I do not know how to solve this differential equation: \frac{a}{x}y'(x) + \frac{1}{2}y''(x) = 0
where $a$ is a constant. Also how can I solve this equation if the right hand side is different than zero?
I do not know how to solve this differential equation: \frac{a}{x}y'(x) + \frac{1}{2}y''(x) = 0
where $a$ is a constant. Also how can I solve this equation if the right hand side is different than zero?
Consider z=y'. Then this is just \frac{a}{x}z(x)+\frac{1}{2}z'(x)=0. That is z'=-\frac{2a}{x}z which is separable.
$\ln|z|=\int \frac{dz}{z} = \int -\frac{2a}{x}dx=-2a\ln|x|+C_0$ and so y'=z=C_1x^{-2a}.
Integrating yields $y=\frac{C_1}{1-2a}x^{1-2a}+C_2$
Substitute : ~y'=u \Rightarrow y''=u'~ , so :
\frac{1}{2}u'+\frac{a}{x}u=0 \Rightarrow \frac{1}{2}u'=\frac{-a}{x}u \Rightarrow \int \frac{du}{u} = -2a\int \frac {dx}{x}
Assuming a solution in an interval not containing $x=0$, lets rewrite the equation as:
2ay' + xy'' = 0
Now put y'=v, y'' = v' to get
2av + xv' = 0
Separating variables produces
\frac{2a}{x} =- \frac{v'}{v}
$2a\log{Cx} =-\log v$
$2a\log{Cx} =-\log v$
{Kx^{-2a}} =y'
$\frac{Kx^{1-2a}}{1-2a}+C =y \text{ and } a \neq \frac{1}{2}$