[Mathematics][MIT 18.03] Proof of a Theory about the Solution to Second-order Linear Homogeneous Dif

Posted 2022-12-22 raymondjiang

Edit: I suddenly found that this theorem is called Abel Identity. And the proof here is exactly the same as what he did.

---

 

　　At first, I‘d like to say thank you to MIT open courses which give me the privilege to enjoy the most outstanding education resources.

　　Okay, come to the point. When I was learning the second-order homogeneous differential equation, the professor quoted a theory in one step to prove that $c_1y_1+c_2y_2$ are all the solutions.

THM: if $y_1,y_2$ are solu‘s to ODE, then either $W(y_1,y_2) = 0$ (i.e. for all x) or$W(y_1,y_2)$ is nonzero (i.e. for all x).

note: W means Wronskian

　　Well, frankly speaking, I got the inspiration from that introduction in wiki too.

Proof:

　　Let‘s say that we have a second-order homogeneous differential equation.

　　$y‘‘ + py‘ + qy = 0$ in which $p$ and $q$ are some functions of x.

　　And $y_1,y_2$ are two solutions to this equation, which means that:

　　　　$y_1‘‘ + py_1‘ + qy_1 = 0$

　　　　$y_2‘‘ + py_2‘ + qy_2 = 0$

　　Last, we get our $W$, $W(y_1,y_2) = y_1y_2‘ - y_2y_1‘$.

　　Step 1. Differential $W$, and we get $W‘ = y_1y_2‘‘ - y_2y_1‘‘$.

　　Step 2. From above, we get $y_1‘‘ = -py_1‘ -qy_1 $ and $y_2$ is the same case.

　　Step 3. Plug these in, we get $W‘ = (-p)(y_1y_2‘ - y_2y_1‘) = (-p)W$, and that is $W‘ + pW = 0$. This is the first solvable linear equation.

　　Step 4. So we get $W = ce^-\int p(x)\,dx$, and $c$ is an arbitrary constant. As a result, if $c$ equals to zero, then $W$ is always zero. If $c$ doesn‘t equal to zero, then $W$ is a exponential function, which means that it always $>0$.

Done.