verify the solution of the differential equation
solution: y=C1 e^(-x)cosx + C2 e^(-x)sinx
Differential equation: y'+y'+2y=0

Question

verify the solution of the differential equation
solution: y=C1 e^(-x)cosx + C2 e^(-x)sinx
Differential equation: y'+y'+2y=0

accessdenied · Answer

When we say that y is a solution to a differential equation, then we should be able to substitute y and its derivatives into the equation and get a unanimously true statement such as 0=0.

accessdenied · Answer

So if we find the derivatives of y first (this requires some product rule), and then plug it into the equation:
  y " + y' + 2y = 0
The left-hand side should cancel itself out in the end!

accessdenied · Answer

The hardest part of this may simply be keeping track of all the negative signs.  (Derivative of cos x = - sin x,  derivative of e^(-x) = - e^(-x).  By the second derivative, there is quite a bit to keep track of.)  So you should be careful finding the derivatives and it should work out.