# Can't get rid of integrals solving this differential equation.

Assuming $\mathbf{A}\equiv \vec A$ , $\dot q\equiv \frac{d}{dt}q$ ,and $\ddot q\equiv \frac{d^2}{dt^2}q$ ,

And Using the Right-hand Cartesian coordinate system with base vectors $\mathbf{\hat i\, \hat j \, \hat k}$ corresponding to the axes $x \, y\, z$,

This is a derived given equation:

$$\mathbf{\ddot r}(t)=\cos(t)\mathbf{\hat i}-\dot r_z(t) \sin(t) \mathbf{\hat i}+\dot r_x(t) \sin(t)\mathbf{\hat k}$$

Having the initial conditions such that $\mathbf{r}(0)=\mathbf0$ and $\mathbf{\dot r}(0)=\mathbf{0}$.

The method is to integrate both sides twice. But after the first integration, due to having product of two t-dependant functions in the second term, an integral will remain unsolved. (using integration by parts).

If you couldn't get what I mean, try integrating the second term of the right side of the equation using this method:

Then you'll find out what I mean by "how to get rid of the integral".

The question is that: is there any alternative to solving this? (Other than integration by parts, I mean).

If not, then what is my mistake? or how should my approach be revised to get the solution?

EDIT: The approach to the equation. This isn't a problem or homework.

I just wondered how the motion of a charged particle in a combination of electric and magnetic field (with the specified function) would be. picture

• Just to be clear: did you use integration by part for second term in form $\dot{r}_z (t) \sin t = \sin t \cdot d(r_z(t))$ ? – Evgeny Jul 6 '16 at 10:42
• I considered $\dot r_z(t)$ as $u$ and $\sin(t)$ as $v$. then substituted in the picture shown. – AHB Jul 6 '16 at 10:50

The $y$ component is not coupled to the other two components and has the trivial solution $r_y(t)=0$.

For the other two components, since $\mathbf r(t)$ itself doesn't appear in the equation, we can solve for $\mathbf v(t)=\dot{\mathbf r}(t)$ and then integrate. So we want to solve

$$\pmatrix{\dot v_x(t)\\\dot v_z(t)}=\pmatrix{0&-\sin t\\\sin t&0}\pmatrix{v_x(t)\\v_z(t)}+\pmatrix{\cos t\\0}\;.$$

These equations are decoupled if we introduce $v_\pm(t)=v_x(t)\pm\mathrm iv_z(t)$ to obtain

$$\dot v_+(t)=\hphantom -\mathrm i\sin tv_+(t)+\cos t\;, \\ \dot v_-(t)=-\mathrm i\sin tv_-(t)+\cos t\;.$$

To solve the homogeneous versions of these equations, divide by $v_\pm(t)$, yielding

\begin{align} \frac{\dot v_\pm(t)}{v_\pm(t)}=\pm\mathrm i\sin t\;, \end{align}

and integrate to obtain

$$\log v_\pm(t)=\mp\mathrm i\cos t+C_\pm\;,\\ v_\pm(t)=c_\pm\mathrm e^{\mp\mathrm i\cos t}\;,$$

and thus

$$v_x(t)=\hphantom -c\cos\cos t+d\sin\cos t\;,\\ v_z(t)=-c\sin\cos t+d\cos\cos t\;.$$

A particular solution of the inhomogeneous equations can in principle be obtained by varying the constants, but in the present case the resulting integral,

$$c_\pm(t)=\int_0^t\mathrm e^{\pm\mathrm i\cos\tau}\cos\tau\,\mathrm d\tau\;,$$

can't be expressed in terms of elementary functions. The same is then also true for $r_x(t)$ and $r_z(t)$, which result from integrating $v_x(t)$ and $v_z(t)$.

You may also be interested in this treatment of the problem.

• So was my approach correct? Yay, I am happy. And I'll have to read your answer carefully once more to understand it. Thank you. – AHB Jul 6 '16 at 20:24
• @AHB: Well, as you noticed yourself, your approach leads to an integral that you can't get rid of. Generally speaking, if you have coupled linear differential equations, you need to first decouple them (as I did here by introducing $v_\pm$), and if you have an inhomogeneous equation, you usually need to solve the corresponding homogeneous equation first. It's rare that you can solve a differential equation simply by integrating it as it stands. – joriki Jul 6 '16 at 20:29