General solution of $\frac{d^2x}{dt^2}+a\frac{dx}{dt}+bx=0$ for identical roots Let the trial solution of a differential equation of the form $\frac{d^2x}{dt^2}+a\frac{dx}{dt}+bx=0$ ($a,b$ are constants), be $x=e^{mt}$. If it turns out that the two roots of $m^2+am+b=0$ are the same i.e. $m_1=m_2$, the general solution cannot be obtained from $$x=c_1e^{m_1t}+c_2e^{m_2t}$$ by setting $m_1=m_2(=m)$. Instead the solution is $$x=(c_1+c_2t)e^{mt}$$ Why is that? How do we justify this solution or explain this?
 A: Write the linear combination slightly differently
$$
x(t)=c_1e^{m_1t}+c_2\frac{e^{m_2t}-e^{m_1t}}{m_2-m_1}
$$
and check that for $m_1\ne m_2$ this indeed still parametrizes the full set of solutions. Now for $m_1\approx m_2$ the second term, by the mean value theorem or definition of the derivative, is close to $c_2te^{m_1t}$ as long as $|m_2-m_1|\,|t|\ll 1$.
In the limit case $m_1=m_2$ the above approximation turns into the actual solution.

To get such a parametrization more naturally, start from the symmetry that is inherent in the solution formula for the quadratic characteristic equation and set $m_{1,2}=m\pm n$. Then
$$
x(t)=e^{mt}(c_1e^{nt}+c_2e^{-nt})=e^{mt}((c_1+c_2)\cosh(nt)+(c_1-c_2)\sinh(nt))
$$
It is the last term going to zero for $n\to 0$ that gives wildly varying coefficients. So scale it to something that is nearly invariant close to $t=0$ and switch parameters to
$$
x(t)=e^{mt}\left(u\cosh(nt)+v\frac{\sinh(nt)}{n}\right).
$$
This again has a good approximation by $e^{mt}(u+vt)$ for $|nt|\ll 1$.
A: HINT:
Say your DE is $\frac{d^2 x}{d t^2} = 0$. The solutions are
$c_1 + c_2 t$.
Now say your DE is $(\frac{d}{dt} - m)^2 x = 0$. Note that we are dealing with the differential operator $\frac{d}{dt} - m$ and its square. Now, we have
$$\frac{d}{dt} (e^{m t} x) = e^{m t} \frac{d x}{dt} + m e^{m t} x$$
that is
$$(\frac{d}{dt} - m) (e^{m t} x) = e^{m t} \frac{d}{d t} x $$
and from here
$$(\frac{d}{dt} - m)^k (e^{m t} x) = e^{m t} (\frac{d}{d t})^k x $$
for all $k\ge 1$.
Basically, the differential operators $\frac{d}{dt} - m$ and $\frac{d}{dt}$ are conjugate.
A: Tracking what happens to the $te^{mt}$ term over the ODE,
$$\frac{d^2x}{dt^2}te^{mt} = \frac{dx}{dt}(mte^{mt} + e^{mt}) = m^2te^{mt} + 2me^{mt}$$
$$\frac{dx}{dt}te^{mt} = mte^{mt} + e^{mt}$$
Collecting the terms by $te^{mt}$ gives
$$m^2+am+b=0$$
and collecting the terms by $e^{mt}$ gives
$$2m+a=0$$
because $m$ is a double root, therefore the differential at $m$ is zero.
To finish, examine the cases for $t^ke^{mt}, k\ge2$ to conclude that this is all the solutions.
