# Purpose of complementary function

I understand that adding the complementary function with the particular integral provides a valid general solution given by the proof here.

However, I don't quite understand why the complementary function is necessary. In my shallow understanding, we are adding on the complementary function simply to give the solution more detail with the rationale 'the more the merrier'.

I'm new to differential equations and only understand that we are trying to find a $$y(x)$$ that satisfies the differential equation. And the particular solution has done that already.

• You don’t just want to find one function which satisfies the ODE; often we want to find them all. So there is the question of uniqueness. Suppose $y_1,y_2$ satisfy $L[y_1]=f$ and $L[y_2]=f$. Does that automatically imply $y_1=y_2$? No. All that implies is that $L[y_1-y_2]=0$, i.e $y_1-y_2$ satisfies the homogeneous equation. Commented Sep 10, 2023 at 2:19
• @peek-a-boo's description is accurate, but let me add that when you start trying to fit boundary conditions and initial conditions, the complementary solutions become essential. Commented Sep 10, 2023 at 2:42
• Right. There are a lot of functions that satisfy, say, $y'=y$, but only one of them also satisfies $y'(0)=17$. Commented Sep 10, 2023 at 3:02
• @peek-a-boo Okay, so there is more than one solution. I must've overlooked that. My understanding now is that the complementary function gives room for more than one solution, since it often if not always contains arbitrary constants (correct me if I'm wrong but they originate from the constants of integration). Since we want as many solutions as possible, we include the complementary function. This is my loose understanding atm. Please correct me if I'm wrong
– Tca
Commented Sep 10, 2023 at 8:23
• You can get some arbitrary constants via integration, and when you write "$+C$" that's probably where they come from. But consider this: If your equation is $\dot y - y =0$, then $y(x) = e^x$ is one complimentary solution, but in fact $y(x) = C_1 e^x$ is also a solution for any $C_1$. And this arb. const. comes from the linear nature of $L[y] = \dot y - y$. Commented Sep 22, 2023 at 3:56