Now, let's introduce one more interesting equation, So-called Ricatti differential equation. It's a nonlinear first order differential equation of the form y prime = p(x) + q(x)y + r(x)y squared, okay? We are assuming r(x) is not identical to 0, okay? Otherwise, In other words, if r(x) is equal to identical to 0, then the remaining equation is a simple linear first order differential equation, right? Now, assume that somehow we find one particular solution, say y1, okay, y sub 1. Assume that y sub 1 is the solution of this Ricatti differential equation. Then, y = y1 + u is also a solution of the same equation if and only if u prime- (q + 2 ry sub 1) times u is = ru squared. Can you recognize this due equation for you? Okay? This is a Bernoulli equation, where n is equal 2, right? And we know how to solve the Bernoulli equation. So starting from this first order, non-linear differential equation, so called the Ricatti differential equation. If we happen to know any one particular solution, and by setting y = y1 + u, we can solve the given Ricatti equation completely, okay? Let's look at the example, okay. Here's the one, the nice example of the Ricatti differential equation, say y prime = 2x squared + 1 over x y- 2y squared, And I claimed that this Ricatti differential equation has a solution given by constant k times x. We don't know what the constant k is. Find the k first, then using this fact, solve this Ricatti equation completely. So, said y1 is = k of x, plugging that expression into this one. What is the y prime? That is ky is equal to 2x squared plus 1 over x times k of x, that is k,- 2y1 squared, that is- 2k squared. k and k cancelled out, you are going to get 2- 2k squared times x squared = 0, okay? Are you following me? Right? So k and k canceled out. You are going to get 0 = 2x squared- 2k squared x squared, so 2x squared is a common, and 1- k squared, and that is equal to 0, okay? That is a possible when k is equal to + or- 1, right, okay? We do not need the both of them, okay? We do not need the both of them. So I'm taking k is equal +1, okay, that means what? y is equal to x, is a solution. If you wish then, you can confirm it. y is equal to x, is a solution to this Ricatti differential equation. Then what I said before, okay, if we know one particular solution of this equation, then there is another solution of the type y is equal to the known solution plus unknown u, right? The known solution is now x, so I said, y = x + u, okay? Plugging that expression into the differential equation. And to simplify it, then you are going to get a first order differential equation for the new unknown u, which is u prime + (4x minus 1 over x) u, that is equal to negative 2 times u squared. This is a Bernoulli equation, okay, with n is equal 2. So set w = u to the -1, okay? And by the simple computation, The differential equation for w becomes w prime + (1 over x- 4x) times w = 2, okay? Which is then linear first order differential equation for w, and which has an integrating factor, okay? Exponential of integral 1 over x- 4x dx, right? That is, in fact, x times exponential negative 2x squared. So multiply this integrating factor on both sides of this equation, you will get derivative of, right, x times e to negative 2x squared times w, that is equal to 2 times of x times exponential minus 2x squared, right? So through the integration, okay, x times exponential -2x squared times w, which is the integral of 2x exponential minus 2x squared dx, right, okay? Can you see it right then? Okay, here, I'm writing it. x times e to the negative 2x squared times w, this is equal to interval of 2xe to the -2x squared and dx, and + some arbitrary constant c, right. okay? What is the anti-derivative of this one? Right, okay? That is equal to, right, e to the negative 2x squared, And negative one-half, right? Okay, let me check it, derivative of it, Okay, + c, so divided through by x times exponential, x times e to the negative 2x squared, okay? You are going to get, okay? w = 1 over x times (c times e to the 2x squared- one-half), right? So that means, what is w? w = u to the negative 1, right? So u =, okay, okay, x over c times e to the 2x squared- one-half. And what is u? U is equal to, what is y? y = x + u, right? So that y = x plus u, x plus, This, that is the same as 2x over, okay? 2c times e to the 2x squared- 1. But we can replace 2c by just c, because when c is an arbitrary constant, 2c is another arbitrary constant, right? So I can use the c down there, okay? So that's the general solution.
