Xt = xo + ut + Bt
Letting t → 0 and using the fact that Xt ~ Xo = x when t is small, we arrive at
x + Gx = 0.
(0.1)
This is the ODE that v should satisfy.
iv. What you need to do: Solve the ODE (0.1). To solve an ordinary differential equation like this, we need to find a particular solution and the general solution. Try a particular solution of the form
fp(x) = Ax + B
where A and B are some constants that you will need to determine. The general solution should be of form
fg(x) = C1e^(B1x) + C2e^(B2x)
The coefficients C1 and C2 are free constants that we will determine in the next part, but you should be able to determine
B1 and B2 by solving the associated characteristic equation (a quadratic equation). Therefore, your final answer for this part should be of form
v(x) = fp(x) + fg(x) = Ax + B + C1e^(B1x) + C2e^(B2x) where A, B, 1, 2 are constants that you can determine. V. What you need to do: Use the appropriate growth condition to argue that C1 = C2 = 0. Suppose it is given that v(x) grows at most linearly when x as x → 0 (see Remark 0.1). Use this to argue C1 = C2 = 0. You are not asked to write a 1-million-page-proof, just one or two sentences to explain your line of thoughts are good enough vi. The solution v(x) = Ax + B you get here should be the same as the solution you get from Part (a).
v(xo)
X0
a^2
