Second order linear differential equation is given below.


where $k$ is constant and $x\neq 0$

I already know that the basis are $y_1=\frac{e^{-ikx}}{x}$ and $y_2=\frac{e^{ikx}}{x}$. (from book)

But i don't know how to find the basis of that ODE...

  • $\begingroup$ So... Do you want a method to find two independent solutions of this ODE? $\endgroup$ – user228113 Jun 24 '15 at 12:53
  • $\begingroup$ let $y=e^{rx}$ and after substitution in :$y''+\frac{2}{x}y'+k^2y=0,$ we get: $r²e^{rx}+\frac{2}{x}re^{rx}+k²e^{rx}=e^{rx}(r²+\frac{2}{x}r+k²)=0$ may this substistion will work .try it out $\endgroup$ – Zeraoulia rafik Jun 24 '15 at 14:19

$$xy''+2y'=-k^2xy$$ $(xy)'=xy'+y$


$$(xy)''=-k^2(xy)$$ This is the wellknown ODE $\quad Y''=-k^2Y \quad\to\quad Y=c_1\cos(kx)+c_2\sin(kx)$ $$xy=c_1\cos(kx)+c_2\sin(kx)$$ $$y=c_1\frac{\cos(kx)}{x}+c_2\frac{\sin(kx)}{x}$$


You may reduce the given DE into another with first derivative removed as follows:

$1$.Put $y=u(x)v(x)$ in the given DE

$2$.Equate the coefficient of $v'(x)$ to zero to obtain $u(x)$.

$3$. Now solve the reduced DE for $v(x)$ with its first derivative term missing by usual methods of CF and PI.

$4$.The solution is $y(x)=u(x)v(x)$.

  • $\begingroup$ ok. I solved the problem from your idea. Thanks! $\endgroup$ – Donghee Park Jun 25 '15 at 4:14

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