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The first is that, given the complementary solution, [itex]y_{c}(x)=c_{1}y_{1}(x)+c_{2}y_{2}(x)[/itex], to some 2nd-order inhomogeneous ODE: $$a_{2}(x)y''(x)+a_{1}(x)y'(x)+a_{0}(x)y(x)=f(x)$$ we assume that the particular solution $y_{p}(x)$ has the form $$y_{p}(x)=u_{1}(x)y_{1}(x)+u_{2}(x)y_{2}(x)$$ where [itex]u_{1}(x)[/itex] and [itex]u_{2}(x)[/itex] are arbitrary functions.

Is the motivation for this ansatz that the homogeneous equation can be viewed as a special case of the inhomogeneous one, i.e. with [itex]f(x)=0[/itex], and as such it is reasonable to assume that the particular solution to the inhomogeneous equation will be of a similar form to the complementary solution?!

The second is that, starting from this ansatz we note that we require that [itex]y_{p}[/itex] is a solution to the inhomogeneous equation, and upon inserting this into the ODE (and doing a little algebra), this leaves us with the equation $$a_{2}\frac{d}{dx}\left[u'_{1}y_{1}+u'_{2}y_{2}\right]+a_{1}\left[u'_{1}y_{1}+u'_{2}y_{2}\right]+ a_{2}\left[u'_{1}y'_{1}+u'_{2}y'_{2}\right]= f(x)$$ which applies a single constraint on the forms of [itex]u_{1}(x)[/itex] and [itex]u_{2}(x)[/itex]. However, in order to find solutions for both [itex]u_{1}(x)[/itex] and [itex]u_{2}(x)[/itex] we require two equations (just a single equation would enable us to find a solution of one in terms of the other, but this other function is still arbitrary and thus we need a further equation to determine its form [I'm a bit unsure whether my argument is correct here?!]). As such, we have one constraint (that the LHS equals the RHS [which has a fixed form [itex]f(x)[/itex]]), and this leaves us with one degree of freedom that we are free to constrain. Thus, we choose that $$u'_{1}(x)y_{1}(x)+u'_{2}(x)y_{2}(x)=0$$ such that $$\left[u'_{1}(x)y'_{1}(x)+u'_{2}(x)y'_{2}(x)\right]= \frac{f(x)}{a_{2}(x)}$$ Is this the correct reasoning?