corrigé

6
R
χ(λ) = λ3+λ2+λ=λλ2+λ+ 1λ0:= 0 λ±:= 1
2−1±i√3
x(t) = c0+c−exp (λ−t) + c+exp (λ+t), c0, c±∈C.
c0c±
˙x(t) = c−λ−exp (λ−t) + c+λ+exp (λ+t) ¨x(t) = c−λ2
−exp (λ−t) + c+λ2 +exp (λ+t)
t:= 0
λ−λ+
λ2
−λ2 + c−
c+=1 1.
c±=−1
diag (λ1, λ2)n=diag (λn
1, λn 2) exp M=P∞
n=0 Mn
n!
A0 5 A A
P:= 2 1
−1 2 P−1=1
52−1
1 2 
P−1AP =D=diag (0,5)
A=P DP −1exp (tA) = Pexp (tD)P−1
exp (tA) = 1
52 1
−1 2  1 0 0 exp (5t) 2−1 1 2 =1
52 1
−1 2  2−1 exp (5t) 2 exp (5t)
=1
54 + exp (5t) 2 (−1 + exp (5t))
2 (−1 + exp (5t)) 1 + 4 exp (5t).
Rexp (tA)X0X0∈C2

x0 = x−x2=x(1 −x)
(x, I)λ∈C(λx, I)
x:= 1 I:= Rλ /∈ {0,1}
f(x) := 1
x−x2g(t) := 2
t
f(x(t)) ˙x(t) = g(t)f g
IfIg(t0, x0)∈Ig×Iff(x0)6= 0
J3t0(x, J)x(t0) = x0
F(x(t)) = G(t)
F f F (x0)=0 G g G (t0) = 0
ln 
x(t)
1−x(t)×1−x0 x0
= ln t2
t2
0
F(x) = ´dx
x(1−x)=´1
x+1 1−xdx= ln 
x
1−x
x0x
1−x
x0
1−x0
x0t26=t2
0(x0−1)
x0−1
x0<0x0∈]0,1[ Imax =R
t∈n±t0qx0−1
x0oImax =i−∞ , t0qx0−1 x0ht0<0x0>1Imax =
it0qx0−1
x0,+∞ht0>0x0>1Imax =i−|t0|qx0−1 x0,|t0|qx0−1 x0hx0<0
t0 0 = x(0) −x(0)2˙x(0) ∈R
t0∈I\{0}x0:= x(t0)t
0x0/∈ {0,1}x(0) = 0 x0= 0 x= 0
x0= 1 t0x= 1
(0,1) x0<0
0 0