Now here is another theorem in which I introduce the general solution of linear homogeneous differential equation, okay? And this theorem explains why do we call, a certain family of solutions to a linear homogeneous differential equation to be fundamental, okay? So let assume that we have n solutions and they for a fundamental set of a solution of order and linear, homogeneous differential equation 4 on the interval I, okay? The order of differential equation must coincide to the number of linear independent solutions, okay? So I'm assuming that we a fundamental set of solutions say from y sub 1 to y sub n, okay? Of the nth order of linear homogeneous differential equations, okay? And my claim is any solution of this differential equation, L(y) = 0 on the interval I must be given in following form, okay? Any of the solution y(x) must be linear combination of the members of the fundamental set of a solution say y sub i, okay? Where the c sub i are arbitrary constants, okay? Before giving the proof, let's note the following simple [INAUDIBLE] Here I claimed that any solution of differential equation 4 right here, differential equation 4, okay? L(y), this is a sub n of x, n as derivative of n, and + a1(x) y prime, and + a0(x) and times y, and that is = 0. This is differential equation 4 right? Look at the conclusion again, any solution of this differential equation 4 must be of this form, right? This right-hand side is a linear combination of end solutions. Is the solution of this differential equation simply by the superposition principle, right, okay? Any of such expression must be a solution simply by superposition principle, okay? But the claim of the point of this theorem is not just that bug, any solution should be of this form, okay? There is no solution other than of this type. So that means the order of differential equation is n and then we have a family of solutions, okay? Containing an arbitrary constant, so we call this as general solution of this differential equation 4, right? Trivially, moreover, the family of a solutions is given by that expression, right? Linear combination where the coefficient ci's can be arbitrary constant, right? This is exactly the set of all possible solutions of differential equation 4, okay? All possible solutions of differential equation 4 in the interval I, okay? In particular, any linear homogeneous differential equation, it can not have any singular solution, okay? Can you remember what we mean by a singular solution? Singular solution of any differential equation is a solution which cannot be obtained from, Its general solution, right? So here we claim then, this family is a general solution, and all other solution must be of this type, that means there is no singular solution. Let's give a proof again only for little n = 2, right? In other words, I'm considering the second of the differential equation, right. Second of the linear homogeneous differential equation, and then we're assuming we have a fundamental set of solutions. Say y1 and y2, okay? And choose arbitrary solution of the same differential equation, which I denoted by phi of x, right? And consider the following system of equations for two unknowns. Linear system of equations for two unknowns, c sub 1 and c sub 2, given by this matrix expression, okay? Where the x sub 0 is an arbitrary any point in the interval I, okay? And consider this matrix form of simultaneous equation, right? Since y1 and y2 is a fundamental set of solutions, they must be linearly independent and that means the rules can never vanish. In other words, these two y2 matrix Has a nonzero determinant, in other words, 2 by 2 matrix is non singular, right. This coefficient matrix is non-singular, and that implies this system of equation should have a unique set of solutions for c sub 1 and c sub 2, right? So whatever it is, choose those unique solutions for c1 and c2, right? And consider initially given solution phi of x and solution given by the linear combination, right? c1 times y sub 1 + c2 times y sub 2, right? Then both these functions of phi of x and this linear combination, they satisfy trivial given the differential equation, right? Given differential equation. And they satisfy y of x0 = phi of x0, and the y-prime x0 = phi prime over x0 right? The initial condition is to trivial satisfied when y is equal to phi this trivial, right? Phi of x0 = phi of x0, phi prime (x0) = phi prime(x0), right? My claim is, this linear combination also satisfied these two initial condition, why? Because c1, c2 is the solution of this simultaneous equation, okay? What does that mean? That means c1 times the y1(x0) plus c2 times the y2 (x0) that is the phi of (x0), right? Then from the second line c1 times y1 prime (x0) plus c2 times y2 prime (x0) that must be equal to phi prime of (x0), right? So, both the phi of (x) and this linear combination, satisfy really the given initial value problem, okay? Again up here to the theorem 4.2 which is uniquely the distance of solution 2 Certain initial value problem in particular by the uniqueness, okay? The two functions or two solutions, phi of x and this, they must coincide, okay? On the whole interval I, okay? That's the end of the proof, right? Here we have a two linear independent solution, in other words, a fundamental set of solution. I'm taking arbitrary another condition, and I'm showing that the arbitrary solution of that differential equation must be a linear combination of the members of the fundamental set of the solutions, right? That's the claim, okay? As an example, let's the following. Let's try to find a general solution of this variable coefficient, Linear homogeneous second under differential equation on the interval zero to infinity by finding particular solutions of the form x to the r with suitable constant r, okay? So, Set y = x to the r in the differential equation, okay? Plug in y= x to the r, what is the y-prime? This is i times x to the r- 1 times x recover the x to the r, okay? What is the y double prime? That is r times (r- 1) times x to the r- 2, right? But here you have x squared, right? So again, you can recover x to the r, right? Okay, so do the simple computation then, okay? You will get, okay? If you set y = x to the r in the left-hand side of this differential equation, then you would get [r(r- 1)- r- 3 times x to the r, you notice that is r squared- 2r- 3 times x to the r = 0, if this is a solution, right? And that can be true when all the roots of this quadratic equation, right? You can factorize this quadratic equation very easily, two roots of it are, one is -1 and the other one is 3. That already means what? y sub 1 which is x to the -1 and y sub 2 which is x cubed right? They are solutions of the given differential equation on the interval from 0 to infinity, right? Compute the Wronskian, simple exercise, compute the Wronskian of x to the -1 and x cubed Which is equal 4 times x which is never 0 on this interval. In other words, these two solutions, y1 and y2 are linearly independent on this interval 0 to infinity. In other words, because the order of the differential equation is 2, these two solutions say x to the -1, right here and x cubed, they form a fundamental set of solutions, okay? So their linear combination with arbitrary constants c1 and c2 is the general solution. And, in fact, they are whole family of solutions, right? No other solution is possible, okay?