Lista Atividade 2 EDO

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  • 1. Equa¸oes Diferenciais Atividade - 2 c˜ ‡ Filipe Miguel Ribeiro Universidade de Brasilia delta3249@gmail.com ‡ 13/10/2011 1
  • 2. ´SUMARIO 2Sum´rio a1 Exercicio 1 32 Exercicio 2 5
  • 3. 1 EXERCICIO 1 31 Exercicio 1a.Resolu¸˜o 1. ca 2 y 2dy x + 3y 2 2 dy 1 + 3y2 x 1+3 x = . Temos que 2xy = 0 ⇒ x = 0 ou y = 0. = y = y ∴dx 2xy dx 2x 2. xConsiderando que y du = u ⇒ y = (u.x) = u x + u = .x + u x dx du 1 + 3u2 du 1 + 3u2 − 2u2 du 1 + u2Deste modo: .x + u = ⇒ .x = ⇒ .x = ⇒ dx 2u dx 2u dx 2u 1 2u 1 2u⇒ dx = du ⇒ dx = du x 1 + u2 x 1 + u2 I1Calculando a Integral I1 temos:  2u  1 + u2 = v I1 = du = = ln |u2 + 1|. Logo: 1 + u2  2udu = dv  y2ln |x| = ln |u2 + 1| + C ⇒ ln |x| = ln |u2 + 1| + ln eC ⇒ x = eC (u2 + 1) = eC +1 ⇒ x2 x2 (x − eC )x3 = (y 2 + x2 )eC ⇒ y 2 = eC x − eCy = ±x. . eCb.Resolu¸˜o 2. ca
  • 4. 1 EXERCICIO 1 4 4ydy 4y − 3x dy −3 = = x ydx 2x − y dx 2− x du 4u − 3 du 4u − 3 − u(2 − u)Conforme a equa¸ao 1 temos: c˜ .x + u = = .x = = dx 2−u dx 2−u4u − 3 − 2u + u2 u2 + 2u − 3 = ⇒ 2−u 2−u 1 2−u 1 2−u⇒ dx = 2 du = dx = du x u + 2u − 3 x u2 + 2u − 3 I2Utilizando o conceito de fra¸oes parciais da integral I2 temos: c˜  2−u A B A(u + 3) + B(u − 1)  A + B = −1 I2 = du = + = ⇒ (u − 1)(u + 3) (u − 1) (u + 3) (u − 1)(u + 3)  3A − B = 2  A = 1/4 , B = −5/4 1 5 1 5 I2 = − du = ln |u − 1| − ln |u + 3| 4(u − 1) 4(u + 3) 4 4 1 = (ln |u − 1| − ln |u + 3|5 ) 4 1 |u − 1| = ln 4 |u + 3|5Voltando a EDO: ` |u − 1| |u − 1| C |u − 1|.eCln |x| = ln + C ⇒ ln |x| = ln + ln e = ln |x| = ln . |u + 3|5 |u + 3|5 |u + 3|5Sendo que u = y/x, temos |y/x − 1|.eCln |x| = ln ⇒ |x|.|y/x + 3|5 = |y/x − 1|.eC |y/x + 3|5
  • 5. 2 EXERCICIO 2 5c.Resolu¸˜o 3. ca y dy x + 3y 1 + 3x du u2 + 2u + 1 1 1−u = = y ⇒ vide 1 ⇒ .x = ⇒ dx = 2 du dx x−y 1− x dx 1−u x u + 2u + 1 1 1−u dx = 2 + 2u + 1 du x u I3Resolvendo a I3 segue da mesma forma que foi resolvida no exercicio 2: 1−u −1 2 2 du = + du = − ln |u + 1| − u2 + 2u + 1 u + 1 (u + 1)2 u 2 −2Portanto, ln |x| = − ln |u + 1| − + C ⇒ ln = |x| = ln (u + 1)−1 .e u .eC = u 1 −2xx = ±y .e y +C x +1d.Resolu¸˜o 4. ca A resolu¸ao ´ semelhante as demonstradas nos exerc´ c˜ e ıcios anterioresa unica diferen¸a ´ que essa EDO est´ na forma diferencial. Basta deixa-la na forma ´ c e ady = f (x, y) e resolve-la pela substitui¸ao u = y/x. c˜dx Resposta: x2 = (x2 + y 2 )1/2 .eC2 Exercicio 2Resolu¸˜o 5. ca Para n = 0 temos: y + p(t)y = q(t)
  • 6. 2 EXERCICIO 2 6. Utilizando o fator integrante exp( p(t)dt), temos: y . exp( p(t)dt) + exp( p(t)dt).p(t)y = exp( p(t)dt).q(t) d Observa que: exp( p(t)dt).p(t) = exp( p(t)dt). Ou seja: dt d d y. exp( p(t)dt) = exp( p(t)dt).q(t) ⇒ y. exp( p(t)dt) = dt dt exp( p(t)dt).q(t) = y. exp( p(t)dt) = exp( p(t)dt).q(t) ⇒ 1 ⇒y= exp( p(t)dt).q(t)dt + C exp( p(t)dt)