[MUSIC] Sometimes, the people call a solution of an ordinary differential equation, an integral, or an integral curve of the given differential equation since integration is done inverse operation of differentiation, okay? By solving an nth order differential equation for, like this one, okay? We usually expected to have n parameter family of solutions, say little phi(x, C1 and the Cn, okay? As a very extreme example, okay? As a very extreme example, consider, Answer the difference or denote the differential equations, say, n is derivative of y = 0, right? Can you solve it then? Okay? Integrate both sides then, you're going to get, integrating both sides, okay, n- 1 derivative = some constant C. Cn, right, Cn. Take one more integration again. You are going to get, okay, y(n- 1) derivative of y = y, okay? Cn of x times x and + Cn- 1, right, okay? Keep going, okay? Take integration on both sides, okay, then through the same pulley induction, okay, you can conclude, okay, y = C0 + C1x + and so on C n- 1 times x to the n- 1, okay? Where the all Cs are arbitrary constant. This family, okay, satisfied the originally given sample and thought the differential equation say, n's derivative of y = 0, okay? Can you see it? Through the n times of integration, you can confirm that this polynomial of degree and -1, okay? Solves this is sample differential equation, right? How many arbitrary constants do we have down there? One, two, and so on, anyway, we have n arbitrary constants. Likewise, okay, in solving any such nth order ordinal differential equation, okay? We usually expect it to have a solution, which depends on n arbitrary constant, okay? Such a solution, if you [INAUDIBLE] we call it as a general solution of this given differential equation, okay? So, for example, this is general solution of that differential equation. A solution, which is free of arbitrary parameters is called a particular solution. For example, If we have, General solution, Say phi of x, C1 and Cn of this nth order or DE, right? This is a general solution, right? Where the old Ci is arbitrary constant, that you may specify the values of those Cis, right? For example, C1, let's take in- 1. How about C2? How about 10? Right? So, how about Cn, negative 3, right? Okay, so now this equation has no arbitrary parameter, right, then we call it as a particular solution, okay? That's what I mean by a particular solution, okay? Our solution is called the singular solution of differential equation. If that solution cannot be obtained by specializing the free parameters in each general solution. Our C, such notion through the examples, okay? Let's look at the following example. Given differential equation is a forced to order, y prime- x times the y to the one-half = 0. Because of this power of y which is not equal to 1, okay? This is a non linear forced to order the differential equation way, I claim its general solution, in other words, the solution containing one free parameter, okay? It's given by x squared over 4 + a C and the square, that's the general solution, okay? I will look to the simple computation to confirm it, okay? We just take a derivative of it, plugging that expression into this equation, and to confirm it to be equal to identical 0, okay? It's a general solution on the whole real line, okay? What if you choose the C = 0? Initially seize an arbitrary constant way. So, you may take C = 0, then you get, x square over 4 and a square, right? In other words, you get x to the 4 over 16, right? This is a particular solution. If you plug in c = 1, then you get another particular solution. There are infinitely many particular solutions depending on each choice of C. On the other hand, By inspection, you can confirm very easily that constant function y = 0 is also solution, right? When y = 0, its derivative is 0, right? When y = 0, y to the one-half = 0. So, when y = 0, the left-hand side of differential equation is equal to 0. So this constant function, okay, satisfy the given differential equation. So this is a solution, right? Can you obtain this as a solution by specifying the value of C in this general solution? Think about it, right? What I mean is, can you get y is equal to identical 0 by taking some special value of C? It's impossible, right? Can you see it, right? There is no choice of C, which makes this expression to be identical 0, right? What does that mean? This is a solution but a singular solution, okay? As I will define now, this is a so called trivial solution. The identical 0 function. If this is a solution then, we call it as trivial solution. Here, if the constant function phi of x is equal to identical 0 on some interval I, if this is a solution to the differential equation 4, right, okay? In other words, what I mean is that if f of x, right, 0 and so, 0. If is identical 0 on the interval I, right? That's what I mean. Then, we call this phi[x] is equal to identical 0, constant 0, our trivial solution, okay? For example, the previous example, right? This one has a trivial solution way which is also singular solution, okay? For the examples, right, look at this differential equation, okay? Differential equation for computing species, right, y times 2- 3xdx + x times the 3y- 1dy is equal to 0, right? If y is equal to identical 0, then second term will be 0. And because of this, y down there, the first term is equal to 0 again. So that means this differential equation is satisfied by this choice. In other words, this given differential equation has a three wheel solution. Let's look at another differential equation down here, very simple one, xdx + ydy = 0. When y = zero, second term is equal to 0, but there is no reason for fist term x times dx to be equal to 0. So, this differential equation has no trivial solution, okay? Moreover, not every differential equation has a solution at all, right? For example, consider this first order non-linear differential equation say, y prime squared + 1 = 0, right? As you know well, any real quantity to square is not negative, so add 1 to that. What I mean is, okay? Y prime square + 1, this quantity is not negative, add 1 then is greater than or equal to 1, right? So there is no chance for this to be equal to 0, okay? Unless we allow y prime to be complex quantity, okay? But we are concerned here with only real-valued functions. So, okay, the given differential equation here has no real-valued solution on any interval, okay?