Mathématiques : applications 2005 Université de Technologie de Belfort Montbéliard

Description

Examen du Supérieur Université de Technologie de Belfort Montbéliard. Sujet de Mathématiques : applications 2005. Retrouvez le corrigé Mathématiques : applications 2005 sur Bankexam.fr.

Sujets

• 2005

Informations

• Publié par: bankexam
• Publié le: 18 août 2008
• Nombre de lectures: 39
• Langue: Français

Extrait

D1
2D = (x,y)∈R |x|+|y|6 21
D1
A D1 1
I
1(x+y)
2I = e dx dy
D1
D2
2 2 2D = (x,y)∈R |x +y 6 1, |y|6x, x> 02
D2
A D2 2
M (R)3  −4 −6 0 A = 3 5 0
3 6 5
A
A
−1D Q A = Q D Q
   
0 1 −2 1 2 1
−1   Q = 0 −1 1 Q = −1 −2 0
1 1 0 −1 −1 0












 x˙ (t)−4x (t)−6x (t) = 8t+11 1 2
(S) x˙ (t)+3x (t)+5x (t) =t2 1 2
x˙ (t)+3x (t)+6x (t)+5x (t) = 63 1 2 3
 x (0) = 01
(CI ) x (0) = 0x 2
x (0) = 03
(S)
˙X(t)+A X(t) = Φ(t),
     
x˙ (t) x (t) 8t+11 1
˙      X(t) = x˙ (t) X(t) = x (t) Φ(t) = t2 2
x˙ (t) x (t) 63 3
(S)
˙Y(t)+D Y(t) = Ψ(t),
−1 −1Y(t) = Q X(t) Ψ(t) = Q Φ(t)
 y (0) = 01
(CI ) y (0) = 0y 2
y (0) = 03
y˙ (t)+5y (t) = 10t+71 1
y˙ (t)+2y (t) = −10t−12 2
y˙ (t)−y (t) = −9t−13 3
(CI )y
y y y1 2 3
Y
X (S)




D
x> 0 y > 0 x+y6 2 ⇒ y6 2−x
x> 0 y < 0 x−y6 2 ⇒ x−26y
x< 0 y > 0 −x+y6 2 ⇒ y6 2+x
x< 0 y < 0 −x−y6 2 ⇒ −x−26y
06x6 2 ⇒ x−26y6 2−x
−26x6 0 ⇒ −x−26y6 2+x
y
−2 x2
D
A(D) = 8
A(D) = dx dy
D
A(D) = dx dy
D
0 2+x 2 2−x
= dy dx+ dy dx
−2 −x−2 0 x−2
0 2
= (2x+4) dx+ (4−2x) dx
−2 0
0 2
= 2 (x+2) dx+ (2−x) dx
−2 0
= 8







































1(x+y)
2I = e dx dy
D
0 2+x 2 2−x
1 1(x+y) (x+y)
2 2= e dy dx+ e dy dx
−2 −x−2 0 x−2
0 2
1 1 1 1 1 1x x+1 − x−1 x 1− x x−1
2 2 2 2 2 2= 2 e e −e dx+2 e e −e dx
−2 0
0 2
x+1 −1 x−1= 2 (e −e ) dx+ (e−e )dx
−2 0
0 2x+1 −1 x−1
= 2 e −e x +[ ex−e
−2 0
−1 −1
= 2e−6e +2e+2e
1
= 4(e− )
e
1
I = 4(e− )
e
πD A =2 2 4
y
x
D
π 7π2 2 2D = (x,y)∈R |x +y 6 1, |y|6x, x> 0 ⇔U = (r,θ) 06r6 1 θ∈ [0, ]∪[ ,2π]2 2
4 4
A = dx dy2
D2
= r dr dθ
U2
π 1 2π 1
4
= r dr dθ+ r dr dθ
7π0 0 0
4
π 1 12π2 2
4 r r
= dθ+ dθ
2 7π 20 0 0
4
π
=
4






1
x = x dx dyG
A2
D2
1 2= r cosθ dr dθ
A2
U2
π 1 2π 1
44 2= r cosθdr dθ+ cosθ dr dθ
7ππ 0 0 0
4
π13 2π
44 r
= cosθ dθ+ cosθ dθ
7ππ 3 00 4√
4 2
=
3π
1
y = y dx dyG
A2
D2
1 2= r sinθ dr dθ
A2
U2
π 1 2π 1
44 2
= r sinθdr dθ+ sinθ dr dθ
π 7π0 0 0
4
π1 2π3
44 r
= sinθ dθ+ sinθ dθ
π 3 7π00
4√
4 2
=
3π
M (R)3  −4 −6 0 A = 3 5 0
3 6 5
−4−λ −6 0
3 5−λ 0 = (5−λ)(−(5−λ)(4+λ)+18 = (5−λ)(2−λ)(1+λ))
3 6 5−λ

λ = 5
λ = 2
λ = −1
A
λ = 5 x = (x,y,z)  −4x−6y = 5x −9x−6y = 5x x = 0  
Ax = 5x ⇔ 3x+5y = 5y ⇔ 3x = 0 ⇔ y = 0  
3x+6y+5z = 5z 3x+6y = 0 z
3E = (x,y,z)∈R |(0,0,z)5





λ = 2 x = (x,y,z)   −4x−6y = 2x  −6x−6y = 0  y = −x
Ax = 2x ⇔ 3x+5y = 2y ⇔ 3x+3y = 0 ⇔ z = x  
3x+6y+5z = 2z 3x+6y+3z = 0 x
3E = (x,y,z)∈R |(x,−x,x)5
λ =−1 x = (x,y,z)   −4x−6y = −x  −3x−6y = 0  x = −2y
Ax =−x ⇔ 3x+5y = −y ⇔ 3x+6y = 0 ⇔ z = 0  
3x+6y+5z = −z 3x+6y+6z = 0 y
3E = (x,y,z)∈R |(−2y,y,0)5
λ = 5  
0 e = 01
1
λ = 2  
1 e = −12
1
λ = 2  −2 e = 13
0
   
5 0 0 0 1 −2   D = 0 2 0 Q = 0 −1 1
0 0 −1 1 1 0
A
−1A A = Q D Q
˙X(t)+A X(t) = Φ(t)
−1˙X(t)+Q D Q X(t) = Φ(t)
−1 −1 −1 −1˙Q X(t)+Q Q D Q X(t) = Q Φ(t)
−1 −1 −1˙Q X(t)+D Q X(t) = Q Φ(t)
˙Y(t)+D Y(t) = Ψ(t)
y˙ (t)+5y (t) = 10t+71 1
y˙ (t)+2y (t) = −10t−12 2
y˙ (t)−y (t) = −9t−13 3








−5ty (t) =c e1 1
−2ty (t) =c e2 2
−ty (t) =c e3 3
par
y (t) =at+bi
par
y (t) = 2t+11
pary (t) =−5t+22
par
y (t) = 9t+103
−5ty (t) = c e +2t+11 1
−2ty (t) = c e −5t+22 2
−ty (t) = c e +9t+103 3
 
−5tc e +2t+11
−2t Y = c e −5t+22
−tc e +9t+103
(CI ) cy i
 
−5t− e +2t+1
−2t Y = −2 e −5t+2
−t−10 e +9t+10
−1X Y(t) = Q X(t) X(t) = QY(t)  
−5t0 1 −2 c e +2t+11
−2t  X(t) = 0 −1 1 c e −5t+22
−t1 1 0 c e +9t+103
 
−2t −tc e −2c e −23t−182 3
−t −2t c e −c e +14t+8X(t) = 3 2
−5t −2tc e +c e −3t+31 2