Hooke's law and air resistance - differential equation

Question

I am trying to solve this problem:
The force acting on a moving object is proportional to its displacement. In the same time, the object experiences the force of air resistance proportional to its speed. Find displacement as a function of time.

My problem is that I have never solved differential equations - I am doing some leisurely reading about Physics and this problem is giving me a hard time.

$$m \frac{d^2x}{dt^2} + b \frac{dx}{dt} + kx = 0$$ This is the equation I am trying to solve. Now, if we neglected the middle term, I know that the auxiliary function $x(t) = Ce^{at}$ could be used, but what can be done in this scenario?

Answer 1

Try to substitute $\exp(\lambda t)$ as well. The difference is that $\lambda$ will have a complex and a real part this time. (The real part will cause the loss of velocity).

So let's rename the variables as I've mentioned in the comments to get $$\ddot x + 2\beta \dot x +\omega_0^2x=0$$ Now substitute $x=\exp(i \omega t)$ (you can use $\lambda$ instead of $i \omega$, it just depends on you), and simplify with the $\exp$ terms: $$(i \omega)^2 + 2i \beta \omega + \omega_0^2 = 0$$ $$-\omega^2 + 2i \beta \omega + \omega_0^2 = 0$$ $$\omega^2 - 2i \beta \omega - \omega_0^2 = 0$$ Now use the quadratic formula: $$\omega=\frac{2i\beta \pm \sqrt{-4\beta^2+4\omega_0^2}}{2}$$ $$\omega=i\beta \pm \sqrt{\omega_0^2-\beta^2}$$ Now we need to separate the solution based on the value of $\omega_0^2-\beta^2$. First, let's assume that it's positive. Then we can call it $\Omega^2=\omega_0^2-\beta^2$, so we have that $$\omega=i\beta \pm \Omega$$ and the solution is in the form $$x(t)=C_1e^{i(i\beta+\Omega )t}+C_2e^{i(i\beta-\Omega )t}$$ $$x(t)=C_1e^{(-\beta+i\Omega )t}+C_2e^{-(\beta+i\Omega )t}$$ And you can make it a bit nicer with Euler's identity.
If $\omega_0^2-\beta^2$ is negative, then we can call it $\omega_0^2-\beta^2=-\Omega^2$, so we have that $$\omega=i\beta\pm i\Omega$$ And the solution is $$x(t)=C_1e^{i(i\beta+i\Omega )t}+C_2e^{i(i\beta-i\Omega )t}$$ $$x(t)=C_1e^{-(\beta+\Omega)t}+C_2e^{-(\beta-\Omega )t}$$ In the third case, $\omega_0^2-\beta^2=0$. Then $$\omega=i\beta$$ So the solution is in the form $$x(t)=C_1e^{i(i\beta)t}+C_2te^{i(i\beta)t}$$ $$x(t)=C_1e^{-\beta t}+C_2te^{-\beta t}$$

Answer 2

You can treat the differential operator as a variable:

(mD$^2$+bD+k)x=0

This may seem like an abuse of notation, but it works.

If you treat what's in the parentheses as a quadratic polynomial in D, and apply the quadratic formula to find roots $r_1$ and $r_2$, then you have $(D-r_1)(D-r_2)x=0$. You can then take x = $c_1x_1+c_2x_2$ where $(D-r_1)x_1=0$ and $(D-r_2)x_2=0$, or $\frac{dx_1}{dt}-r_1x_1=0$ and $\frac{dx_2}{dt}-r_2x_2=0$. Now you have separable first order differential equations.