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Capitolul 1 Ecuatii diferentiale de ordinul ˆıntâi rezolvabile prin ...

Capitolul 1 Ecuatii diferentiale de ordinul ˆıntâi rezolvabile prin ...

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82 CAPITOLUL 3<br />

Rezultatul este imediat în baza propozit¸iilor (3.1.1, 3.1.2, 3.1.3).<br />

Teorema 3.1.7 Solut¸iile sistemului (3.2) sunt funct¸ii <strong>de</strong> forma:<br />

e λkt Pqk−1(t) +<br />

l<br />

e µkt<br />

[Qrk−1(t) cosνkt + Rrk−1(t) sin νkt] , i, j = 1, n<br />

k=1<br />

un<strong>de</strong> λ1, . . .,λp sunt valorile proprii reale ale lui A cu ordin <strong>de</strong> multiplicitate<br />

respectiv q1, . . .,qp; µk + iνk, k = 1, l sunt valorile proprii complexe ale lui A<br />

cu ordin <strong>de</strong> multiplicitate rk; Pqk−1, Qrk−1 ¸si Rrk−1 sunt vectori coloanǎ ai<br />

cǎror elemente sunt polinoame <strong>de</strong> grad qk − 1 respectiv rk − 1.<br />

Exercit¸ii<br />

1. Rezolvat¸i urmǎtoarele sisteme:<br />

a)<br />

<br />

x1 ˙ = −x1+8x2<br />

x2 ˙ = x1+ x2<br />

R:<br />

b)<br />

c)<br />

d)<br />

e)<br />

f)<br />

<br />

x1 ˙ = −3x1+ 2x2<br />

x2 ˙ = −2x1+ x2<br />

<br />

x1 ˙ =2x1− x2<br />

x2 ˙ = x1+ 2x2<br />

⎧<br />

⎨<br />

⎩<br />

⎧<br />

⎨<br />

⎩<br />

⎧<br />

⎨<br />

⎩<br />

x1 ˙ = 3x1+12x2− 4x3<br />

x2 ˙ = −x1− 3x2+ x3<br />

x3 ˙ = −x1− 12x2+6x3<br />

x1 ˙ = x1+x2− 2x3<br />

x2 ˙ =4x1+ x2<br />

x3 ˙ =2x1+x2− x3<br />

x1 ˙ = 2x1− x2− x3<br />

x2 ˙ = 3x1− 2x2− 3x3<br />

x3 ˙ = −x1+ x2+ 2x3<br />

x1(t) = c1 · e 3t + c2 · e −3t<br />

x2(t) = 1<br />

2 ·c1 · e 3t − 1<br />

4 ·c2 · e −3t<br />

<br />

x1(t) = c1 · e<br />

R:<br />

−t + c2 · t · e−t x2(t) = c1 · e−t + 2t+1<br />

2 · c2 · e−t <br />

x1(t) = c1 · cost · e<br />

R:<br />

2t + c2 · sin t · e2t x2(t) = c1 · sin t · e2t −c2 · cost · e2t ⎧<br />

⎨<br />

R: x2(t) = −<br />

⎩<br />

3<br />

x1(t) = c1 · e 2t + c2 · e t + c3 · e 3t<br />

8 c1 · e2t −1 2 c2 · et −1 3 c3 · e3t x3(t) = − 7<br />

8 c1 · e2t − c2 · et − c3 · e3t ⎧<br />

⎨ x1(t) = −<br />

R:<br />

⎩<br />

1<br />

⎧<br />

⎨<br />

R:<br />

⎩<br />

2 c1 · e−t + + 1<br />

4 c3 · et x2(t) = c1 · e−t + c2 · et + c3 · t · et x3(t) = + 1<br />

2 c2 · et + 1<br />

2 c3 · t · et x1(t) = c2 + c3 · e t<br />

x2(t) = c1 · e t +3 c2<br />

x3(t) = −c1 · e t − c2 + c3 · e t

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