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SE
t7→ xpart(t)
SHA
(H)A
A
x0+a(t)x=b(t) (E)
a b I R K =R C x
tC1IK
x0+a(t)x= 0 (H)
t7→ x(t)I
x0(t) + a(t)x(t) = 0 ⇐⇒ x0(t)
x(t=−a(t)
A a I
ln |x(t)|=−A(t) + K|x(t)|=eKe−A(t)
x(t) = λe−A(t)λ=±eK

A a I
t7→ x(t)⇐⇒ ∃λ∈K∀t∈I, x(t) = λe−A(t)
⇐=
=⇒x t 7→ x(t) = e−A(t)z(t)∀t∈I e−A(t)6= 0 t7→ z(t)
I
SHSH
C1(I, K)
SE
xpart :t7→ xpart(t)
t7→ x(t)⇐⇒ t7→ x(t)−xpart(t)
SE=xpart +SHC1(I, K)
t7→ xpart(t) + λe−A(t)
t7→ xpart(t)t7→ A(t)t7→ a(t)I
t7→ xpart(t)
t7→ λ(t)e−A(t)
λ0(t) = b(t)eA(t)
t7→ λ(t)
1
x0+a(t)x=b(t) (E)
a, b ∈ C(I, K) (t0, x0)∈I×Kt7→ x(t)E I
x(t0) = x0
t7→ xpart (t)t7→ A(t)t7→ a(t)I
t7→ x(t) = xpart (t) + λe−A(t)
λ λ
λ=x0−xpart (t0)eA(t0)
y0+ 2y=1
1 + exRy(0) = 1
tx0−2x=t3x(1) = 1 ]0,+∞[

y0+ tan(t)y= sin(2t)y(0) = 1 ] −π/2, π/2[
(E)x00 +a(t)x0+b(t)x=c(t)
IRa, b, c I K=R C t7→ x(t)
IK
(H)x00 +a(t)x0+b(t)x= 0
SH(H)C2(I, K)
H
H
(t2+ 2t)x00 −2(t+ 1)x0+ 2x= 0
t7→ x(t) ]0,+∞[
1 2
xpart :t7→ xpart(t) (E)
SE=xpart +SH
C2(I, K)
(E)
xpart +α1ϕ1+α2ϕ2
xpart (E)ϕ1, ϕ2SHα1, α2
K

2
t0,(x0, x1)∈I×K2(E)x:I→K
x(t0) = x0x0(t0) = x1
(t2+ 2t)x00 −2(t+ 1)x0+ 2x= 1
x x(1) = x0(1) = 1
xpart (E)
(E)x00 +a(t)x0+b(t)x=c(t) (H)x00 +a(t)x0+b(t)x= 0
a, b, c ∈ C0(I, K)
t7→ ϕ(t) (H) 0 I
x(E)t7→ ϕ(t)×z(t)
z0
z(t) = x(t)
ϕ(t)ϕ
z00 (t) + z0(t)ha(t)+2ϕ0(t)
ϕ(t)i=c(t) ϕ(t)
(H)
(E) (t2+t)x00 + (t−1)x0−x=t2e−t
(H) (E)
(H) ]1,+∞[
t7→ e−t(E) ]1,+∞[
(E) ]1,+∞[

(E)
(H)
x(t) =
+∞
X
k=0
aktk, t ∈]−R, R[
akk∈N K R
(E)xy00 −y0+ 4x3y= 0
(E) ]0,+∞[ (E)
0y7→
+∞
X
n=0
anxnR
an+4 an
p∈Na4p+1 a4p+3
p∈Na4pa0p a4p+2 a2p
y
(E) ]0,+∞[
(H)
x0
1(t) = a11x1(t) + · · · +a1nxn(t)
· · ·
x0
n(t) = an1x1(t) + · · · +annxn(t)
aij K=R C t7→ x1(t), . . . , t 7→ xn(t)R K
(H)X0(t) = AX(t)
A∈ Mn(K)XR Kn
SH(H)
F(R,Kn)R Kn
A
R
A∈Mn(K)