General solution of $\frac{d^2x}{dt^2}+a\frac{dx}{dt}+bx=0$ for identical roots

Question

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?

Answer 1

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 $|nt|\ll 1$.

Answer 2

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.

Answer 3

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.