Question

3. Solve the equation
(i) (D^2+ 5D + 4)y = e^-x
sin2x
(ii) (D^4– 1)y = cos2x coshx.​

Answer 1

Answer:

Ok Good question

Step-by-step explanation:

$= -e^-xsin2x + 2e^-xcos2x \\ D = dy/dx = -e^-xsin2x + e^-x(2cos2x) \\ y = e^-xsin2x \\ 4e^-xsin2x \\ D^2 = d^2y/dx^2 = \\ e^-xsin2x + 2(-e^-x)cos2x - 2e^-xcos2x - \\ D^2 + 5D + 4 \\ = - 3e^-xsin2x - 4e^-xcos2x \\ - 3e^-xsin2x - 4e^-xcos2x - 5e^-xsin2x + 10e^-xcos2x + 4 \\ - 8e^-xsin2x + 6e^-xcos2x + 4$

Simple