## MIT 1803 lecture 11

This lecture covered a study of linearity in  y”, y’, y, etc. on the second order homogeneous differential equations.  He addressed the questions

1. Why are $c_1y_1 + c_2y_2$ solutions of a second order linear ODE?
2. What are they all the solutions?

By answering carefully answering these questions elegantly, the theory can be developed to also handle  higher order ODEs without any extra work.

To insure the work is elegantly extensible we first prove the superposition principle using operator notation.  Let $D$ be the differentiation operator that operates on or applies to $y$, and let $L$ be the linear operator.  The definition of a linear operator is an operator that obeys the rules

$L(u_1 + u_2) = L(u_1) + L(u_2)$

$L(cu) = cL(u)$

(where $c$ is a constant).   Now we can observe that $D$ is linear because

$(u_1 + u_2)' = u_1' + u_2'$

$(cu)' = cu'$

Notes that the second order ODE $y'' + py' + qy = 0$ becomes

$D^2y + pDy + qy = 0$

$(D^2 + pD + q) y = 0$

$Ly = 0$

Imagine $L$ as a black box that takes in a function $u(x)$ and outputs a function $v(x)$.  Finding the solution to the homogeneous linear ODE is equivalent to asking “if we want $v(x)=0$, so what $u(x)$ must we put into $L$?”

Proof of 2) starts by letting $y= c_1y_1 + c_2y_2$ and giving initial conditions $y(x_0) = a$ and $y'(x_o) = b$, then

$c_1y_1(x_0) + c_2y_2(x_0) = a$

$c_1y_1'(x_0) + c_2y_2'(x_0) = b$

this set of (two) linear equations is solvable iff the Wronskian $W(y_1,y_2)(x)$ is

$W(y_1,y_2) = \left| \begin{matrix} y_1 & y_2 \\ y_1' & y_2' \end{matrix} \right| \neq 0$

note that if $y_2 = cy_1$ (the solutions are not linearly independent) then $W(y_1,y_2) = 0$ (but note that the reverse is not strictly true – the Wronskian can be zero for other reasons).

Now, somehow, we are meant to find $Y_1$ and $Y_2$ which are normalized solutions, which are better than other solutions because their initial values are nicer.  Why are normalized solutions so good?  Because they allow us to instantly solve the initial conditions problem.  This can be seen from

$y = y_1 Y_1 + y'_1Y'_2$