characteristic polynomial based on differential equation

Question

I have this differential equation:

$$\ddot{x}-\frac{1}{6} \dot{x} - \frac{1}{6}x = e^t$$

When I DSolve it looks like this:

DSolve[x''[t] - 1/6 x'[t] - 1/6 x[t] == Exp[t], x[t], t]

Which gives me the correct answer. With that, we know I have the characteristic polynomial:

z^2 - 1/6 z - 1/6

I also find the roots:

Root[%,1]
Root[%%,2]

So all the results are correct. My question is, how can make this more easy for me, so I don't need to type manually the characteristic polynomial with its roots? I have tried with the function StringReplace to rewrite my differential equation into the characteristic polynomial, but it was messy and not functioning properly.

Answer 1

Let the ODE

ode=x''[t] - 1/6*x'[t] - 1/6*x[t] == Exp[t]

to get the characteristic polynomial we follow the literature:

cpoly = Simplify[(ode[[1]]/.x->(Exp[r*#]&))/Exp[r*t]]

now we calculate the roots:

{Root[cpoly,1], Root[cpoly,2]}

or

Solve[cpoly==0, r]

Also for the odes of the form

 ode=t^2*x''[t] - 1/6*t*x'[t] - 1/6 x[t] == Exp[t]

the only change is on the substitution:

cpoly2 = Simplify[(ode[[1]]/.x->(#^r&))/t^r]

Finally, if you want to have the multiplicity of the root handy you can use Tally:

Tally[Solve[cpoly==0, r]//Flatten]

Answer 2

First,

t1 = x''[t] - x'[t]/6 - x[t]/6 /. {Derivative[k_Integer][x][__] :> z^k, x[__] :> 1}
   -1/6 - z/6 + z^2

t2 = x'''[t] - 5 x'[t]/6 - 3 x[t] /. {Derivative[k_Integer][x][__] :> z^k, x[__] :> 1}
   -3 - 5 z/6 + z^3

Then,

z /. Solve[t1 == 0, z]
   {-1/3, 1/2}


z /. Solve[t2 == 0, z, Cubics -> False]
   {Root[-18 - 5 #1 + 6 #1^3 &, 1], Root[-18 - 5 #1 + 6 #1^3 &, 2],
    Root[-18 - 5 #1 + 6 #1^3 &, 3]}

Remove Cubics -> False to see the actual Cardano solution.