PSYC 221 : Review Final Test New 14 3

• School: Southeastern Louisiana University
• Department: Psychology
• Course: PSYC 221
• Professor: All

Document Summary

Use variation of parameters to solve these: find the general solution of y + 3y = e 3t, first, nd the solutions of the homogeneous equation. The characteristic equation is r2 + 3r = 0, or r(r + 3) = 0, so r = 0, 3. Thus, the two linearly independent solutions of the homogeneous equation are y1(t) = 1, y2(t) = e 3t: next, write the general solution as y(t) = u1(t)y1(t)+u2(t)y2(t), where u1, u2 are unknown. 2 = e 3t, or in terms of matrices, we solve. 1 e 3t ] det[ y1 det[ y1 y2. 2 = e 3t e 3t e 3t det[ 0 det[ 1 det[ 1 det[ 1. + c2: finally, write down the general solution: y(t) = u1(t)y1(t) + u2(t)y2(t) = c1 + c2e 3t e 3t e 3t e 3t t. 3 where c2 is some other arbitrary constant: find the general solution of y .