(0^2 - 1) \times 1 = -1$, so the segment slopes downward. At $(t,x) = (2,1)$, slope is $(2^2 - 1) \times 1 = 3$, so the segment slopes steeply upward.\n\n6. This pattern matches the description: slopes point downward below the $t$-axis for $x>0$ when $-1 < t < 1$, and upward for $t < -1$ or $t > 1$.\n\nFinal answer: The direction field is given by line segments at each point $(t,x)$ with slope $$m = (t^2 - 1)x.$$