Does Reduction of Order Always Solve Non-Homogeneous Second Order ODEs?

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Hi, say I have a second order ODE a_1 \left( x \right)\frac{{d^2 y}}{{dx}} + a_2 \left( x \right)\frac{{dy}}{{dx}} + a_3 \left( x \right)y = f\left( x \right) and I have y_H = h\left( x \right) which satisfies the DE for case of f(x) = 0. Can I always, in principle, find the general solution to the DE for the case where f(x) is not necessarily zero by using the substitution y\left( x \right) = u\left( x \right)h\left( x \right). I know that there are various substitutions for second order ODEs but I would like to know if the one I mentioned always works. Any help would be good thanks.
 
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Benny said:
Hi, say I have a second order ODE a_1 \left( x \right)\frac{{d^2 y}}{{dx}} + a_2 \left( x \right)\frac{{dy}}{{dx}} + a_3 \left( x \right)y = f\left( x \right) and I have y_H = h\left( x \right) which satisfies the DE for case of f(x) = 0. Can I always, in principle, find the general solution to the DE for the case where f(x) is not necessarily zero by using the substitution y\left( x \right) = u\left( x \right)h\left( x \right). I know that there are various substitutions for second order ODEs but I would like to know if the one I mentioned always works. Any help would be good thanks.

Absolutely. The problem of course is effecting the integration: After letting y=uh and substituting into the ODE, then let:

w=u^{'}

resulting in:

w^{'}+\left(\frac{2h^{'}}{h}+\frac{a_2}{a_1}\right)w=\frac{f}{a_1h}
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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