First Order Linear Equations

Consider the fol lowing equation :



with initial conditions . We would like to find the general solution.
First, we consider the homogeneous problem:



This can be solved by sepa ration of variables :



which is integrated leading to



so we get



where . The unknown constant is found by setting and
using the initial data. For example:



we get a(t) = 1/(1 + t) and A(t) = ln(1 + t) so that



so that x_H(t) = 1 + t.

Now, how do we solve the general system . We introduce a useful trick called
the integrating factor . We can rewrite the equation as



We call the integrating factor. Note that



Let x(t) = y(t)F(t). Let's differentiate this :

since dF/dt = a(t)F = 0. Thus, we see that since dx/dt-a(t)x = b(t), we have



so that



This can be integrated:



and since x(t) = y(t)F(t), we get



is the general solution.
Example.
Solve

x' = x/(1 + t) + 2t x(0) = 3.

a(t) = 1/(1 + t). Thus A(t) = ln(1 + t) and . So



The initial data yields, K = 3 so that

x(t) = 3 + 5t - 2 ln(1 + t).

Homework Solve the following:

1. x' = -x/t + t with x(1) = 0

2. x' = -4x + sin(3t) with x(0) = 0.

3. x' = -t²x + t with x(0) = 1.

4. x' = -2x/t + sin(t) with x( π) = 1/π

For problems 2 and 4, describe the behavior as t → ∞.
Note the third one should convince you of the worthlessness of exact methods
of solution in general .