If you plan to deal with the atomic world, the movement of the planets, chemical processes, electrical circuits, weather forecasts, or with the spread of a virus, then you will eventually encounter so-called differential equations. Once you understand how differential equations work and how to solve them, you will be able to see into the past and into the future. In this video, I will teach you the basics for it. Let's look at Hooke's law as a simple example. F is equal to minus D times Y. This law describes the restoring force F on a mass attached to a spring. The mass experiences this force when you displace it by the distance Y from the equilibrium position. D is a constant coefficient that describes how hard it is to stretch or compress the spring. The mass M of the ball attached to the spring is hidden in the force. We can write the force according to Newton's third law as M times A. A is the acceleration that the mass experiences when it is displaced by the distance Y from its rest position. As soon as you pull on the mass and release it, the spring will start swinging back and forth. Without frictionness in this case, it will never stop swinging. While the mass oscillates, the displacement Y changes. The displacement is therefore dependent on the time t. Thus also the acceleration A depends on the time t. The mass of course remains the same at any time, no matter how much the spring is displaced. This is also true in good approximation for the spring constant D. If we now bring M to the other side, we can use this equation to calculate the acceleration experienced by the mass at each displacement Y. But what if we are interested in the question at which displacement Y will the spring be after 24 seconds? To be able to answer such a future question, we must know how exactly Y depends on the time t. We only know that Y does depend on time, but not how. And exactly when dealing with such future questions, differential equations come into play. We can easily show that the acceleration A is the second time derivative of the distance traveled, so in our case it is the second derivative of Y with respect to time t. Now we have set up a differential equation for the displacement Y. You can recognize a differential equation in short dq by the fact that in addition to the surged function Y, it also contains derivatives of this function, like in this case the second derivative of Y with respect to time t. A differential equation is an equation containing a surged function Y and derivatives of this function. You will certainly encounter many notations of a differential equation. We have written down our differential equation in the so-called Leibniz notation. You will often encounter this notation in physics. We can also write it down a bit more compactly without mentioning the time dependence. If the function Y depends only on the time t, then we can write down the time derivative even more compactly with the so-called Newton notation. One time derivative of Y corresponds to one point above Y, so if there is a second time derivative as in our case, there will be two points. Obviously, this notation is rather unsuitable if you want to consider the tenth derivative, for example. Another notation you are more likely to encounter in mathematics is the Lagrange notation. Here we use primes for the derivatives. So for the second derivative, two primes. In Lagrange notation, it should be clear from the context with respect to which variable the function is differentiated. If it is not clear, then you should write out explicitly on which variables Y depends. Each notation has its advantages and disadvantages. However, remember that these are just different ways of writing down the same physics. And rearranging and renaming does not change the physics under the hood of this differential equation. To answer our previous question at which displacement Y will the spring be after 24 seconds, we must solve the post-differential equation. Solving a differential equation means that you have to find out how the function Y you are looking for exactly depends on the variable t. For simple differential equations like the one of the oscillating mass, there are solving methods you can use to find the function Y. Keep in mind, however, that there is no general recipe for how you can solve an arbitrary differential equation. For some differential equations, there is not even an analytic solution. Here the word analytic means that you cannot write down a concrete equation for the function Y. The only possibility in this case is to solve the differential equation on the computer numerically. Then the computer does not spit out a concrete formula, but data points which you can represent in a diagram and then analyze the behavior of the differential equation. Once you encounter a differential equation, the first thing you need to figure out is which one is the function you are looking for and which variables it depends on. In our differential equation of the oscillating mass, the function we are looking for is called Y and it depends on the variable t. As another example, look at the wave function that describes the electric field of an electromagnetic wave propagating at the speed of light c. What is the function you are looking for in this differential equation? It is the function E because its derivatives occur here. On which variables does the function E depend? The dependence is not explicitly given here, but from the derivatives you can immediately see that E must depend on X, Y, Z and on T. That is on a total of four variables. Let's look at a slightly more complex example. This system of differential equations describes how a mass moves in a three-dimensional gravitational field. Here you have a so-called coupled differential equation system. In this case, a single differential equation is not sufficient to describe the motion of a mass in the gravitational field. In fact, three functions are searched here, namely the trajectories X, Y and Z, which determine a position of the mass in three-dimensional space. Each function describes the motion in one of the three spatial directions. And all three trajectories depend only on the time t. What does it even mean if we have coupled differential equations? The word coupled means that, for example, in the first differential equation for the function X, there is also the function Y. So we cannot simply solve the first differential equation independently of the second one because the second equation tells us how Y behaves in the first equation. In all three differential equations, all of the searched function X, Y and Z occur, which means that we have to solve all three differential equations simultaneously. After you have found out what function you are searching for and which variables it depends on, you should answer some more basic questions to get to know the differential equation better. Is the differential equation ordinary or partial? Partial differential equations describe multi-dimensional problems and are significantly more complex. Of which order is the differential equation? Most order differential equations are usually easy to solve and describe, for example, exponential behavior such as radioactive decay or the cooling of a liquid. Differential equations of second order, on the other hand, are somewhat more complex and also often occur in nature. Maxwell's equations of electrodynamics, Schrodinger's equations of quantum mechanics, these are all second-order differential equations. Only starting from the second order, a differential equation can describe nosolation. And only starting from the third order, a differential equation can describe chaos. Is the differential equation linear or nonlinear? The superposition principle applies to linear differential equations, which is incredibly useful, for example, in the description of electromagnetic phenomena. Nonlinear differential equations are much more complex and occur, for example, in nonlinear electronics in the description of superconducting currents. Moreover, chaos can only occur in nonlinear differential equations of third order and higher. When you encounter such an equation sometimes, the only thing you can do is throw away your pen and paper and solve the equation numerically on the computer. Only nonlinear differential equations cannot even be solved analytically. Is the linear differential equation homogeneous or inhomogeneous? Homogeneous linear differential equations are simpler than inhomogeneous ones and describe, for example, an undisturbed oscillation, while inhomogeneous differential equations are also able to describe externally disturbed oscillations. First let's learn how to answer these questions. After you have classified a differential equation, you can then specifically apply an appropriate method to solve the equation. Even if there is no specific solving method, you will know how complex a differential equation is based on the classification. Our equation for the oscillating mass is an ordinary differential equation. Partial means that the function y we are looking for only depends on one variable, in this case on the time t. The wave equation, on the other hand, is a partial differential equation. Partial means that the searched function e depends on at least two variables and derivatives with respect to these variables occur in the equation. In this case, e depends on four variables, t, x, y and z, and in the differential equation also derivatives with respect to t, x, y, z appear. Furthermore, our equation for the oscillating mass is a differential equation of second order. The order of the differential equation is the highest occurring derivative of the searched function. Partial means in our equation the second derivative of y is the highest one. This is therefore the second order differential equation. The differential equation for the radioactive decay law, on the other hand, is a first order differential equation, because the highest occurring derivative of the search function n is the first derivative. Moreover, our equation for the oscillating mass is linear. Partial means that the searched function and its derivatives contain only powers of one, and there occur no products of derivatives with a function, like y squared or y times the second derivative of y. There also occur no composed functions, like sine of y or square root of y. Note that to the power of 2 in the second derivative in the Leibniz notation is not the power of the derivative, but merely a notation, that it is the second derivative. The radioactive decay law is also linear. What about the wave equation? It is also linear. The coupled differential equation system for the motion of a mass and a gravitational field, on the other hand, is non-linear. Here the searched functions x, y and z occur in quadratic form. But even if the squares were not there, there would still be the square root and the fraction, which make the differential equation system non-linear. In the next types of differential equations, the coefficients multiplied by the searched function and its derivatives are important. In some solving methods it is important to distinguish between constant coefficients and non-constant coefficients. Constant coefficients do not depend on the variables on which the searched function depends. Non-constant coefficients do depend on the variables on which the searched function depends. A coefficient must not necessarily be multiplied with the searched function or its derivative. It can also stand alone. In this case we call the single coefficient a perturbation function. In our differential equation for the oscillating mass there is a constant coefficient, which is multiplied by the searched function y, namely d over m. Strictly speaking there is also a coefficient in front of the second derivative, namely 1, and the single coefficient, that is the perturbation function, is 0 here, so it does not exist. If the perturbation function is 0, then we call the linear differential equation homogeneous. So the differential equation for the oscillating mass is a homogeneous differential equation. The wave equation is also homogeneous, because here there is also no single coefficient. The perturbation function is 0. The differential equation for a forced oscillation, on the other hand, is inhomogeneous. Here the external force f corresponds to the perturbation function. As you can see it stands alone without being multiplied by the function y, or its derivatives. Moreover, the perturbation function f is time dependent, so it is a non-constant coefficient. A differential equation alone is not sufficient to describe a physical system uniquely. The solution of a differential equation describes quite a few possible systems that have a certain behavior. For example, the solution of the radioactive decay law describes an exponential behavior. However, the knowledge about an exponential behavior is not sufficient to be able to say concretely how many atomic nuclei have decayed, for example, after 10 seconds. This is exactly why for every differential equation there are usually given constraints. These are additional information which must be given to a differential equation in order to specify the solution of the equation. The number of necessary constraints depends on the order of the differential equation. For a first-order differential equation, a single constraint is necessary, namely a function value of the search function. For the radioactive decay law, for example, it should be stated how many not yet decayed atomic nuclei were present at the time t equals zero, for example, 1000 atomic nuclei at time t equals zero. For a second-order differential equation, two constraints are necessary, a function value of the search function and, for example, a function value of the first derivative. For an oscillating mass, the function value could be y of zero equals one, which sets the initial displacement to one at time zero. And the function value of the first derivative, that is, the velocity, could be y prime of zero is equal to zero, which sets the initial velocity of the mass. For a third-order differential equation, then three constraints would be necessary to describe a system uniquely. A function value of the search function, a function value, for example, of its first derivative, and a function value, for example, of its second derivative. For a fourth-order differential equation, then four constraints would be necessary, and so on. Most of the time, you will come across the so-called initial conditions and boundary conditions. These are just names for constraints that tell you what kind of information you have about the system. Sometimes, for example, you know in which state the system was at a certain time. This could be the initial time at which you displaced and released the mass on the spring. You specify at one certain point in time, for example, at time t equals zero, which value the displacement y had. And since we need two constraints for the oscillating mass, you also specify which value the derivative y prime, that is, the velocity, had at the same point in time, t equals zero. In such a case, we speak of initial conditions. We call a differential equation, together with initial conditions, an initial value problem. If we solve the initial value problem, we can use the solution to predict the future behavior of the system. Sometimes you are unlucky and do not know the velocity of the oscillating mass at a certain initial time. So you don't know the derivative y prime at time zero, at which you also know the displacement y. But you really need two constraints, otherwise you can't calculate concrete numbers. But maybe you know that, for example, after six seconds, the oscillating mass was in the maximum displaced state. So you know the displacement y of six. If you have constraints given to describe the system at two different points in time, then we call them boundary conditions. We call a differential equation, together with two or more boundary conditions, a boundary value problem. If we solve the boundary value problem, we can use the solution to predict how the system behaves within these boundaries. The sentence function values at two different points in time was, of course, just an example. Instead of time, it could be any variable that fixes the system at the boundaries, at different times, at different positions, at different angles, and so on. So far, so good. Now, let's look at four methods with which you can solve simple differential equations. Solving method number one, separation of variables. With this method, you can solve ordinary differential equations of first order, which must also be linear and homogeneous. This type of differential equations has the form y prime plus k times y is equal to zero. The coefficient k must not necessarily be constant, but can also depend on x. Also note that before the first derivative y prime, the coefficient must be equal to one. If this is not the case, then you simply have to divide the whole equation by the coefficient, which is in front of y prime. Then you have the right form. In this solving method, y and x are considered as two variables and separated from each other by bringing y to one side and x to the other side of the equation. The Leibniz notation of the differential equation is best suited for this purpose. Bring k times y to the right-hand side. Multiply the equation by dx and then divide the equation by y. This way you have only y dependence on the left side and only x dependence on the right side. Now you can integrate over y on the left-hand side and over x on the right-hand side. Integrating one over y gives the natural logarithm of y. Also, don't forget the integration constant. Let's call it a, for example. Now you have to solve for the function y. Use the exponential function on both sides. You can split the sum in the exponential term on the left side into a product where e to the power of ln of y is simply y. Bring the constant e to the power of a to the right side and rename the coefficient to a new constant c. As a result, you get a general solution formula that you can always use to solve homogeneous differential equations. You don't have to apply the separation of variables method again and again, but you can use the solution formula directly. For example, let's look at the differential equation for the radioactive decay law. In this case, the searched function y is the number of not-yet-decayed atomic nuclei n and the variable x corresponds to the time t. And the coefficient k in this case is a decay constant lambda. According to the solution formula, you have to integrate the coefficient that is the decay constant lambda over t. Integrating a constant just gives t. And we already have the general solution for the decay law. Now you know the qualitative behavior of the physical process, namely that atomic nuclei decay exponentially. But you cannot say concretely yet how many nuclei have already decayed after a certain period of time. This is because you don't know the constant c yet. After all, in the decay law, c gives the number of atomic nuclei that were present at the beginning of your observation. So you need an initial condition as additional information to the differential equation. It could be for example like this. n of 0 is equal to 1000. That means at the time t equals 0, there were 1000 atomic nuclei. Inserting this initial condition results in this equation. e to the power of 0 is 1, so c must be 1000. Now you can insert an arbitrary point of time and find out how many not-yet-decayed atomic nuclei are still there. Solving method number two, variation of constants. This method is well-suited for ordinary differential equations of first order, which are linear. The differential equation can be homogeneous or inhomogeneous. The inhomogeneous one is more general. You have this type of differential equation if you can bring your differential equation into the following form. y prime plus k times y is equal to s. The inhomogeneous version differs from the homogeneous one only in the single coefficient. That is, the perturbation function s is not 0. Thus this type of differential equation is somewhat more difficult to solve. In this solving method, you make the ansatz that the general solution y is given by a coefficient c that depends on x multiplied by a homogeneous solution that we denote as yh. You have already learned how to find the homogeneous solution yh with the previous method. All you have to do is to set the perturbation function to 0. Then you have the homogeneous differential equation. You solve it with separation of variables method or directly by using the corresponding solution formula. Let us insert this ansatz into the inhomogeneous equation for y. We also want to replace the derivative y prime with our ansatz. To do this, we must first differentiate y with respect to x. Since both c and yh depend on x, we need to apply the product rule. You do this by differentiating c1s leaving yh and then leaving c and differentiate yh. The result is the derivative of our ansatz. Now we insert the derivative y prime into the inhomogeneous differential equation. If you just factor out c of x, you might see why this approach is so awesome. In parentheses is the homogeneous differential equation which is 0, so we can omit this term completely. You can now rearrange the equation for the unknown coefficient c prime. Now to eliminate the derivative c prime, we have to integrate both sides over x. We cannot integrate the right side concretely because s is different depending on the problem. Therefore we leave the right side unchanged. The left side on the other hand can be integrated and additionally yields an integration constant. We put this constant directly on the right side and define it for example as a constant a. If you now just substitute the found coefficient c into the original ansatz, then you get the general solution of an ordinary linear differential equation of first order. Let's take an example from electrical engineering. Consider a circuit consisting of a coil characterized by the inductance l and a resistor r connected in series. Then we take a voltage source which provides a voltage u0 as soon as we close the circuit with a switch. Then a time-dependent current i flows through the coil and the resistor. The current does not have its maximum value immediately but increases slowly due to land's law. Using Kirchhoff's laws we can set up the following differential equation for the current i. Remember that the point above the i means the first time derivative. This is an inhomogeneous linear differential equation of first order. The searched function y corresponds here to the current i. The perturbation function s corresponds to u0 over l and is time independent in this case. We denote the homogeneous solution by ih. First we need to determine the homogeneous solution ih. We can quickly calculate this using the solution formula for the homogeneous version of the differential equation that you learned before. The coefficient k in front of the searched function i corresponds to r over l and is also time independent in this case. So far so good. We may omit the constant ch in the solution formula here because we consider it later anyway in the constant a which we find in the other solution formula. The coefficient r over l is constant and integrating a constant only introduces a variable t. Thus the homogeneous solution is ih is equal to e to the power of minus r over l times t. Let's insert it into the inhomogeneous solution formula. Note that 1 over the exponential function containing a minus in the exponent is simply equivalent to the exponential function without the minus sign. Now we have to calculate the integral. u0 over l is a constant and can be placed in front of the integral and when integrating the exponential function the exponential function is preserved. Only l over r is added as a factor in front of the exponential function. Finally we hide the constant of integration in the constant a and this is our general solution. We can simplify it a bit more by multiplying out the parentheses. Two exponential functions simplify to one. To get a solution specific to the problem we need to determine the unknown constant a. For that we need an initial condition. If we say that the time t equals zero is the time when the current i was zero because we have not yet closed the switch then our initial condition is y of zero is equal to zero. Insert it into the general solution. e to the power of zero is equal to one. Solve for a and you get a is equal to minus u0 over r. Thus we have successfully determined the specific general solution. Solving method number three exponential ansatz. The exponential ansatz is suitable for ordinary differential equations of arbitrary order that are linear and have constant coefficients. Of course the method quickly becomes complex for higher orders. The method is best suited for second order differential equations. The general form of a second order linear differential equation looks like this. y prime prime plus k1 times y prime plus k0 times y is equal to s. Here we assume that the coefficients k1 and k0 as well as the perturbation function s are independent of x. They are a constant. If the perturbation function is not zero then you must first ask a mathematician. Then he will tell you that the general solution y of an inhomogeneous linear differential equation is composed of two parts. A homogeneous solution yh of the homogeneous differential equation and of a particular solution which we denote by yp. The homogeneous solution yh solves the differential equation if you set the perturbation function s equal to zero. In the method of the exponential ansatz as the name suggests we'll make the guess that the homogeneous solution yh has the form of an exponential function. yh is equal to c times e to the power of lambda times x. Since the first and second derivatives of yh occur in the general formula of the differential equation we must differentiate our exponential ansatz twice. The first derivative is yh prime is equal to c times lambda times e to the power of lambda times x. And the second derivative is c times lambda squared times e to the power of lambda x. Now we can insert the exponential ansatz and the corresponding derivatives into the differential equation. Let us factor out c times e to the lambda x. If we divide by this factor then we get the so-called characteristic equation for lambda. When we solve this equation we find the unknown lambda. Since it is a quadratic equation for lambda we get two solutions lambda one and lambda two. We have to consider both of them. Basically you can set up the characteristic equation directly by looking at your differential equation without having to do all these steps. Compare your differential equation with a characteristic equation.  In this case the coefficient is one. The coefficient of lambda is in front of the first derivative of y in this case k1. And the coefficient k0 in front of the function y itself stands alone in the characteristic equation. By the way if you had a homogeneous third order differential equation then the characteristic equation would start with the cubic term and so on. A quadratic equation has two solutions lambda one and lambda two and you can determine these for example with the quadratic formula. Since you get two lambda values we have to consider both. To do this you have to extend the exponential ansatz by another term in which the second lambda value is in the exponent. With the corresponding lambda values this is the solution of the homogeneous differential equation of second order. Depending on the values of the coefficients k1 and k0 the solutions may show different behavior because if k0 is greater than k1 squared over 4 then you are taking the square root of a negative number. In this case you get a solution that describes oscillations. I will show you this in an example. If the differential equation to be solved is inhomogeneous that is the perturbation function is not zero then we still have to add the particular solution yp to the homogeneous one to find a general solution. For the particular solution we have to choose an appropriate approach depending on what form the perturbation function s has. Here we look at the simplest possible form namely when the perturbation function s is constant. Then the particular solution is given by a perturbation function divided by the coefficient k0. yp is equal to s over k0. Now to get a general solution of such an inhomogeneous differential equation we have to add the homogeneous solution and the particular solution. The two unknown constants c1 and c2 are determined by the constraints as you know. Also note that this method of the exponential ansatz is a guess method. This is the case if the actual solution of the differential equation is not of exponential form. Therefore if you use a solving method that contains the word ansatz in its name be sure to check your solution. You do this by substituting the solution you have found in the differential equation and checking if both sides are equal. Let's make an example. Do you remember the differential equation for the oscillating mass? This is a second order differential equation with constant coefficients. The perturbation function is zero. That means we only have to find out the homogeneous solution and we do that with the exponential ansatz we just learned. Let's take the fast way and directly write down the characteristic equation. We expect a quadratic equation because we have a second order differential equation. The second derivative is preceded by the coefficient one so we just write lambda squared. Then the coefficient in front of the first derivative. Since the first derivative is missing in the differential equation the lambda term is absent as well.  This coefficient stands alone in the characteristic equation. Altogether the characteristic equation reads lambda squared plus d over m is equal to zero. For this equation we don't even need the quadratic formula. We get the solution directly if we first bring d over m to the other side and then take the square root. Consider that the inverse of squaring gives two solutions a positive and a negative square root. Also we have an interesting case here when the square root of a negative number is taken. Square root of a negative number is not a real number but an imaginary number. Do you remember what that means? We expect that the system must oscillate. Even if you don't know imaginary or complex numbers yet you can split the term inside the square root into a product of minus one and d over m. According to the square root laws you can split this product into two square roots. Square root of minus one is defined as the imaginary unit a number which we denote by the letter i. That's all you need to know about imaginary numbers. Let us denote square root of d over m shortly as omega. If we insert the lambda values we have just found into the exponential ansatz we get the general solution for the considered differential equation. This solution seems very abstract but i will show you that this solution corresponds to an oscillation. Before that let us determine the unknown constants c1 and c2 with initial conditions for our problem. For example we could have observed that at the time t equals zero the displacement of the spring was one. The spring was displaced to the maximum so the initial condition is y of zero is equal to one. Insert this condition into the solution to determine c1. e to the power of zero is one. Rearrange for c1 and you get c1 is equal to one minus c2. Next step is to determine the unknown constant c2 using the second initial condition. For this we use the fact that at t equal zero the velocity of the mass was zero. In physics you learn that velocity corresponds to the first time derivative of the displacement so our second initial condition is given by y prime of zero is equal to zero. So let's differentiate our general solution with respect to time t.  The exponential functions become one and the factor i omega cancels out. Rearranging for c2 we find out that c2 is equal to c1. So we know that c2 must be equal to c1. Nice. Let's replace c2 with c1 to determine concrete value for the constants. The equation results in c1 is equal to one half. Since c1 and c2 are equal c2 must be also equal to one half. Insert one half into our general solution. Now let's find out what this abstract solution has to do with oscillations. For this we get our friend Euler to help us who tells us his famous Euler formula. e to the power of i omega t is equal to cosine of omega t plus i times sine of omega t. This relation tells us how the complex exponential function is related to cosine and sine functions. So the first complex exponential in our solution becomes cosine and sine with positive omega t and the second complex exponential becomes cosine and sine with negative omega t. We can omit the minus sign in the argument of the cosine function because cosine is symmetric. That means it has the same value for arguments x and minus x. The sine function on the other hand is anti-symmetric. Therefore we cannot omit the minus sign in the argument but rather pull it out in front of the sine function. Very nice. The complex sine function drops out and the cosine can be summed up. And this is our final solution. As you can see the displacement y changes periodically with time. The mass attached to the spring oscillates back and forth and the oscillation is described by the cosine function. Solving method number four separation ansatz. This is sometimes called the product ansatz for reasons we will soon understand. This method is suitable for partial differential equations of arbitrary order. This solving method is only used to transform a partial differential equation into several ordinary differential equations and then to solve them with other methods. Let us illustrate this method directly with an example. Let's take a look at the one-dimensional wave equation for an electric field. The function we are looking for is the electric field e. This depends on the position x and on the time t. Since the function depends on two variables and their derivatives appear in the equation it is a partial differential equation. Here we make a product ansatz for the search solution e. e is equal to r times u. We assume that the solution e can be split into a product of two functions that we call r and u. One function r depends only on x and the other function u depends only on t. In the wave equation we have a second derivative of e with respect to location x and a second derivative with respect to time t. Differentiating the product ansatz with respect to x yields u times the second derivative of r with respect to x since u is independent of x and thus acts like a constant. In contrast, in the derivative of the product with respect to time t the function r acts like a constant because it does not depend on t. The goal now is to separate everything that depends on x from what depends on t. For this we divide this equation by r times u. Thus we have achieved that everything that depends on x is on the left side and everything that depends on t is on the right side. If you manage to separate a partial differential equation this way then the separation ansatz was successful. Now we can vary x on the left side without changing the right side because there is no x on the right side. The same is true for the time t. If we vary the time on the right side the left side remains unchanged because there is no time t. Thus both sides must be constant. So let us set the left side and the right side equal to a constant k. Thus we have transformed a partial differential equation into two ordinary differential equations and the good thing is that the two differential equations are not coupled. That means you can solve them independently and then multiply the solutions like in the product ansatz to get the general solution for the partial differential equation. You can solve the two ordinary differential equations with the exponential ansatz you learned before. So that's it. Now you have learned everything necessary to classify and to solve simple differential equations. Separation of variables for ordinary homogeneous differential equations of first order, variation of constants for ordinary linear differential equations of first order, exponential ansatz for linear differential equations of arbitrary order and separation ansatz for partial differential equations. There are of course more classifications and solving methods. There are entire books devoted to solving differential equations. Some differential equations are so complex that it is better to not even try to solve them by hand such as the navier stokes differential equations. Numerical solving with a computer is the best approach here. But that's another big topic that you will learn in the next video. With this in mind, bye and see you next time.


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What's the word y'all? The end season tournament is now in its knockout phase. We get Lakers versus Suns. We get Kings versus Pelicans, Bucks versus Knicks, Pacers versus Celtics. I gotta admit I've loved every single second of the end season tournament. Not every single aspect of it There's still a lot of courts that I don't mess around with but for the most part the end season tournament for me has been a success I think the NBA is gonna determine whether it's a success or not based on ratings and things like that And I can't give you the behind the scenes of that because I don't know it But just from a fans perspective the end season tournament has been dope going into last night's the last day of the end season tournament group stage The only two teams we knew that had advanced to the knockout phase was the Lakers and the Pelicans all six of the other spots Was up for grabs and it was very confusing to figure it out But it was worth it in my opinion the moment the end season turning was announced I was like hey This is a good idea because at the end of the day we just want more people engaged in basketball earlier When you have a season of 80 plus games It's gonna be hard to get a casual fan to care about it in November and December I know plenty of people in my personal life that really start watching NBA every season and I'll start break because in their opinion That's when the season starts There is no stakes leading up to that point and the end season tournament is a way to add some stakes into it now Everybody values things differently, right? I'm sure there are some of y'all that are watching this that didn't enjoy the end season tournament You don't see real stakes involved and that is completely okay All that matters is that the players the coaches and everybody involved on the court see that there are stakes involved And we saw that last night because we have moments like this where the Bulls are down my 30 plus points in the fourth quarter and Joe Missoula and the bosses Celtes decide to intentionally foul Andre Drummond and you see coach Billy D is looking at Joe Missoula like yo What's going Joe Missoula has his face that he always looks confused He know what he's doing and in this case He's saying hey going into tonight the bosses Celtes need two different things to happen for them to advance and one of those things They could control which is winning by 23 or more points against the Bulls now I want you to see that score and everything you can do the math It's it's over 23, but they just wanted to make sure that they got as much wiggle room as they could Billy Donovan obviously didn't like that and He shouldn't he shouldn't care or we shouldn't care that Billy Donovan doesn't like that They talk yada yada yada Drummond does drumming things He missed two free throws and eventually the bosses Celtes won by a zillion points And as you saw they're advancing to the next stage and everybody has an opinion about this right Everybody has the opinions Billy Donovan the right Joe Missoula in the right at the end of the day I've been a guy that's always said the unwritten rules and not just basketball But sports in general majority of them are just bad Billy if you don't want this to happen How about we don't be down back a zillion points now I just said him second standing up for his guy and drumming who he mentioned or after that He was embarrassed for drumming yada yada yada But whatever but the end of the day these unwritten rules about things that you should have shouldn't do I've always found a little bit weird But it also showed us the true colors of the Boston Celtics because leading into this game Jason Tatum Jason Tatum said this the point differential thing I'm not the biggest fan of it last Friday Where the game was already over and the guys were still trying to score You know it's all about respecting the game and respecting your opponent Then we get to yesterday when it matters the point differential matters Jason Tatum and Joe Missoula They're like hey forget about everything I said because even Joe Missoula has said the day before that he's not even thinking about the end season tournament I don't know about that I saw Jason Tatum in a game with six seconds six minutes ago when you are about 30, you know and I'm again I'm not complaining. I'm okay with these type of things because it adds another element to the regular season We saw earlier where where what Tatum is referring to where teams were up by 10 Instead of dribbling the ball out they try to make it up by 12 because the point differential means so much Another example of this was the Knicks getting their wild card spot They went against the Charlotte Hornets last night And you can see they got this spot because they had the same record as the Magic the Nets and the Cavs But their point differential was 13 above the Cavaliers This might just be a normal game for time But though but if you look at the fourth quarter of their game against the Hornets yesterday at the four-minute mark around the five-minute mark Rogier Brandon Miller Hey, word Mark Williams and Miles Bridges all get subbed out for the reserves. I don't know Yorkside Boy Randall Brunter. He had a 20-20 game shot to Julie's Randall They stayed into the game until the very last I'm sorry until the last minute But they were there because they wanted to secure that they were winning by a ton of points and they did now It's not all perfect, right? There are some things that I would personally change about it and we'll talk about that in a second But for it to be the first one of hopefully many I think it went as good as you could have imagined like even one of the last games of the day the nasty Televised game on TNT we saw Warriors versus King and I think for the Kings to make it They just had to lose by less than 11 points or win the game outright But the game felt like a playoff atmosphere now You could say that's because the Warriors and the Kings have played each other 11 times in the last six months Look that up. That is that is a true number They played each other 11 times in the last six months given their their playoffs and then early in this season They played each other a bunch too. Maybe it was just that they now have built the rivalry But also the Kings went out for it, you know and the NBA is continuing to do different things like on the Dallas Mavericks broadcast they had Dirk Dirk he was there on the Miami. He broadcast. He has Shaquille O'Neal So like they're just incorporating small things here there to get again the casual fan more involved and again I will remind people this is just the first year of things the prize pool as of right now It's half a million dollars to every player that wins But if this continues to be successful and they sell it which is their goal in a year That prize pool can be more money or they might shift the price pool every time I film at this time of day The Sun is like rotating We got to get the content off All right, I've also on the Kenny Beecham podcast said that hey if the money is not enough and I feel like based on what we saw Yesterday the money is enough, but if they determined that the money was enough There wasn't good enough incentive for the players to care then we can maybe change it to say like hey You went into the tournament guaranteed a playoff spot and I guaranteed a home court And I guarantee anything but at least a place a plate in spot and that gives people the incentive to try every single night Or you can pick to add a lottery pick Not the odds it gets you the first overall pick, but let's say you get a matter of fact It's probably better. I'll show you right now There are 14 teams that are in the lottery But under the new thing if we were to change the prize pool you make it 15 teams in the lottery and let's say The bosses Celtics end up winning it right There's a team that will normally have a 30th pick because they're really good But now they have the 15th highest odds and even if we say that the 15th highest odds is a I don't know Let's say one percent chance. It's still a one percent chance or no less than one But it's a zero point two percent chance Which is better than the zero point zero percent chance that you have if you're the boss of Celtics Or if you're a team that I don't know it's already in here is I'm looking at it Every team that made the plan this year as of right now projects to be a playoff team But hypothetically let's say the Utah Jazz had got found their way into the knockout stage and eventually went on to win The end season tournament now their six percent chance of winning the first overall pick is now a seven percent chance because they have Another one on the back end again Just spitballing ideas at the end of the day the NBA is gonna tweet things here or there to figure out What exactly works best the one thing personally I would change is the amount of teams that go into the group stage now I don't want it to turn into the NBA playoffs or more than half of the teams more than two thirds of the teams Is have a chance, but I do like the idea of a knockout states to the knockout stage Does that make sense? Here's one idea that we just add an extra team So it is a 10-man bracket instead of an eight-man bracket where two wild the top two wild card teams are playing against each other to Fight against the one seed Logistically, I don't know how any of this works because you're adding an extra game to discourage you in whatever, but I like the NBA loves revenue Don't it so extra ball is extra ball. You trying to tell me you want to see devil book will go against Did you know the devil could play play against Anthony average and want to go home game Cass versus nicks are ready to hit they think in the playoffs last year So like boom play that game to go to the next stage and then we get to where we are today The other idea is a 12-man bracket, which is probably harder to explain to people because it's almost half of the league But again adds more knockout game. I'm not gonna build the diagram But the 12 games we will see that the magic get a knockout game as well And then the Rockets are the other team that advanced so like I don't know the more teams the better until we get to a Certain point. I think 12 would have to be the max But I think 10 makes a lot of sense to and like a nerdy cool thing about the incident tournament so far is that TNT and ESPN Are integrating their personalities for the n-season tournaments in my finals on Thursday So it's gonna have Charles Barkley Kenny to Jess Smith's Shaquille O'Neal with Stephen a Smith and Michael will ball Like that is our Ernie Johnson. I skip over your name my father.  These are like some of the best personalities in the game potentially working together like These are competitors at the end of the day, but the n-season turn is bringing us all together and well I can't say us. I'm not a part of this but like next year The NBA come a phone, you know, I'm here for it. Again, I feel like the n-season tournament is like a Lost none experiment, you know, I don't think anybody lost any interest I don't think it had negative value to the NBA viewing experience for the first time And I think that at the end of the day, that's a plus don't like the courts But it got people talking don't like the jerseys, but it got people talking and the ball that was played was good Let me know what you think about the n-season tournament so far Has it has it been a double L as always I'll be in the comment section