Question

1. (D2+4D-5)y = e3x+4cos4x​

Answer 1

Answer:

For the equation (D^2 - 4D - 5)y = e^3x + 4cos3x the characteristic equation is

m^2 - 4m - 5 = (m + 1)(m - 5) =0 . The solution to the homogeneous equation is

yh = C1e^-x + C2e^5x .For the particular integral use the undetermined coefficients

approach taking yp = Acos3x + Bsin3x + Ce^3x .Substituting in the equation gives

A = - 14/85 , B = - 12/85 , C = -1/8 . Then , the solution is

y = C1e^-x + C2e^5x - (2/85)[7cos3x + 6sin3x] - (e^3x)/8 .