sujet omsi 12072546681

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y0+a(t)y=b(t)
a b I R
ϕ:I→Rϕ
ϕ0(t) + a(t)ϕ(t) = b(t).
y0+a(t)y= 0 E0
ϕ:I→R(E0)t∈I
ϕ0(t) + a(t)ϕ(t) = 0
a I A a I
ϕ0(t) + a(t)ϕ(t) = 0 ⇐⇒ ¡ϕ0(t) + a(t)ϕ(t)¢eA(t)= 0 ⇐⇒ ³ϕ(t)eA(t)´0 = 0.
I ϕ(t)eA(t)λ∈R
∀t∈I, ϕ(t) = λe−A(t)
s y0+a(t)y=b(t)
t→s(t) + λe−A(t)λ∈R
•t→s(t) + λe−A(t)(E)
•ϕ(E) (ϕ−s)0+a(t)(ϕ−s) = b(t)
(ϕ−s) (E0)λ
ϕ(t)−s(t) = λe−A(t)⇐⇒ ϕ(t) = s(t) + λe−A(t).
i∈J1, nK
y0+a(x)y=gi(x) (Ei)
yi(Ei)
n
X
i=1
yiy0+b(x)y=
n
X
i=1
gi(x) (E)

t→λ(t)e−A(t)
λ0(t)e−A(t)=b(t)⇐⇒ λ0(t) = eA(t)b(t)
λ t →eA(t)b(t)I b I
a
•b(t) = P(t)emt P m 6=−a
b(t) = Q(t)emt Q P
•b(t) = P(t)emt P m =−a
b(t) = tQ(t)emt Q P
•b(t) = λcos(ωt) + µsin(ωt)αcos(ωt) +
βsin(ωt)
i)y0−2y= 2t2;ii)y0−3y=et;iii)y0−y=et;iv)½y0+y= cos t
y(0) = 0 .
y0−2y= 0 t→λe2t
yp2yp(t) = a+bt +ct2
y0
p−2yp=t2b+ 2ct = 2a+ 2bt + 2ct2+t2−2a+b+ (−2b+ 2c)t−2ct2= 2t2
−2a+b= 0
−2b+c= 0
−2c= 2
a=−1/4, b =−1/2, c =−1yp=−1/4−1/2t−t2
y(t) = λe2t−1/4−t/2−t2, λ ∈R
y0−3y= 0 t→λe3t
ypyp(t) = µety0
p= 3yp+et
µet= 3µet+et(2µ+ 1)et= 0 µ=−1/2yp=−1
2et
y(t) = λe3t−1
2et, λ ∈R
y0−y= 0 t→λet
ypyp(t) = µtety0
p=yp+et
µet+µtet=µtet+etµet=µetµ= 1 yp=tet
y(t) = λet+tety(0) = 1 λ= 1 y(t) = (1 + t)et
y0+y= 0
t→λe−typyp(t) = acos t+bsin t
y0
p+yp= cos t−asin t+bcos t+acos t+bsin t= cos t(a+b) cos t+(−a+b) sin t= cos t
½a+b= 1
−a+b= 0
a=b= 1/2yp=1
2(cos t+ sin t)y(t) = λe−t+1 2(cos t+ sin t), λ ∈Ry(0) = 0
λ+1
2= 0 λ=−1 2y(t) = 1 2(−e−t+ cos t+ sin t)

•(a, b, c)∈R3a6= 0 IRg:I→C
ay00 +by0+cy =g(x) (E)
•
ay00 +by0+cy = 0 (E0)
(E)
•(E)I ϕ :I→CI
∀x∈I, aϕ00(x) + bϕ0(x) + cϕ(x) = g(x).
S S0(E)E0
y:R→Cx7→ erx r∈Cy0=
rerx y00 =r2erx ay00 +by0+cy = (ar2+br+c)erx y∈ S0⇐⇒ ∀x∈R, ar2erx +brerx +cerx = 0
⇐⇒ (ar2+br +c)erx = 0 ⇐⇒ ar2+br +c= 0.
ar2+br +c=0r1r2
•(α, β)∈C2y:x7→ αer2x+βer1x(E0)
ay00 +by0+c=α(ar2
1+br1+c)er1x+β(ar2 2+br2+c)er2x= 0.
•y(E0)y(x) = z(x)er1x
y0(x) = (z0(x) + r1z(x))er1x;y00(x) = (z00 (x)+2r1z0(x) + r2
1z(x))er1x;
y∈ S0⇐⇒ ∀x∈R, ay00(x) + by0(x) + cy(x) = 0
⇐⇒ ∀x∈R, z0(x)(b+ 2r1a) + az00 (x) = 0
⇐⇒ ∃λ∈C∀x∈R, z0(x) = λe−b+2ar1
ax
⇐⇒ ∃(λ, µ)∈C2∀x∈R, z(x) = −λa
b+ 2ar1
e−b+2ar1
ax+µ
⇐⇒ ∃(α, β)∈C2∀x∈R, z(x) = αe−b+2ar1
ax+β
⇐⇒ ∃(α, β)∈C2∀x∈R, y(x) = αe−(b
a+r1)x+βer1x
⇐⇒ ∃(α, β)∈C2∀x∈R, y(x) = αer2x+βer1x
ar2+br +c=0r
y∈ S0⇐⇒ ∀x∈R, ay00(x) + by0(x) + cy(x) = 0
⇐⇒ ∀x∈R, az00 (x) = 0 ( b+ 2ar = 0)
⇐⇒ ∃(α, β)∈C2∀x∈R, z(x) = αx +β
⇐⇒ ∃(α, β)∈C2∀x∈R, y(x) = (αx +β)erx

ar2+br +c=0r1r2
S0={x7→ αer2x+βer1x,(α, β)∈R2}.
ar2+br +c=0r
S0={x7→ (αx +β)erx,(α, β)∈R2}.
ar2+br+c= 0 r1=a+ib
r2=a−ib
y ay00 +by0+cy = 0
z ay00 +by0+cy = 0 y= Re(z)
E0
E0
y∈ S0⇐⇒ ∃(α, β)∈C2∀x∈R, y(x) = αer2x+βer1x= eax(αeibx +βe−ibx).
y
yE0⇐⇒ ∃(λ, µ)∈R2∀x∈R, y(x) = eax(λcos(bx) + µsin(bx)).
(a, b, c)∈R3a6= 0 IR
ay00 +by0+cy = 0 (E0)
ar2+br +c= 0
r1r2
E0
S0={x7→ αer2x+βer1x,(α, β)∈R2}.
r
E0
S0={x7→ (αx +β)erx,(α, β)∈R2}.
r1=a+ib
r2=a−ib E0
S0={x7→ eax(αcos(bx) + βsin(bx)),(α, β)∈R2}.
Ry00 + 6y0+ 25y= 0
r2+ 6r+ 25 = 0 ∆0=−16 <0r1=−3 + 4i r2=−3−4i
S0={ϕ:R→R, ϕ(x) = e−3x(λcos 4x+µsin 4x),(λ, µ)∈R}.
Ry00 + 2y0+y= 0
r2+ 2r+ 1 = 0 r=−1
S0={ϕ:R→R, ϕ(x) = (λx +µ)e−x,(λ, µ)∈R}.
Ry00 −3y0+ 2y= 0

S S0
ay00 +by0+cy =g(x) (E)
(E)
ay00 +by0+cy = 0 (E0)
i∈J1, nK
y00 +ay0+by =gi(x) (Ei)
yi(Ei)
n
X
i=1
yiy00 +ay0+by =
n
X
i=1
gi(x) (E)
(E) (Ei)
P(x)emx
y00 +ay0+by =P(x)emx P n
m
y(x) = Q(x)emx Q
P m
y00 +ay0+by =P(x)emx
x7→ tkQ(x)emx Q P
•k= 0 m
•k= 1 m
•k= 2 m
y(x) = Q(x)emx y0(x)=(Q0(x) + mQ(x))emx y00(x) = (Q00 (x) +
2mQ0(x)+m2Q(x))emx y(E)Q00 + (2m+a)Q0+(m2+am+b)Q=
P(x)Q
Ry00 + 2y0+y=x2(ex−e−x)
(E0)y00 + 2y0+y= 0
r2+ 2r+ 1 = 0 r=−1 (E0)
S0={x7→ (λx +µ)e−x,(λ, µ)∈R2}
(E)
(E1)y00 + 2y0+y=x2ex
(E2)y00 + 2y0+y=−x2e−x