When solving a linear ODE of the form $$y' + p(x)y = q(x)$$ does using an integrating factor, $e^{\int p dx}$ provide us a method to get a general solution to the ODE, or is it a particular solution? Why / why not?
When using them to solve a problem, such as finding the solution to $$y' + 3(x+1)^{-1}y = 4(x+1)^{-2}$$ when using the integrating factor method, i.e exploiting the property that the integrating factor $A$ produces the identity $$(Ay)' = A(y' + py) = Aq$$ then integrating, I arrived at the solution $$y = \frac{2}{x+1} + \frac{C}{(x+1)^3}$$ Where $C$ is some constant.
I also tried to solve the homogeneous version, ending up with $ y = K(x+1)^{-3}$, K is some constant. Furthermore, $y = \frac{2}{x+1}$ is a particular solution to the inhomogeneous version. So, it seems that the integrating factor method has the particular solution + homogeneous solution = general solution built in, but why, and how?
Best Answer
Short answer: the integrating factor method furnishes a general solution.
Long answer:
In first order linear ODE, the Green's function $G$ for a given $t_0$ satisfies the homogeneous ODE with $G(t_0)=1$. A particular solution to the inhomogeneous equation with $y_p(t_0)=0$ is $y_p(t)=\int_{t_0}^t G(t-s) f(s) ds$. (This is non-obvious but standard, you can look up a proof.) A solution to the IVP with $y(t_0)=y_0$ is then given by $y=y_p+y_0 G$.
Now $G$ is really the reciprocal of an integrating factor (chosen with a suitable integration constant). In view of that you see that integrating after multiplying by $1/G$ results in $y/G=y_p/G+y_0$ and then you multiply by G to finally solve for $y$. This is the connection.
This Green's function idea generalizes to higher order linear ODE even though the integrating factor method does not.