"Don't panic!"
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I've been reviewing my knowledge on the technique of variation of parameters to solve differential equations and have a couple of queries that I'd like to clear up (particularly for 2nd order inhomogeneous ODEs), if possible.
The first is that, given the complementary solution, y_{c}(x)=c_{1}y_{1}(x)+c_{2}y_{2}(x), 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 u_{1}(x) and u_{2}(x) 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 f(x)=0, 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 y_{p} 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 u_{1}(x) and u_{2}(x). However, in order to find solutions for both u_{1}(x) and u_{2}(x) 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 f(x)]), 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?
The first is that, given the complementary solution, y_{c}(x)=c_{1}y_{1}(x)+c_{2}y_{2}(x), 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 u_{1}(x) and u_{2}(x) 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 f(x)=0, 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 y_{p} 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 u_{1}(x) and u_{2}(x). However, in order to find solutions for both u_{1}(x) and u_{2}(x) 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 f(x)]), 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?