y^(5) - 3y ^ (4) + 3y''' - 3y'' + 2y' = 0
I know the solution is : y = c1 + c2* e^x + c3*e^2x + c4 cos x + c5 sin x
I am able to get the c1 solution. After that I am confused as to how the other solutions come about.
y^(5) - 3y ^ (4) + 3y''' - 3y'' + 2y' = 0
I know the solution is : y = c1 + c2* e^x + c3*e^2x + c4 cos x + c5 sin x
I am able to get the c1 solution. After that I am confused as to how the other solutions come about.
Using the ansantz gives
Now we just need to factor the above polynomial
By the rational roots theorem if there are other rational roots they must be either
now by either long or synthetic division you will find that 1 and 2 are both roots giving the factorization
Putting all of this back in gives the solution