* Consider the initial value problem*

*
(a) Find the solution u(t) of this problem.
(b) Find the smallest T such that for all t > T.
*

**Part (a) **

The MATLAB module Ch03Sec03Prob24 will solve a second order linear initial value problem and plot the solution in the case of constant coefficients and complex roots of the characteristic equation. The differential equation is entered as a vector that contains the coefficients of the characteristic equation (enclosed in square brackets). The program also prompts the user for the domain of the function for the purpose of producing a graph. Here is a transcript of the MATLAB session that produced the graph below:

>> Ch03Sec03Prob24 Enter the characteristic polynomial as a vector [a b c] => [5 2 7] We will solve the homogeneous equation: 5 u'' + 2 u' + 7 u = 0. Enter the initial value of u => 2 Enter the inital value of u' => 1 Enter the domain of the function: Minimum t-value => 0 Maximum t-value => 20 The solution to the IVP is u(t) = 2*exp(-0.2*t).*cos(1.16619*t) + 1.20049*exp(-0.2*t).*sin(1.16619*t).

**Part (b) **

From the graph in part (a), we can see that the graph is oscillating, but approaches zero. It seems that for *t* around 15, the function values
are between -0.1 and 0.1, but this is a really rough guess. Let's use MATLAB
to get a better approximation of *t* so that |*u(t)*| < 0.1. The
data points are stored in arrays titled 'tpts' and 'upts'.

>> size(upts) % Remind ourselves that there are 1000 data points ans = 1 1000 >> for i=1:1000 j=1001-i; % Look backwards through data points if (abs(upts(j)) > 0.1) % Quit looking when |u| > 0.1 and then exit loop disp(sprintf('=> When t = %.5g, u = %.5g is less than 0.1 in absolute value.', tpts(j+1),upts(j+1))) return end % End if statement end % End for loop => When t = 14.515, u = -0.099663 is less than 0.1 in absolute value.

Thus, for all *t* > 14.515, we have -0.1 < *u(t)* < 0.1.