How to solve differential equations of the form: $$\ddot{r}+A(t)r=B(t)$$

In particular I would like to solve the following differential equation: $$\ddot{r}-\omega^2r=\frac{V}{t}$$ Where $\omega$ and $V$ are constants. Till now I've learned how to solve homogeneous ODE's with constant coefficients, so I've no idea how to even approach this problem. Any hints or references will be helpful.


Problem: A straight smooth tube revolves with angular velocity $\omega$ in a horizontal plane about one extremity which is fixed. If at $t=0$ a particle inside it be at a distance $a$ from a fixed end and moving with a velocity $V$ along the tube, show that its distance at time $t$ is $$r=a\cosh(\omega t)+\frac{V}{t}\sinh(\omega t)$$

My attempt: The radial acceleration is given as follows: $$\ddot r-r\dot{\theta}^2=\frac{V}{t}$$ Since $\dot{\theta}=\omega$, we get: $$\ddot{r}-\omega^2r=\frac{V}{t}$$

  • 1
    $\begingroup$ That's looks like harmonic equation, but in Russian physic books this is called "pseudo-harmonic " equation. $\endgroup$ – openspace Jul 5 '16 at 12:41
  • $\begingroup$ Should I write the problem from where I derived the equation ? Maybe, it'll provide some context. $\endgroup$ – nls Jul 5 '16 at 12:47
  • $\begingroup$ yes, I guess it will be helpful $\endgroup$ – openspace Jul 5 '16 at 12:47

Notice that, the general solution will be the sum of the complementary solution and particular solution.

  • The complementary solution:


Assume, that the solution will be proportional to $e^{\lambda t}$ for some constant $\lambda$.

Substitute $r(t)=e^{\lambda t}$ and $\frac{\text{d}^2}{\text{d}t^2}\left(e^{\lambda t}\right)=\lambda^2e^{\lambda t}$

$$\frac{\text{d}^2}{\text{d}t^2}\left(e^{\lambda t}\right)-\omega^2e^{\lambda t}=0\Longleftrightarrow$$ $$\lambda^2e^{\lambda t}-\omega^2e^{\lambda t}=0\Longleftrightarrow$$ $$e^{\lambda t}\left(\lambda^2-\omega^2e^{\lambda t}\right)=0\Longleftrightarrow$$

Since $e^{\lambda t}\ne0$ for a finite $\lambda$, zeros must come from the polynomial.

$$\lambda^2-\omega^2e^{\lambda t}=0\Longleftrightarrow\lambda=\pm\omega$$

So, for the complementary solution we found:

$$r_c(t)=\text{C}_1e^{\omega t}+\text{C}_2e^{-\omega t}$$

  • Determine the particular solution by variation of parameters:

First of all list the basis solutions in $r_c(t)$ so: $r_{p_1}(t)=e^{\omega t}$ and $r_{p_2}(t)=e^{-\omega t}$.

Now, compute the Wronskian of $r_{p_1}(t)$ and $r_{p_2}(t)$:

$$\mathcal{W}(t)=\left|\begin{matrix} e^{\omega t} & e^{-\omega t} \\ \frac{\text{d}}{\text{d}t}\left(e^{\omega t}\right) & \frac{\text{d}}{\text{d}t}\left(e^{-\omega t}\right) \end{matrix}\right|=\left|\begin{matrix} e^{\omega t} & e^{-\omega t} \\ \omega e^{\omega t} & -\omega e^{-\omega t} \end{matrix}\right|=2\omega$$

Now, let:

  • $$y(t)=\frac{\text{V}}{t}$$
  • $$q_1(t)=-\int\frac{r_{p_1}(t)y(t)}{\mathcal{W}(t)}\space\text{d}t=-\int\frac{\text{V}e^{\omega t}}{2\omega}\space\text{d}t$$
  • $$q_2(t)=\int\frac{r_{p_2}(t)y(t)}{\mathcal{W}(t)}\space\text{d}t=\int\frac{\text{V}e^{-\omega t}}{2\omega}\space\text{d}t$$

The particular solution will be given by:


Now, we can set the general solution:

$$r(t)=r_c(t)+r_p(t)=\text{C}_1e^{\omega t}+\text{C}_2e^{-\omega t}-e^{-\omega t}\int\frac{\text{V}e^{\omega t}}{2\omega}\space\text{d}t+e^{\omega t}\int\frac{\text{V}e^{-\omega t}}{2\omega}\space\text{d}t$$


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