Second order linear differential equation is given below.
$y''+\frac{2}{x}y'+k^2y=0,$
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…
Best Answer
$$xy''+2y'=-k^2xy$$ $(xy)'=xy'+y$
$(xy)''=(xy''+y')+y'=xy''+2y'$
$$(xy)''=-k^2(xy)$$ This is the wellknown ODE $\quad Y''=-k^2Y \quad\to\quad Y=c_1\cos(kx)+c_2\sin(kx)$ $$xy=c_1\cos(kx)+c_2\sin(kx)$$ $$y=c_1\frac{\cos(kx)}{x}+c_2\frac{\sin(kx)}{x}$$