In the last video, we solved the Airy's equation, y double prime minus xy equals zero. We did a series solution on satz, and we obtained two independent series. The first one multiplied by the free constant a0, 1 plus x cubed, over 3 times 2, plus x to the sixth, over 6 times 5, times 3 times 2, et cetera. The second series multiplied by the free constant a1, x plus x to the fourth, over 4 times 3, plus x to the seventh, over 7 times 6, times 4, times 3. We can define these two independent series then as two functions, they are called Airy's functions. The first one, we call y-naught of x, the second one we call y1 of x. Okay. That's all very nice, but when you want to see a function, what you really want to see is a graph of the function. So, what does y-naught of x look like when you plot it versus x? What does y1 of x look like when you plot it versus x? You can use the power series to plot, but it's accurate when x is small, but very quickly becomes inaccurate when x is large, you need more and more terms. It's not a very efficient numerical method to plot. Instead, you can use built-in ODE solvers. In MATLAB, you would use something called ODE 45, and it would be very quick solution to plot y-naught and y1. But you need some information to plot these two functions. What you need is two initial conditions. You cannot do the numerical solution of a second order differential equation unless you have two initial values, y of zero and y prime of zero. The nice thing about these series solutions is that these have very specific then initial value. So, this y-naught of x satisfies this differential equation with the following initial values, right? y-naught of zero, if you plug in x equals zero, this is one, right? y-naught prime of zero, if you take the derivative of this, the leading term will be x squared, so that will be zero. So, if you solve this second-order differential equation with the initial conditions y of zero equals one and y prime of zero equals zero, the solution you get is this solution of this function y-naught of x represented by this power series. For this one, you have y1 of zero plugging in x equals zero all terms have an x in them, so that would be zero, and y one prime of zero. Taking the derivative the derivative of x is one, all the other terms will go to zero, so that will be one. So, we use the initial conditions y1 of zero equals zero, and y one prime of zero equals one, and you solve this differential equation numerically you will get the y1 of x represented by this power series. Let's have a look at the graph then of the solution. The upper graph is y-naught of x versus x. If you look at x equals zero, you see that y-naught of zero equals one, and the derivative of the function at x equals zero is zero. The bottom graph is y1 of x versus x. If you look at x equals zero, you'll see y1 of zero equals zero, and the derivative of that function at x equals zero is one. The most compelling feature of these graphs is that there are oscillatory for negative x and growing exponentially for positive x. Let's go back and look at the equation and see why that's the case. Okay. So, let's have a look at the equation, y double prime minus xy equals zero. When x is negative, this has y double-prime plus something times y equals zero. We know that y double prime plus y equals zero, that's our sine and cosine function. So, this is oscillatory which is why for negative x, the Airy's functions are oscillatory because of this plus sign x is negative. If x is positive, the equation would look something like y double prime minus y equals zero. The coefficient is varying, but the sign is correct. This type of equation has exponentially growing and exponentially decaying solutions, so this is exponential. What you see in the numerical solution of the Airy's equation is the exponentially growing solution.Okay. So, let me summarize and review. We did solve the Airy's equation a non-constant coefficient equation using a series solution method. This was our solution. We have analytically the formal the areas function near x equals zero, but it's very difficult to use these series solutions when x is much different than zero. So instead, we go to a numerical solution, but the numerical solution is aided by knowing what the right initial conditions are to pick up either this first solution or the initial conditions to pick up the second solution. We can also say something qualitatively about the behavior of this equation by considering what is the sign in front of the y term. A plus sign would be oscillatory solutions here valid for negative x, and a minus sign would be exponential like solutions here valid for positive x. I'm Jeff Jasnoff. Thanks for watching, and I'll see you in the next video.