# Does Using an Integrating Factor give a General Solution? (Lin. ODE)

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?

• $y=C_1-3ln(x+1)-\frac{4}{x+1}$ Jul 28 '18 at 5:25
• Though the answer below seems sufficient I would like to point out just one case that is of first order linear differential equation in which I.E makes the equation homogeneous and allows partial differential solutions... Jul 28 '18 at 5:42
• @Leekboi Integrating factor method provids a general solution,not a particular solution. Particular solution can be found in Initial value problems.$:-)$ Jul 28 '18 at 5:43
• Sometimes general solutions can be manipulated to via change of arbitrary constant but that doesn't mean it isn't the general solution. In fact doesn't the general solution come built in with all the particular solutions? Jul 28 '18 at 5:47

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.