How can i find the basis solutions of homogeneous linear ODE?
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...
$\endgroup$ 22 Answers
$\begingroup$$$xy''+2y'=-k^2xy$$ $(xy)'=xy'+y$
$(xy)''=(xy''+y')+y'=xy''+2y'$
$$(xy)''=-k^2(xy)$$ This is the wellknown ODE $\quad Y''=-k^2Y \quad\to\quad Y=c_1\cos(kx)+c_2\sin(kx)$ $$xy=c_1\cos(kx)+c_2\sin(kx)$$ $$y=c_1\frac{\cos(kx)}{x}+c_2\frac{\sin(kx)}{x}$$
$\endgroup$ $\begingroup$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 $y(x)=u(x)v(x)$.
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