0
$\begingroup$

i need to solve the following second order differential equation $$\ddot{x}(t)+\frac{1}{t^\frac{1}{4}}x(t)=0\,.$$ I don't succeed in guessing a particular solution. Any suggestion?

$\endgroup$
2
$\begingroup$

$$\ddot{x}(t)+\frac{1}{t^\frac{1}{4}}x(t)=0\,.$$

This is a generalized form of Bessel ODE : $$y''+\lambda^2 z^{\frac{1}{\nu}-2}y(z)=0$$ which solution is known as : $$y(z)=c_1\sqrt{z}\:\text{J}_{\nu} \left(2\lambda\nu z^{\frac{1}{2\nu}}\right) + c_2\sqrt{z}\:\text{J}_{-\nu} \left(2\lambda\nu z^{\frac{1}{2\nu}}\right)$$ In the present case $y=x \;,\: z=t \;,\: \lambda=1 \;,\: \frac{1}{\nu}-2=-\frac{1}{4}\quad\to\quad \nu=\frac{4}{7}$

$$x(t)=c_1\sqrt{t}\:\text{J}_{4/7} \left(\frac{8}{7}t^{7/8}\right)+c_2\sqrt{t}\:\text{J}_{-4/7} \left(\frac{8}{7}t^{7/8}\right)$$ This is consistent with the change of variable suggested by Claude Leibovici, whom I salute and congratulate for his proposal.

$\endgroup$
  • $\begingroup$ Hi Jean ! May I confess that I did not know the generalized form of Bessel ODE ? Thanks for your answer. $\endgroup$ – Claude Leibovici May 11 '17 at 15:18
2
$\begingroup$

Hint

If you first let $x=y \sqrt t$, you should end, after simplifications, with $$ t^2 y''+ t y'+\left( t^{7/4}-\frac 14\right) y=0$$ which "looks close" to a Bessel differential equation.

Now, try to define $t^{7/8}=u$ and rewrite the equation. I suppose that you will end with something more interesting.

$\endgroup$
  • $\begingroup$ Do you mean $u=t^{7/8}y$ or what? $\endgroup$ – user404629 May 11 '17 at 8:56
  • $\begingroup$ @user404629. Thinking more, I "suppose" that it should be something like $t^{7/8}=k u$ where $k$ is a constant to be identified to have a pure Bessel differential equation. But now, the work is going from $y(t), y'(t), y''(t)$ to $y(u), y'(u), y''(u)$. This should work. $\endgroup$ – Claude Leibovici May 11 '17 at 9:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.