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 $\displaystyle y=e^{mt}$ gives
$\displaystyle m^5-3m^4+3m^3-3m^2+2m=0$
Now we just need to factor the above polynomial
$\displaystyle m^5-3m^4+3m^3-3m^2+2m=m(m^4-3m^3+3m^2-3m+2)$
By the rational roots theorem if there are other rational roots they must be either $\displaystyle \pm1 \text{ or } \pm 2$
now by either long or synthetic division you will find that 1 and 2 are both roots giving the factorization
$\displaystyle m(m-1)(m-2)(m^2+1)$
Putting all of this back in gives the solution
$\displaystyle y=c_1+c_2e^{t}+c_3e^{2t}+c_4\cos(t)+c_5\sin(t)$