y" + 2y' + 5y = 0
The roots of D^2+2D+6=0 are -1±2i, so the solution to the DE is
y = e^-x (a sin2x + b cos2x)
y' = e^-x ((2a-b)cos2x - (a+2b)sin2x)
Now plug in your values to solve for a and b
d^2y/dx^2 + 2 dy/dx +5y=0
Very confused by how to do this, I have found the aux equation and it’s roots (-1+-2i) and put the equation into the form:
y=e^ax( C cos bx + D sin bx)
But now I need to find out the solution when y=1, x=0 and dy/dx=1.
But I’m not sure what part I am differentiating. Can anyone guide me a little?
3 answers
Thanks oobleck,
Having done this if it seems that D, C are 0 or 1 then that means the explicit solution is just the same as the solution found just without D and C?
Having done this if it seems that D, C are 0 or 1 then that means the explicit solution is just the same as the solution found just without D and C?
I arrived at 1=-1C+D+1+2D, am I totally of track?
3 answers