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Consider the ODE

$$y'' + y'/x - y / x^2 = 0.$$

If I were the apply the method of reduction of order to this ODE, knowing that $y(x) = x$ is a solution of this hom. ODE.

$$y_g (x) = u x \\ y' = u' x + u \\ y'' = u'' x + u' + u' \\ \Rightarrow u'' x + 3 u' = 0.$$

Setting $w = u'$, $$w' + 3w = 0 \Rightarrow w = e^{-3x} + c_1 \\ \Rightarrow u = e^{-3x} + c_1 x + c_2,$$ so $$y_g (x) = x e^{-3x} + c_1x^2 + c_2 x.$$

However, it is clear that $x^2$ is not a solution of this hom. ODE. Moreover, we have a term $x e^{-3x}$, which does not contain any arbitrary constant, and cannot be generated by the solutions of this hom. ODE.

My question is that, what this method does not work in this case ?

I mean if the method works in the inhomogeneous case, it should also work in the homogeneous case, since the latter is a special case of the former, so what went wrong ?

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  • $\begingroup$ It's already been pointed out that the equation $w'+3w=0 $ was simply wrong, so this really doesn't matter, but you might note for future reference that you also _solved_ $w'+3w=0$ incorrectly. $\endgroup$ Apr 4 '18 at 15:26
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You made a transciption error. The equation in $w$ is $$ xw'+3w=0 $$ which has solutions $x^3w=c$, $u'(x)=w(x)=cx^{-3}$ so that (absorbing constant factors in the integration constant) $$u(x)=Cx^{-2}+D$$ and thus $$y(x)=xu(x)=Cx^{-1}+Dx.$$

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Hint $$ \begin{align} y_g (x) &= u x \\ y' &= u' x + u \\ y'' &= u'' x + u' + u' \\ \Rightarrow u'' x + 3 u' &= 0. \end{align} $$

It should be : $$w'x+3w=0$$ $$w=\frac K{x^3}$$ $$u=K\int \frac {dx}{x^3}=\frac {K_1}{x^2}+K_2$$ $$....$$

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