Bourbaki marks this one as harder than the rest.
Let G be a complete normed algebra with unit e over the field R.
a) Let t↦a(t) be a regulated function on an interval J⊂R with values in G. Show that the integral u of the linear equation dx/dt=a(t)x which takes the value e at a point t0∈J is invertible on J, and that its inverse is the solution of the equation dx/dt=−xa(t) (if v is the integral of this last equation which takes the value e at the point t0, consider the linear equations satisfied by uv and vu). Deduce that, for every x0∈G, the integral of dx/dt=a(t)x which takes the value x0 at the point t0 is equal to u(t)x0.
b) Let a(t),b(t) be two regulated functions on J, and u and v the integrals of the equations dx/dt=a(t)x,dx/dt=xb(t) taking the value e at the point t0. Show that the integral of the equation dx/dt=a(t)x+xb(t) which takes the value x0 at the point t0 is equal to u(t)x0v(t).
c) Let a(t),b(t),c(t),d(t) be four regulated functions on J, and (u,v) a solution of the system of two linear equations
dtdx=a(t)x+b(t)y,dtdy=c(t)x+d(t)y.
Show that, if v is invertible on J, then w=uv−1 is an integral of the equation dz/dt=b(t)+a(t)z−zd(t)−zc(t)z (the "Ricatti equation"); and conversely. Deduce that every integral of this last equation on J, taking the value x0 at the point t0, is of the form (A(t)x0+B(t)e)(C(t)x0+D(t)e)−1, where A(t), B(t), C(t) and D(t) are continuous maps from J into L(E) satisfying the identity U.(xy)=(U.x)y.
d) Let w1 be an integral of dz/dt=b(t)+a(t)z−zd(t)−zc(t)z on J; show that if w is another integral of this equation such that w−w1 is invertible on J then w−w1 can be expressed in terms of integrals of the equations dx/dt=−(d+cw1)x and dx/dt=x(a−w1c) (consider the equation satisfied by (w−w1)−1, and use b) ).