How do you find the solution to this initial value problem? y'' + 2y' + 5y = 4e^(-t)*cos(2t) where y(0) = 1, and y'(0) = 0. Thanks a lot. The answer by the way is y = e^(-t)*cos(2t) + (1/2)*e^(-t)*sin(2t) + t*e^(-t)*sin(2t).
2007-08-01 17:09:53 · 2 answers · asked by Charlie4590 2 in Science & Mathematics ➔ Mathematics
By the nature of the function e^(-t) and the cos(2t) function, one can anticipate a solution of the form
A*exp(-t)*cos(2t) + B*exp(-t)*sin(2t)+Ct^2*exp(-t)*cos(2t)+Dt*epx(-t)*sin(2t). = y. These are all terms that would yield a y", y' or a y of the form K*exp(-t)*cos(2t). Then you take the first and second derivitives and evaluate y"+2y+5y based on the problem and the two initial values to evaluate the constants A, B, C and D. Evidentally, the constant C=0.
2007-08-01 17:28:59 · answer #1 · answered by cattbarf 7 · 0⤊ 0⤋
Do your own homework.
2007-08-01 17:12:58 · answer #2 · answered by da s 2 · 0⤊ 0⤋