All Solutions to 1st-Order Ordinary Differential Equations

Added: 2026-05-12

[00:00:00]Look at this differential equation. My challenge for you is to decide which one is the best method to solve it based on its structure. It has a very common structure, but only a person with trained eyes can recognize the pattern and can tell what the method is. In this video, we'll see the seven most important techniques you can use to solve first order ordinary differential equations like this one. Just as when we go to a restaurant, we got to pick different tools to eat different kinds of dishes and drink different kinds of beverages. When a differential equation is handed to us, we need to match it with the right method so that we can solve it. Let's see the seven different structures. Now, your goal is to use your pattern recognition skills to spot which one best fits this equation.

[00:00:48]The first method is separation of variables.

[00:00:52]That's the general structure that you have to spot.

[00:00:57]Separation of variables is the most natural method to use in order to solve this type of differential equation.

[00:01:03]Let's see an example. Consider this equation.

[00:01:08]Imagine you invest money in the S&P 500, which is basically part of the stock market. On average, your money grows over time proportionally to how much you already have. This differential equation can be solved with separation of variables because it matches the structure.

[00:01:28]So we separate variables and then we integrate both sides.

[00:01:38]K is a constant that represents the average return usually from 7 to 10% every year and C is a constant determined by the initial conditions given to the system. In this case, it represents the initial capital invested.

[00:01:54]The second method is about an integrating factor for linear equations.

[00:01:59]The idea here is to take advantage of the product rule for derivatives.

[00:02:05]We basically want to force the left hand side into a product derivative. In order to do that, we need to create a function called mu of x and then define it this way. The strategy here is to multiply both sides of our differential equation by this factor and then integrate them.

[00:02:20]Let's see an example to illustrate the method. This differential equation represents Newton's law of cooling.

[00:02:27]Imagine we have a hot cup of coffee left on a table. Solutions to this equation describe how the temperature of the object changes over time as it exchanges heat with the surroundings.

[00:02:39]Capital T is the temperature of the cup at time T. TE is the constant ambient temperature and K which is positive is a parameter that measures how quickly the object cools. It depends on the material around it or on the substance itself.

[00:02:56]Once we rewrite this equation accordingly, we can recognize the patterns.

[00:03:01]That's a linear equation. So we can apply the method of multiplying both sides by an integrating factor. This is the integrating factor in our case.

[00:03:11]multiply the differential equation by it.

[00:03:14]Reverse the product rule for derivatives and then integrate both sides with respect to time.

[00:03:21]After all these calculations, we get the solutions in terms of the initial condition. In other words, the initial temperature of our cup of coffee.

[00:03:31]The truth is that you cannot learn just by watching somebody performing all these calculations. As cool as it is to understand all these things for the very first time, I can guarantee you that you will forget almost everything next week if you don't practice. If you're serious about mastering differential equations and you want to see all the methods that we're discussing in this video in details explained in this order, intuition examples, general method for finding solutions and exercises, then download the PDF in the description of this video.

[00:04:04]Third we have homogeneous equations substitution.

[00:04:08]Homogeneous means same structure after scaling. So for example these are homogeneous functions. f(x,y) = x + y.

[00:04:17]Its degree is 1. Let's see why.

[00:04:20]Multiplying each variable by a factor say two for example is the same as multiplying 2 ^ of 1 by the original function. We can confirm this property with a factor of 7.3 as well. or even with any real number lambda for g of xy = x^2 + x y + y^2 we can multiply its variables by a factor like three and the result is always this factor squared times the original function that's why its degree is 2. The function h of xy as y /x is also homogeneous but with degree zero.

[00:04:57]Multiplying any real number lambda by its variables always results in the same function h with a multiplying factor of one.

[00:05:05]But now we will see functions that are not homogeneous x + y + 1. Multiply the variables by a factor lambda and the function spits out something that cannot be written as lambda to some power k multiplied by the original function. At least not for all real numbers lambda.

[00:05:22]That's why this function is not homogeneous. For x^2 + y, we get the same problem. We can't guarantee that its structure will be preserved after scaling.

[00:05:33]And the last counter example is age of xy = y /x + x. Pay attention to this one because it is very similar to the main challenge of this video. Once again, no matter how hard you try, we can't scale it without changing its structure.

[00:05:50]Going back to our challenge, we see that this function on the right hand side is homogeneous because x - y /x can be written as 1 - y /x. And this is clearly homogeneous according to our definition here. So this the third method that we're seeing here is the correct one to solve this differential equation.

[00:06:10]However, there is actually more than just one method to solve it. And we already saw in this video another one that also applies to this particular problem. We just didn't say explicitly, but can you tell which one? Well, let's finish studying the homogeneous method and the next one. And then we're going to reveal it. We'll see how to solve this differential equation and we will also see that actually this is part of a bigger of a larger class of differential equations. But for now, just take your best guess. Let's see an example of a homogeneous differential equation that can be used to solve problems in engineering. Imagine a perfectly idealized fluid like a flow of water in the plane. At each point xy, the fluid has a velocity vector.

[00:06:52]The horizontal component of each velocity vector is a function of the x and y coordinates. And the vertical component is another function of x and y. So the vector field is this with components x + y and x - y.

[00:07:08]Now a streamline is a curve that is everywhere tangent to the velocity field. In other words, if a particle is moving through the fluid, the streamline shows the path that this particle is going to take if you just place it in the vector field. Since the slope of a curve on the xy plane is given by the infinite decimal rate of change of y with respect to the infinite decimal rate of change of x. And since the slope of the velocity vector is the vertical velocity over the horizontal velocity, then the derivative of y with respect to x can be expressed as the function v over the function u or simply x - y / x + y.

[00:07:48]This is a differential equation and therefore there is a differential equation associated with this particular vector field. Solving it will give us the natural path of a particle following the flow of water.

[00:08:01]But not only that, this is a homogeneous differential equation because we can divide the numerator and the denominator by x and then we obtain the differential equation in this form. The right hand side only depends on the ratio y /x which implies that this is a homogeneous function.

[00:08:19]In order to solve it using the method of substitution, we need to follow these steps. First define a variable new as y /x.

[00:08:28]Use the product rule to take the derivative of y with respect to x.

[00:08:35]Substitute it back into the left hand side of the differential equation.

[00:08:43]Separate variables mu to the left and x to the right.

[00:08:48]And finally, integrate both sides to find all solutions y in terms of x and a constant c, which is determined by the initial condition you pick for the problem you're trying to solve.

[00:09:01]These curves represent the streamlines of the vector field. Each value of C gives a different path corresponding to a different initial condition.

[00:09:11]By the way, if you want to have access to all the interactive graphs that we are seeing in this video, check out the PDF link below. We made them in Geogebra and Desmos. We also wrote the step-by-step process of how to solve homogeneous differential equations in general. You can find it all in the PDF below. For the next method, we're going to have Sophia tell us about it.

[00:09:32]The Berni equation.

[00:09:35]This is the general structure.

[00:09:39]We have the first derivative of y functions p of x and q of x and the dependent variable y to the^ of n with n being a real number. Notice that this is almost never linear. I mean only when n is zero or n is 1, this differential equation is linear.

[00:09:57]Take a look at what happens when we divide both sides by y ^ of n and define a new variable v as y ^ of 1 - n.

[00:10:05]We can differentiate the new variable with respect to x and find a new relation that can replace the left hand side of our differential equation.

[00:10:14]After a few more rearrangements, we get this new version of the original differential equation.

[00:10:21]This is actually great news. Can you see why? We basically turned the original nonlinear differential equation into a linear one in the variable v. Now that it's linear, we can use the method of an integrating factor. By the way, the differential equation of our challenge is also a Berni equation. P of X is 1 /X. Q of X is 1 and N is zero.

[00:10:45]And since N is zero, this differential equation is linear and therefore it also matches the second method, integrating factor. So you can use a fork and a knife to eat this dish.

[00:10:57]In fact, using the correct integrating factor, we can find all solutions of y in terms of x.

[00:11:04]The next method is a generalization of Bernoli's equation. Ricardi equation.

[00:11:13]It is nonlinear because of the y to the power of true term. Unlike bernoli equations, ricotti equations are not usually solvable directly. But if you already know one particular solution, let's say it was your lucky day and you guessed it right, call it y1 of x. Then we can assume that the other solutions are of the form y1 + 1 / u for some new variable u. The idea here is that we substitute it into the general recut equation and this way turn it into a linear differential equation. Then just like we did with bernoli we can solve it with an integrating factor. Let me just tell you something real quick. I want to help you see the big picture of all these methods. Notice how there is a very clear pattern for solving first order ordinary differential equations.

[00:11:57]First check if the equation is linear.

[00:11:59]If it is, you can solve it with an integrating factor. If it is not, try to use some kind of substitution or some clever trick so that you can simplify it. You can often turn this differential equation into a linear or separable one.

[00:12:11]Then you can solve this resulting differential equation with one of the methods we've seen. Now, this strategy will not work every single time. I can't promise you that. But you will definitely be surprised by how many differential equations you can solve just by using this strategy. And I would say especially when it comes to differential equations in applied mathematics.

[00:12:31]>> Let's see a concrete example. Now imagine a falling object subject to quadratic air resistance.

[00:12:37]This differential equation can model this process. The independent variable here is time. And we want to find an expression of the dependent variable. So velocity in terms of time. Basically the body's velocity at each instant of time.

[00:12:50]We can rewrite it this way so that we can better spot the ricotti structure.

[00:12:57]The problem with this method is that we need to already know at least one particular solution. Fortunately, we know from physics that when an object falls and experiences air resistance, it tends to have a stage in which it accelerates up to a certain point and then reaches constant velocity. This is called the terminal velocity. So this terminal velocity, call it v1, must be one of the solutions of our differential equation. So in other words, if we plug it in the Ricotti equation, we satisfy it.

[00:13:28]Since the terminal velocity is constant, its derivative is zero. Solving it for V1, we find a particular solution.

[00:13:36]Once again, the details will be added to the PDF below. But the gist of it is that once we substitute the expression V1 + 1 / U for some dependent variable U into the Ricotti equation, we can turn it into a linear differential equation.

[00:13:49]This makes our life way easier when looking for solutions.

[00:13:53]We solved the exercise, found all the solutions, and discussed the physical meaning of each mathematical result in the PDF link in the description below.

[00:14:02]Next, Cleo equation.

[00:14:05]The pattern you need to recognize here is this.

[00:14:09]The dependent variable y is equal to the independent variable x time the derivative of y plus another function that depends on y prime.

[00:14:18]The general method consists of defining the quantity p as the derivative of y and treating it as a fixed parameter so that this differential equation turns into a straight line.

[00:14:28]I know it doesn't look like much but this little trick will make our search for solutions way easier than before.

[00:14:34]Now let's see the application of this type of differential equation in economics. Suppose a company has the total cost of production given by the function C of Q where Q positive is the quantity produced. The derivative of C of Q is usually called the marginal cost and it's the rate at which the total cost changes with respect to the increase or decrease of quantity produced. K is a constant related to market friction. Meaning anything that makes the system less efficient than an ideal market.

[00:15:06]Our goal here is to solve this differential equation so that we have a model that's reasonable that will predict how the cost of production will increase as we vary the quantity of products produced. Start by noticing how it perfectly fits the clear structure.

[00:15:24]We define P as the derivative of the cost function and differentiate both sides with respect to Q.

[00:15:34]This gives us two cases.

[00:15:38]The first results in a straight line.

[00:15:42]The second gives us an expression involving a natural logarithm which indicates that the relative growth of cost with respect to the quantity produced slows down as production increases.

[00:15:54]And the last one is a generalization of the clar equation. The lrangege equation.

[00:16:00]The main difference between Lrange and Claro is that the slope is not just P anymore but a function PH of P instead.

[00:16:08]The strategy is similar to the one for CLO. Differentiate both sides with respect to the independent variable X.

[00:16:15]Rearrange the terms to find the separable linear or just easy to solve version of this differential equation and then solve it with one of the methods we've seen before. Hopefully at this point we find a linear differential equation that is solvable. This is not always possible, but anyway, the problem is much simpler at this point, or at least it's much simpler compared to how we started it. In the PDF below, you're going to find an example of a lrange differential equation related to optics where you have to predict the path of a light ray inside of a specific medium.

[00:16:47]To be very honest, there should be an eighth method to include in this list, and this is exact differential equations.

[00:16:55]I need to admit something. I spent like two days trying to come up with a simplified version like a simple explanation of this kind of differential equation. But I failed. I couldn't do it because you need to know many other things in order to understand how to solve exact differential equations. So instead, I just added a short description of it in the PDF below. But if you guys would like us to make a video dedicated to it since it's so important, especially for physics, then I think we can fit a very good explanation in like 15 to 20 minutes.

[00:17:24]just let us know in the comments section below. The last thing I want to say is that even though we saw seven different methods, probably notice that many of them repeat themselves often. If we boil it down to the absolutely most fundamental ones, I would say that you should focus on these four. Separation of variables, integrating factors for linear equations, substitution or change of variables, and exact equations.

[00:17:49]If you learn these four methods very well and you learn how to interchange them and use them together, I think you're going to be able to solve many first order ordinary differential equations. I can't say all of them or most of them, but at least a lot of them for sure. These methods we've seen here give exact analytical solutions to these differential equations. So we can get a closed form expression like y ofx equals something. However, for some differential equations, we can't find this closed form. And in these situations, we need to look for numerical solutions, which are basically approximations. But this is something for another day. Now, if you want to learn how to read the differential equations, click on this video right here. I'll see you there.