Here's the part in which (7) is derived from (1).

If the functions p and g are continuous on an open interval I = (, ) containing the point t = t0, then there exists a unique function y = (t) that satisfies the differential equation

y' + p(t)y = g(t) (1)

for each t in I, and that also satisfies the initial condition

y(t0) = y0, (2)

where y0 is an arbitrary prescribed initial value.

[describes a formula defined in an earlier section, which leads to:]

(t)y = integral((t)g(t) dt + c), (3)

where

(t) = exp[integral(p(t) dt)]. (4)

The derivation in Section 1.2 shows that if Eq. (1) has a solution, then it must be given by Eq. (3). By looking a little more closely at that derivation, we can also conclude that the differential equation (1) must indeed have a solution. Since p is continuous for [(t)y]' = (t)g(t). (5)

Since both and g are continuous, the function g is integrable, and Eq. (3) follows from Eq. (5). Further, the integral of g is differentiable, so y as given by Eq. (3) exists and is differentiable throughout the interval (t) = exp[integral(p(s)ds) from t0 to t], (6)

and it follows that (t0) = 1. Using the integrating factor given by Eq. (6), and choosing the lower limit of integration in Eq. (3) also to be t0, we obtain the general solution of Eq. (1) in the form

y = 1/(t)[ integral((s)g(s)ds+c) from t0 to t]. (7)

Question:

(a) Show that the solution(7) of the general linear equation (1) can be written in the form y=cy_1(t) + y_2(t) where c is an arbitrary constant. (b) Show that y_1 is a solution of the differential equation y'+p(t)y=0 corresponding to g(t)=0. (c) Show that y_2 is a solution of the full linear equation y'+p(t)y=g(t)