How to solve $\ddot{x}=-x-2\dot{x}$

I am currently dealing with the harmonic oscillator and for the case of critical damping, i need to solve the equation $$\ddot{x}=-x-2\dot{x}.$$ I can write the problem as a system of equations $$\begin{pmatrix}\dot{x}\\\dot{v}\end{pmatrix}=\begin{pmatrix}0&1\\-1&-2\ \end{pmatrix}\begin{pmatrix}x\\v\end{pmatrix},$$ but the matrix is not diagonalizable, so this doesn't bring me any further.

Is there some way to simplify the problem?

My goal is to determine the set of all functions $$x$$ (for example defined on an open interval $$I$$) with $$\ddot{x}=-x-2\dot{x}$$.

• This is a second order ODE with constant coefficient. The method to solve them is very well known... characteristic equation is given by $0=r^2+2r+1=(r+1)^2$, and thus, the general solution is given by $(A+Bx)e^{-x}$.
– Surb
Aug 15 '20 at 9:26

Using the differential operator, the equation can be written

$$(D+1)^2x=0$$ and you can proceed by solving two first order equations. With $$y:=(D+1)x$$,

$$(D+1)y=0$$ is separable and is easily found to have the general solution $$y=Ce^{-t}.$$

Now

$$\dot x+x=Ce^{-t}$$

and by variation of the constant, this amounts to

$$\dot xe^t+xe^t=\dot{(xe^t)}=C=\dot{(Ct+C^*)}$$

and

$$x=(Ct+C^*)e^{-t}.$$

• Perfect! One alternative way to derive the last equation: Set $f(t)=x(t)e^t$ (1), then your penultimate equation is equivalent to $\dot{f}=C$, which is equivalent to $f(t)=Ct+b$ (2) (if $x$ is defined on an open interval). Putting (1) and (2) together, we obtain the result $x(t)=(Ct+b)e^{-t}$. Aug 15 '20 at 12:25
• @Filippo: theory says that in case of a multiple root, the solution will be a polynomial times the exponential.
– user65203
Aug 15 '20 at 13:29

The easiest way to sovle equations like this is by making an exponential anstaz: $$x(t) = e^{at}$$. Substitution gives you a quadratic polynomial in $$a$$ with roots $$a_1,a_2$$. The combination $$x(t) = c_1 e^{a_1t} + c_2 e^{a_2t}$$ will give the general solution.

Note: There are exceptions when you have a double root or complex roots, but the principle is the same.

EDIT: The case of critical damping is exactly when you get a double root. In this case, the common trick is to multiply one of the solutions by $$t$$ until you get something that works. Hence, the general solution for critical damping is of the form $$x(t) = (c_1 + c_2t) e^{at}$$, where $$a$$ is the single root of the quadratic.

Hint: $$\ddot{x}(t)=-x(t)-2\dot{x}(t)$$ $$\ddot{x}(t)+2\dot{x}(t)+x(t)=0$$ Is equivalent to: $$(xe^t)''=0$$ Integrate twice.

• Wow! This is great, thank you :) But how did you come up with this? Is this an example of a systematic (more general) approach? Aug 15 '20 at 10:39
• Multiply by $e^t$ the differential equation then factorize and you get the compact form of the differential equation I posted. @Filippo Aug 15 '20 at 10:50