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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?

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    what do you mean 'right hand side different than zero'? Do you mean a constant term or a function?2012-02-15

3 Answers 3

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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 $y=\frac{C_1}{1-2a}x^{1-2a}+C_2$

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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}

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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'

$\frac{Kx^{1-2a}}{1-2a}+C =y \text{ and } a \neq \frac{1}{2}$