Repeated roots of a characteristic equation of third order ODE

[Hall]

Homework Statement

$y^{'''} - 6y^{''} +12y^{'} - 8 y=0$

Relevant Equations

Let the solution be $y = e^{mx}$. Then, the characteristic equation is:
$m^3 -6m^2 + 12m -8 = 0$

The characteristic equation $m^3 -6m^2 + 12m -8 = 0$ has just one single, I mean all three are equal, root $m=2$. So, one of the particular solution is $y_1 = e^{2x}$. How can we find the other two? The technique $y_2 = u(x) e^{2x}$ doesn't seem to work, and even if it were to work how would have we got the the third particular solution?

 

[Orodruin]

The technique y2=u(x)e2x doesn't seem to work, and even if it were to work how would have we got the the third particular solution?

What do you obtain when you attempt this approach?

Edit: I obtain the expected general solution.

 

[Hall]

What do you obtain when you attempt this approach?

Edit: I obtain the expected general solution.

Yes, I have done it again and got $u= x^2$. So, $y_2= x^2 e^{2x}$.

 

[Orodruin]

Yes, I have done it again and got $u= x^2$. So, $y_2= x^2 e^{2x}$.

Please show your reasoning. The solution you have found is one of the solutions, but the approach should give you all solutions.

 

[Hall]

Please show your reasoning. The solution you have found is one of the solutions, but the approach should give you all solutions.

I put $y = u(x) e^{2x}$ in the original differential equation and after doing everything I got:
$$
u’’’ =0
$$
I think, $u=x$ will also work, will it?

 

[Orodruin]

I put $y = u(x) e^{2x}$ in the original differential equation and after doing everything I got:
$$
u’’’ =0
$$
I think, $u=x$ will also work, will it?

What do you obtain if you integrate $u'''$ three times?

 

[Hall]

What do you obtain if you integrate u‴ three times?

$c_1 + c_2x + c_3x^2$

 

[Orodruin]

$c_1 + c_2x + c_3x^2$

Indeed. So what can be concluded about the general form or the solution to your ODE?

 

[Hall]

Indeed. So what can be concluded about the general form or the solution to your ODE?

$\text{General Solution} ~ y= e^{2x} (c_1 + c_2x + c_3 x^3)$

 

[Hall]

I was trying to solve
$$
y^{(5)} + 2 y^{‘’’} + y’=0$$
The characteristic equation:
$$
m^5 +2m^3 +m =0$$
Gives three particular solutions:
$y_1 = 1, ~ y_2= \cos x , ~ y_3= \sin x$

I assumed $y_4 = u(x) \cos x$, but replacing this $y_4$ didn’t give me $u$.

 

[Hall]

Here is what I’m getting:

 

[Orodruin]

So, first of all, when your differential equation does not contain the function itself, but only derivatives, it is usually simpler to rewrite it in terms of a differential equation for the lowest derivative. In this case, the lowest derivative is $y'$ and therefore the easiest thing is to set $f = y'$ and obtain
$$
f^{(4)} + 2 f'' + f = 0.
$$
The characteristic polynomial has two double roots $\pm i$. It will generally be easier to work with the complex exponentials $e^{\pm ix}$ than with the cos and sin functions. Doing so you should not have any problems identifying the solutions.

Note that setting $f(x) = u(x) e^{ix}$ should still give you the general solution, not just one extra solution, just as in the previous problem we could identify all solutions from $u^{(3)} = 0$.

It is also worth remembering the general theory. If we consider a linear differential operator $\mathcal L$ of order $n$ with constant coefficients, then it may be written on the form
$$
\mathcal L = \prod_{i = 1}^\ell (d_x - m_i)^{k_i}
$$
where $d_x = d/dx$ and $m_i$ is the $i$th root of the characteristic polynomial, which has multiplicity $k_i$, and $\sum_i k_i = n$. Note that all of the factors in $\mathcal L$ commute and so any function $e^{m_i x}$ is going to solve $\mathcal L y = 0$ because $(d_x - m_i)e^{m_i x} = (m_i - m_i)e^{m_i x} = 0$. If any of the roots $m_i$ are multiple ($k_i > 1$), then we need to identify the remaining linearly independent solutions. In order to do this, suppose that $m$ is a $k$-tuple root of the characteristic polynomial. Without loss of generality we can order our roots such that the last $k$ roots are $m$ and rewrite $\mathcal L$ according to
$$
\mathcal L = \mathcal P (d_x - m)^k
$$
where
$$
\mathcal P = \prod_{i=1}^{\ell-1} (d_x - m_i)^{k_i}.
$$
With the ansatz $y = u(x) e^{mx}$ we have
$$
(d_x - m) y = (u' + mu - mu)e^{mx} = u' e^{mx}
$$
and continuing in the same fashion
$$
(d_x - m)^k y = u^{(k)} e^{mx}.
$$
It therefore holds that
$$
\mathcal L y = \mathcal P u^{(k)} e^{mx} = 0
$$
where one obvious solution is $u^{(k)}(x) = 0$ and therefore $u(x)$ could be any polynomial of degree $k-1$ or lower.

Now, we could press on and assume that $h(x) = u^{(k)} \neq 0$ in order to solve the differential equation $\mathcal P he^{mx} = 0$, which is a linear differential equation of order $n-k$ with constant coefficients, but that will just give us the solutions related to the other roots of the original differential equation.

All in all, if $\mathcal L$ is on the form above, then the general solution takes the form
$$
y(x) = \sum_{i = 1}^\ell p_i(x) e^{m_i x},
$$
where $p_i(x)$ is an arbitrary polynomial of order $k_i-1$ or lower.

Can you apply this to your problem?

 

[Hall]

@Orodruin after putting $f = u e^{ix}$ I’m getting
$$
u^{(4)} + 4i u^{(3)} -4 u’’=0
$$

 

[Orodruin]

@Orodruin after putting $f = u e^{ix}$ I’m getting
$$
u^{(4)} + 4i u^{(3)} -4 u’’=0
$$

Right. So this is a fourth order equation where the lowest derivative is a second derivative. Any first order polynomial will therefore be a solution. You can go on and try to find the remaining solutions but they will correspond to the other roots of the characteristic polynomial.

To be specific, if you let $h = u''$ then $h'' + 4i h' - 4h = 0$ has the characteristic polynomial $m^2 + 4im - 4 = (m+2i)^2$ and therefore has the solution $h(x) = (Ax+B) e^{-2ix}$ so $f(x) = (ax+b)e^{-ix} + (cx+d)e^{ix}$.

However, you do not need to do this to find the remaining solutions as you can already find them from the same type of argumentation for the other roots of the original differential equation.

 

[Mark44]

$\text{General Solution} ~ y= e^{2x} (c_1 + c_2x + c_3 x^3)$

Typo? The third term in the parentheses should have an exponent of 2.

 

[Hall]

Any first order polynomial will therefore be a solution.

Can we do like this:

Given: $y^{(5)} + 2y^{'''} +y' =0$

By characteristic equation we get three specific solutions: $y_1 = x, y_2 = e^{ix}, y_3 = e^{-ix}$.
Put $y_4 = u(x) e^{ix}$ and assume it to be a solution, we would get
$$
u^{(4)} +4iu^{'''} -4 u^{''} = 0
$$
Thus, $u= x$ will satisfy the things above, hence $y_4 = x e^{ix}$.
Put $y_5 = u(x) e^{-ix}$, and assume it to be solution. Going through same steps (I'm assuming by symmetry that all the odd powers of $\iota$ will get negative and even powers will remain same) will produce $u = x$, therefore, $y_5 = x e^{-ix}$.

So, we have got five independent solutions:
$y_1 = x, y_2 = e^{ix}, y_3 = e^{-ix}, y_4 = xe^{ix}, y_5 = x^{-ix}$.
General Solution: $y = c_1 x + c_2 e^{ix} + c_3 e^{-ix} + c_4 xe^{ix} + c_5 xe^{-ix}$

?

 

[Mark44]

Can we do like this:

Given: $y^{(5)} + 2y^{'''} +y' =0$

By characteristic equation we get three specific solutions: $y_1 = x, y_2 = e^{ix}, y_3 = e^{-ix}$.
Put $y_4 = u(x) e^{ix}$ and assume it to be a solution, we would get
$$
u^{(4)} +4iu^{'''} -4 u^{''} = 0
$$
Thus, $u= x$ will satisfy the things above, hence $y_4 = x e^{ix}$.
Put $y_5 = u(x) e^{-ix}$, and assume it to be solution. Going through same steps (I'm assuming by symmetry that all the odd powers of $\iota$ will get negative and even powers will remain same) will produce $u = x$, therefore, $y_5 = x e^{-ix}$.

So, we have got five independent solutions:
$y_1 = x, y_2 = e^{ix}, y_3 = e^{-ix}, y_4 = xe^{ix}, y_5 = x^{-ix}$.
General Solution: $y = c_1 x + c_2 e^{ix} + c_3 e^{-ix} + c_4 xe^{ix} + c_5 xe^{-ix}$

?

You're close, but $y = c_1x$ is not a solution. The characteristic equation is $r(r^2 + 1) = 0$, so the solutions are r = 0, r = i (repeated), and r = -i (repeated). The solution basis is $\{e^0 = 1, e^{ix}, xe^{ix}, e^{-ix}, xe^{-ix} \}$.
The general solution is
$y = c_1 \cdot 1 + c_2 e^{ix} + c_3 e^{-ix} + c_4 xe^{ix} + c_5 xe^{-ix}$.

Of course, $c_1 \cdot 1$ can be written more simply as just $c_1$.

Also, the solution can also be written in terms of $\cos(x)$ and $\sin(x)$ and multiples by $x$ of these two functions.

 

[Orodruin]

Can we do like this:

You can, but it is a bit long winded. The faster approach is to use the completely general formula right off the bat:

All in all, if $\mathcal L$ is on the form above, then the general solution takes the form
$$
y(x) = \sum_{i = 1}^\ell p_i(x) e^{m_i x},
$$
where $p_i(x)$ is an arbitrary polynomial of order $k_i-1$ or lower.