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2E2 Tutorial Sheet 13 Second Term1 1 February 2004 1. (2) Find the general solution to y 0 − 2y = −t (1) Solution: This follows from the general solution to y 0 + ry = f (t) which is y = Ce −rt +e −rt Z ert f dt so here r = −2 and f (t) = −t so, using integration by parts Z 2t 2t te−2t dt y = Ce − e Z 1 −2t 1 −2t 2t 2t e dt = Ce − e − te + 2 2 1 −2t 1 −2t 2t 2t = Ce − e − te − (e ) 2 4 1 t = Ce2t + + 2 4 (2) (3) (4) 2. (3) Find the general solution to y10 = 5y2 − 23 y20 = 5y1 + 15. (5) with y1 (0) = −3 and y2 (0) = 5. Solution: First of all rewrite the equation in matrix form 0 5 −23 0 y = y+ . 5 0 15 (6) Now, the matrix A= 1 0 5 5 0 (7) Conor Houghton, houghton@maths.tcd.ie and http://www.maths.tcd.ie/~houghton/ 2E2.html 1 has eigenvalue λ1 = 5 with eigenvector x1 = 1 1 (8) and eigenvalue λ1 = −5 with eigenvector 1 x2 = −1 (9) so if we write y = f 1 x1 + f 2 x2 (10) and subsituting this into the differential equation gives −23 0 0 (f1 − 5f1 )x1 + (f2 + 5f2 )x2 = . 15 (11) Now to seperate the equation lets decompose the inhomeneous part, sometimes called the forcing term, over the two eigenvectors: −23 = g 1 x1 + g 2 x2 (12) 15 or, writing it out, −23 15 = g1 + g 2 g1 − g 2 (13) and, hence, g1 = −4 and g2 = −19. Putting this back into the equation leads to (f10 − 5f1 )x1 + (f20 + 5f2 )x2 = −4x1 − 19x2 (14) f10 − 5f1 = −4. (15) Hence Thus, this is of the form y 0 + ry = f with r = −5, f (t) = −4 and so Z 5t 5t f1 = C1 e − 4e e−5t dt and so f1 = C1 e5t + 4 5 (16) (17) Similarly, 2(f20 + 5f2 ) = (1 − 1) −23 15 = −38 (18) or f20 + 5f2 = −19. 2 (19) Thus, r = −5, f (t) = 19 and using integrating gives above f2 = C2 e−5t − 19 . 5 The general solution is therefore 4 19 1 1 5t −5t y = C1 e + + C2 e − . 1 −1 5 5 If y1 (0) = −3 and y2 (0) = 5 then we get 19 4 1 1 −3 + C2 − . = C1 + 1 −1 5 5 5 (20) (21) (22) and hence −3 = C1 + C2 − 3 23 5 = C 1 − C2 + 5 so C1 = −C2 = 1/5 and 1 −5t 19 1 5t 4 1 1 + − e − . y= e + 1 −1 5 5 5 5 (23) (24) 3. (3) Find the solution to y10 = y1 + 2y2 + et y20 = 2y1 + y2 Solution: Here we have et 0 (26) 1 2 2 1 , (27) y = Ay + where A= this has eigenvalue λ1 = 3 with eigenvector 1 x1 = 1 and eigenvalue λ1 = −1 with eigenvector 1 x2 = . −1 3 (25) (28) (29) Once again, we split the forcing term over the two eigenvectors: t et et e = x1 + x2 0 2 2 We get (30) 1 f1 − 3f1 = et 2 Z 1 3t 3t e−2t dt. f1 = C 1 e + e 2 so and so, (31) (32) 1 f1 = C1 e3t − et 4 In the same way (33) 1 f2 + f 2 = et 2 Z 1 −t −t e2t dt. f2 = C 2 e + e 2 and so Integrating gives (34) (35) 1 f2 = C2 e−t + et 4 (36) This means y= 1 C1 e − et 4 3t 1 1 4 + C2 e −t 1 + et 4 1 −1 (37)