y = e^-0.4x(C cos0.2x + Dsin0.2x)
y' = -0.4e^-0.4x(-0.2C sin0.2x + 0.2 D cos0.2x)
This from trying to work out a particular solution to a inhomogeneous equation
Paul B
I am a bit confused now so i think Ill post whole problem and maybe you could tell me where I am going wrong.
Starting with:
5(d^2y/d^2x) + 4(dy/dx) + y = 0
I have find a particular solution to the initial value problem where
y(0) = -2, y'(0)=3
I know a general solution is
y = e^-0.4x(C cos0.2x + Dsin0.2x)
after finding the roots.
So substituting in the values into the derivative
y' = -0.4e^-0.4x(-0.2C sin0.2x + 0.2 D cos0.2x)
ends with 0.4C+0.2D =3
So C = 3 and D = 9
Doing the same with the general solution with x=-2
y = e^-0.4x(C cos0.2x + Dsin0.2x)
I end up with 1( C x 1 + D x 0) in other words C = -2
My confusion comes from the fact that C should be the same in both instances??
Cheers for the help