(a) Suppose Y1(t) is a solution of an autonomous system dY/dt = F(Y). Show that Y2(t) = Y1(t + t0) is also a solution for any constant t0. (b) What is the relationship between the solution curves of Y1(t) and Y2(t)?

To show that Y2(t) = Y1(t + t0) is also a solution for the autonomous system dY/dt = F(Y)

we'll have to prove that Y2(t) satisfies the same differential equation as Y1(t).

Let's go through the steps:

Step 1: Define Y2(t) in terms of Y1(t).

We have Y2(t) defined as Y1(t + t0).

Step 2: Differentiate Y2(t) with respect to t.

We will use the chain rule of differentiation which states that if we have a function g(h(t)), then the derivative dg/dt is dg/dh times dh/dt.

Applying this rule, we get

dY2/dt = dY1(t + t0)/dt = dY1/d(t + t0) * d(t + t0)/dt.

Since t0 is a constant, d(t + t0)/dt equals 1.

Thus the equation simplifies to dY2/dt = dY1/d(t + t0).

Step 3: Substitute the expression for dY1/dt.

We were given that Y1(t) is a solution to the differential equation dY/dt = F(Y), so dY1/dt = F(Y1).

Substituting this into the equation for dY2/dt we have

dY2/dt = F(Y1(t + t0)).

Step 4: Evaluate F at Y2.

Because F is a function of Y only and does not explicitly depend on t (hence the term 'autonomous'), we can write F(Y1(t + t0)) as F(Y2(t)).

Step 5: Combine the results to confirm that Y2(t) satisfies the differential equation.

We've shown that dY2/dt = F(Y2(t)), which means Y2(t) satisfies the same differential equation as Y1(t), thereby proving that Y2(t) is indeed a solution for any constant t0.