Question

Solve the given initial value problem. dx dt 4t 4x+y- e x(0)1 y(0)3 dt The solution is x)-and y()

Please retain EXACT values, wherever they appear. Such as, radicals/square roots, e^n, and fractions. This is very important in being able to understand the process. Thank you.

Answer 1

dx/dt = ux + y - e^ut rightarrow (1), x(0) = 1 dy/dt = x + uy (2) , y(0) = -3 by elimination method (D - 4)x - y = -e^4t (1) (D - 4)y - x = 0 (2) so link this 2 equation (1) time 1 implies x(D - 4) - y = 0 (2) times (D - 4) implies - x (D - 4) + (D - 4) (D - 4)y = 0/ - y + (D - 4)^2 y = 0 (- 1 + D^2 + 16 - 8 D)y = 0 D^2 - 8D + 15 = 0 auxiliary equation is m^2 - 8m + 15 = 0 m^2 - 5m - 3m + 15 = 0 m(m - 5) - 3 (m - 5) = 0 (m - 5)(m - 3) = 0 therefore m = 5, 3 C.F = c_1 e^5t + c_2 e^3t P.I = 1/D^2 - 8D + 15 [- e^4t] = -1/(D - 5)(D - 3) e^4t