Okay, here is the fourth theorem, concerning to the linear homogeneous differential equation, okay? Which is so called the superposition principle for homogeneous equation. Homogenous means what? The right hand side b(x) = 0, like this one, 0. Then we call this linear differential equation is homogeneous, okay? So, consider the [INAUDIBLE] order homogeneous differential equation say (4), and we assume that there are finitely many solution of this linear homogeneous differential equation, okay? From y sub 1 to y sub k, okay? Then our claim is the arbitrarily linear combination, okay? Sum of c sub i times y sub i, where c sub i is arbitrary constant. Is also solution of the same differential equation, okay? That claim I recorded as a superposition principle for homogeneous differential equation, okay? Its proof is very simple, by the linearity 3 way, okay. What is the linearity? If you apply such a finite linear combination of functions, and apply the linear operator L, then we have this is the same as sum of c sub i times L[y sub i]. We were assuming at the beginning that y sub i, they're all solutions of l over y is equal to 0, right? So this quantity is a 0, right? L of yi = 0 times any number is a 0? Some of them will be equal to 0, right? That's the end of the proof, right? Okay, so very simple observation, okay? For example, consider this, Initial value problem for variable coefficient homogenous second order ordinary differential equation. x squared time y double prime- x times y prime + y = 0. And satisfying two initial conditions, say y(1) = 1 and y prime of 1 = -1, okay? After confirming that t sub 1 = x and y sub 2 = x times the log of x. Both are solutions of this differential equation on the interval from zero to infinity, okay? Showing that both with y sub 1 and y sub 2 down there, we'll solve these solutions of these equation. That's an easy computation, I'll leave it to you, okay? Then what we can say by the superposition principle that I've just introduced, any of the linear combination, say, y = c sub 1 times x + c sub 2 times x times log of x, is also a solution of the same differential equation, okay? Any of this functions for arbitrary two constants c sub 1 and the c sub, they satisfy this a second order homogeneous differential equation part of a super position principle. Now for this family of solution, apply the two given boundary condition y(1) = 1 and y prime of 1 = -1, right? So first 1 = y(1) when x = 1, plug in 1 into this equation then, you have a c sub 1 only, right? Because log of 1 = 0, okay? That already means the c sub 1 must be = 1, okay? Now look at the initial condition, y prime of 1 = -1 so you should have -1 = y prime(1), okay? From this y, what is the derivative? So let's compute it here okay? So now here we have y = c sub 1 times x + c sub 2 times x logx. We need to revert this y'= c1 + c2 logx + c2 times x times derivative of log of x, that is 1(x), right? Then plug in x = 1, plugging from here, compute it when x = 1, okay? How much is this one? Log 1 = 0 so, you get = c1 + 0 + when x is equal to 1, you get c2, right? So we get this equation, -1 = y prime of 1 and that should be = c1 + c2 and we already know that c1 = 1. So that means that c2 = -2, right? So when c1 = 1 and the c2 = -2, you get function x- 2x times log of x, okay? It must satisfy both differential and these two initial conditions, right? So this is a solution, right? This is a solution to the given initial variable problem. In fact this function is the only one possible solution to the given initial very problem as the next theorem shows, okay?
