Second order differential equation with variable coefficients

Question

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?

Answer 1

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

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.

Answer 2

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.